ch08

Chapter 8

Kinematics and Kinetics of RigidBodies in Three-dimensional Motion

8.1 Spherical coordinates

8.2 Angular Velocity of Rigid Bodies in Three-Dimensional Mo-tion

8.3 Angular Acceleration of Rigid Bodies in Three-Dimension Mo-tion

8.4 General Motion Of and On Three-Dimensional Bodies

87

8.4.1GOAL: Determine the angular velocity and acceleration of one arm of the illustrated mechanism.GIVEN: Constant angular velocity of inner arm, and constant angular velocity of outer arm withrespect to inner armDRAW: The figure shows the mechanism with the original coordinate axes and some newly at-tached unit vectors. Unit vectors ı , ,

k are aligned with the ground-fixed X,Y, Z axes. Unit

vectorsb1,

b2,

b3 are attached to the inner arm OA.

FORMULATE EQUATIONS: We’ll use the expressions for angular velocity and accelerationon a rotating body.SOLVE: The angular velocity of arm AB is equal to the angular velocity of arm OA plus therelative angular velocity of arm AB with respect to arm OA. The angular velocity of OA is ψ

k ,

and the relative angular velocity of AB with respect to OA is −θb2. Thus:

ωAB = ψk − θ

b2 (1)

This can be written is theb -frame as:

ωAB = ψb3 − θ

b2

To determine the angular acceleration of AB we can differentiate (1):

αAB =d

dtωAB =

d

dt

(ψ

k − θ

b2

)= ψ

k︸︷︷︸

=0

+ ψ

(d

dt

k

)︸︷︷︸

=0

− θb2︸︷︷︸

=0

−θ(d

dt

b2

)

Since the angular speeds are constant, andk is fixed in space, the only term remaining is −θ

(ddt

b2

).

The tip of unit vectorb2 sweeps in the −

b1 direction with speed ψ. So we have:

αAB = −θ(d

dt

b2

)= −θψ(−

b1)

αAB = θψb1

88

Alternatively, we could have used the expression:

d

dtωAB =

d

dt

∣∣∣∣S

ωAB + ωOA×ωAB

= 0 + ψk ×(ψ

k − θ

b2) = θψ

b1

89

8.4.2GOAL: Determine the angular velocity of a rotating disk.GIVEN: Inner shaft’s angular velocity and the angular velocity of the disk with respect to theshaft.DRAW:

FORMULATE EQUATIONS: We’ll use the expression for angular velocity on a rotating body.SOLVE: The angular velocity of the disk D is equal to the angular velocity of inner shaft AB plusthe relative angular velocity of disk D with respect to shaft AB. The angular velocity of AB is−ω

1 , and the relative angular velocity of disk to shaft is ω

2ı . Thus:

ωD = ω2

ı − ω1

90

8.4.3GOAL: Determine the angular acceleration of a rotating caster.GIVEN: Angular velocity of the caster’s frame is ω

1

b3 and the angular velocity of the caster with

respect to the frame is −ω2

b1.

DRAW:

FORMULATE EQUATIONS: We’ll use the expression for acceleration on a rotating body:d

dt

∣∣∣∣∣N

ωC =d

dt

∣∣∣∣∣F

ωC + ωF×ωC (1)

SOLVE: The caster C has a constant rotation rate with respect to the frame F and so (1) simplifiesto

d

dt

∣∣∣∣∣N

ωC = ωF×ωC = ω1

b3×(−ω

2

b1) = −ω

1ω

2

b2

αC = −ω1ω

2

b2

91

8.4.4GOAL: Determine the angular velocity and acceleration of one wheel of the illustrated mechanism.GIVEN: Angular velocity of the three-armed body is given by ωT = 2 rad/s. h = 0.1 m andr = 0.004 m.DRAW: The figure shows the mechanism with ground-fixed ı , ,

k unit vectors as well

b1,

b2,

b3

(fixed to the rotating body T ).

ASSUME: We’ll assume that the wheels roll without slip on the bottom of the microwave oven.FORMULATE EQUATIONS: We’ll use the expressions for angular velocity and accelerationon a rotating body.SOLVE: The angular velocity of the wheel A is equal to the angular velocity of arm three-armbody (T ) plus the relative angular velocity of wheel A with respect to the body. The angularvelocity of the three arm body is 2 rad/s = 2

b2 rad/s ( and

b2 are identical for this problem).

The wheel rolls without slip and can find the wheel’s angular velocity by determining the speed ofthe wheel A’s center (GA) and then using

vC = vGA+ ωA×rC/GA

Applying the no-slip constraint that vC = 0 will get us the angular velocity information we need.We’ll assume that the wheel A is rotating with respect to T with angular speed φ.

vA = (2b2 rad/s)×(0.1

b1 m) = −0.2

b3 m/s

vC = vA + vC/A

= −0.2b3 m/s + ωA×(−r

b2)= −0.2

b3 m/s + (φ

b1 + 2

b2)×(−r

b2)= −0.2

b3 m/s− rφ

b3

Applying the no-slip constraint that vC = 0 gives us

φ = −0.2 m/sr

= − 0.2 m/s0.004 m

= −50 rad/s

Thus we haveωA = (−50

b1 + 2

b2) rad/s

αA =d

dt(ωA) =

(50d

dt

b1 + 2

d

dt

b2

)rad/s =

(50(−2

b3 rad/s)

)rad/s

αA = −100b3 rad/s2

92

8.4.5GOAL: Compute the angular acceleration of a bicycle wheel as a bicyclist travels a circular path.GIVEN: Radius of wheel, its inclination from vertical, time for the bicyclist to complete one fullcircle, and forward speed of bicyclist.DRAW:

The x, y, z coordinate frame, with corresponding unit vectorsb1,

b2,

b3 is fixed to the frame of the

bicycle. The ı , ,k unit vectors are fixed to the inertial ground frame. The relationship between

these two sets of unit vectors is:ı

k

b1 cosφ 0 sinφb2 0 1 0b3 − sinφ 0 cosφ

ASSUME: We assume that the given 25 mph speed is that of the contact between the wheel andground as it travels a circular path along the ground. This neglects the small variations in forwardvelocity that result from the inclination angle of the bike and the bicyle/rider’s dimensions. (e.g.Since the bicyclist’s center of mass travels a smaller circle than the center of the wheel, the forwardvelocities of each would be slightly different.)FORMULATE EQUATIONS: We’ll use the expressions for angular velocity and accelerationof a rotating body.SOLVE: First we need to determine the two angular rates; that of the bicyclist about the centerof the circular path, and that of the wheel about its center. Since the bicyclist completes one fullcircle in 6 s, we have:

θ =2π rad

6 s=π

3rad/s

The angular speed of the wheel about its center will be the forward speed of the wheel divided byits radius:

ψ =25 mph13 in

=440 in/s

13 in= 33.8 rad/s

These angular rates are constant.The angular velocity of the bicycle about the center of the path is −θı , and the angular velocityof the wheel with respect to the frame of the bicycle is −ψ

b3. The total angular velocity of thewheel is the sum of these:

ω = −θı − ψb3

93

The wheel’s angular acceleration may be computed from:

α =d

dt

∣∣∣∣N

ωBody =d

dt

∣∣∣∣S

ωBody + ωS×ωBody

=d

dt

∣∣∣∣xyz

(−θı − ψb3) + (−θı )×(−θı − ψ

b3)

= 0 + θψ(ı ×b3)

From the transformation matrix, the unit vector ı is

ı = cos θb1 − sin θ

b3

Thus

α = θψ(cos θb1 − sin θ

b3)×

b3 = θψ cos θ

b2

α = −(π

3rad/s

)(33.8 rad/s) cos 30

b2

α = −30.7b2 rad/s2

94

8.4.6GOAL: Determine the angular acceleration of a person’s forearm.GIVEN: Time for the arm segments to reach given configurations. Body geometry.DRAW:


dt

∣∣∣∣∣N

ωBC =d

dt

∣∣∣∣∣F

ωBC + ωF×ωBC (1)

where F indicates a rotating frame of reference that is rotating at the same speed as AB.SOLVE:

ωBC = ω1

b1 + ω

2

b2

We’re given that the two arm segments change orientation by π2 rad in 1 s and so have

ω1

= −π2

rad/s, ω2

=π

2rad/s

The rotation rates are constant so our angular acceleration expression simplifies to

αBC = ddt

∣∣∣∣∣N

ωBC = ωF×ωBC

= ω1

b1×(ω

1

b1 + ω

2

b2)

= ω1ω

2

b3

αBC = (−π2 rad/s)(π2 rad/s)b3 = −π

2

4b3 rad/s2

95

8.4.7GOAL: Determine a forearm’s angular acceleration for 0 ≤ t ≤ 1 s.GIVEN: Rotational information with regard to the arm segments and body geometry.DRAW:


dt

∣∣∣∣∣N

ωBC =d

dt

∣∣∣∣∣F

ωBC + ωF×ωBC (1)

where F indicates a rotating frame of reference that is rotating at the same speed as AB.SOLVE: The arm segment AB alters orientation by π

2 rad in 1 s and so we haveωAB = ω

1

b1 = −π

2b1 rad/s

BC rotates with a constant angular acceleration relative to AB such that after starting from restit has moved π

2 rad after 1 s. Denoting its relative rotation angle with φ, we have

∆φ =π

2rad =

12φt2 ⇒ φ =

2(π/2)(1 s)2

⇒ φ = π rad/s2

We can now use our formula for the derivative of a vector in a rotating body:

αBC = ddt

∣∣∣∣∣F

ωBC + ωF×ωBC

= φb2 + ω

1

b1×(ω

1

b1 + (φt)

b2)

= φb2 + ω

1φt

b3

αBC = πb2 rad/s2 −

(π2

2b3 rad/ s3

)t

96

8.4.8GOAL: Determine the angular velocity and angular acceleration of the forearm segment during agiven motion of the arm.GIVEN: Position of the forearm and upper arm at times t = 0 and t = 1 s, and the fact that theangular speeds of the segments are constant.DRAW: The figures show the arm at times t = 0 and t = 1 s. The unit vectors

b1,

b2,

b3, fixed to

the upper arm, are shown along with the inertial x, y, z axes. At time t = 1 s, φ = 90 and ψ = 45.

FORMULATE EQUATIONS: We’ll use the formulas for the angular velocity and accelerationof a rotating body.SOLVE: Since the angular speeds are constant, we can determine the values by dividing the changein angle by the change in time:

ψ =∆ψ∆t

=π/4 rad

1 s=π

4rad/s

φ =∆φ∆t

=π/2 rad

1 s=π

2rad/s

The angular velocity of the forearm is the angular velocity of the upper arm plus the angularvelocity of the forearm with respect to the upper arm:

ωBC = ωAB + ωBC/AB

= −φb1 + ψ

b3

ωBC = −π2

b1 + π

4

b3 rad/s

The angular acceleration is thus:

αBC = ωBC =d

dt

(−π

2b1 +

π

4b3

)= −π

2b 1 +

π

4b 3 = 0 +

π

4

(π

2b2

)

αBC = π2

8

b2 rad/s2

97

Alternatively,

αBC =d

dt

∣∣∣∣AB

ωBC + ωAB×ωBC

=d

dt

∣∣∣∣AB

(−π

2b1 +

π

4b3

)︸︷︷︸

=0

+[−π

2b1×

(−π

2b1 +

π

4b3

)]

=π2

8b2 rad/s2

98

8.4.9GOAL: Find the angular velocity of the central box C in a differential.GIVEN: vA = −2

k m/s, vB = −3

k m/s. r

1= 0.2 m, r

2= 0.3 m, d = 0.5 m.

DRAW:

FORMULATE EQUATIONS: We’ll use the equations for finding velocities on rotating bodies.SOLVE: All bodies in the differential have two rotational degrees of freedom, which are in the ı

and directions. No rotation is allowed in thek direction, as both wheels remain in contact with

the ground. LetωA = ωA1

ı + ωA2

ωB = ωB1ı + ωB2

ωC = ωC1

ı + ωC2

ωG3= ωG31

ı + ωG32

ωG4= ωG41

ı + ωG42

By the geometric constraints, ωA2 = ωB2 = ωC2 , and ωG31 = ωG41 = ωC1 .We begin by computing the angular velocities of wheels A and B from the given linear velocitiesand the rolling constraint. vA, vB refer to the velocity of the centerpoints of the two wheels,respectively, and ωA, ωB refer to the angular velocity of the wheels themselves. The wheels havetwo angular velocity components, one in the ı direction and another in the direction (due to thelefthand turn).

vA = ωA×rA/G

vB = ωB×rB/H

vAk = (ωA1

ı + ωA2 )×r2 vB

k = (ωB1

ı + ωB2 )×r2

⇒ ωA1 = vA/r2 ⇒ ωB1 = vB/r2

The above yields:

ωA1 =−2 m/s0.3 m

= −6.6 rad/s

ωB1 =−3 m/s0.3 m

= −10 rad/s

To determine the components of the wheel angular velocities we’ll use the relative motion of Bwith respect to A (in this case point A can be considered an extended point of wheel B):

vB = vA + ωB×rB/A

vBk = vA

k + (ωB1

ı + ωB2 )×2dı

⇒ ωB2 =−vB + vA

2d=

3 m/s− 2 m/s2(0.5 m)

= 1.0 rad/s

99

Thus we haveωC2 = ωB2 = ωA2 = 1.0 rad/s (1)

To find ωC1 we need to determine the rate at which gears G3 and G4 are rotating about axle AB.Gears G3 and G4 roll against gears G1 and G2, and it is this rolling, in addition to the rolling ofthe wheels, that determines the ı component of ωC . Examining the relative velocities of the gearswith respect to each other and with respect to point O will allow us to find ωC1 .

vF/O= ωC×rF/O

= (ωC1ı + ωC2

)×r1

= ωC1r1k

vF/O= ωC1r1

k (2)

Taking a different path to F from O:

vF/O= vE/O

+ vF/E

= [(ωA1ı + ωA2

)×(−r1 ı + r1 )] + [(ωG31

ı + ωG32 )×r1 ı ]

= (ωA1r1 + ωA2r1 − ωG32r1)k

vF/O= (ωA1r1 + ωA2r1 − ωG32r1)

k (3)

Combining (2) and (3), we see that

ωC1 = ωA1 + ωA2 − ωG32 (4)

Taking yet a different path:

vF/O= vJ/O

+ vF/J

= [(ωB1ı + ωB2

)×(r1 ı + r1 )] + [(ωG31

ı + ωG32 )×−r1 ı ]

= (ωB1r1 − ωB2r1 + ωG32r1)k

vF/O= (ωB1r1 − ωB2r1 + ωG32r1)

k (5)


ωC1 = ωB1 − ωB2 + ωG32 (6)

Combining (4) and (6), with ωA2 = ωB2 , yields:

ωC1 =ωA1 + ωB1

2(7)

Thus,ωC = ωC1

ı + ωC2 =

−6.6 rad/s− 10 rad/s2

ı + 1.0 rad/s

ωC = −8.33ı + 1.0 rad/s

100

8.4.10GOAL: Find the angular velocity of gear G

2.

GIVEN: vA = 10k m/s, vB = −10

k m/s. r

1= 0.2 m, r

2= 0.3 m, d = 0.5 m.

DRAW:




ı + ωA2

ωB = ωB1ı + ωB2

ωC = ωC1

ı + ωC2

ωG3= ωG31

ı + ωG32

ωG4= ωG41

ı + ωG42


vA = ωA×rA/G

vB = ωB×rB/H

vAk = (ωA1

ı + ωA2 )×r2 vB

k = (ωB1

ı + ωB2 )×r2

⇒ ωA1 = vA/r2 ⇒ ωB1 = vB/r2

The above yields:

ωA1 =10 m/s0.3 m

= 33.3 rad/s

ωB1 =−10 m/s

0.3 m= −33.3 rad/s


vB = vA + ωB×rB/A

vBk = vA

k + (ωB1

ı + ωB2 )×2dı


2d=

10 m/s + 10 m/s2(0.5 m)

= 20.0 rad/s

101

Having both ωB1

and ωB2

, and knowing that wheel B is rigidly attached to gear G2, we haveωG2

= ω1

ı + ω2

= (−33.3ı + 20.0 ) rad/s

102

8.4.11GOAL: Find the angular velocity of Gear G3.GIVEN: vA = 10

k m/s, vB = 10

k m/s. r

1= 0.2 m, r

2= 0.3 m, d = 0.5 m.

DRAW:




ı + ωA2

ωB = ωB1ı + ωB2

ωC = ωC1

ı + ωC2

ωG3= ωG31

ı + ωG32

ωG4= ωG41

ı + ωG42


vA = ωA×rA/G

vB = ωB×rB/H

vAk = (ωA1

ı + ωA2 )×r2 vB

k = (ωB1

ı + ωB2 )×r2

⇒ ωA1 = vA/r2 ⇒ ωB1 = vB/r2

The above yields:

ωA1 =10 m/s0.3 m

= 33.3 rad/s

ωB1 =10 m/s0.3 m

= 33.3 rad/s


vB = vA + ωB×rB/A

vBk = vA

k + (ωB1

ı + ωB2 )×2dı


2d=

10 m/s− 10 m/s2(0.5 m)

= 0

103

Thus we haveωC2 = ωB2 = ωA2 = 0 (1)

From the figure it is clear that gear G3 will have the same angular velocity as the carrier C as wellas an angular velocity with respect to C. Gears G3 and G4 roll against gears G1 and G2, and itis this rolling, in addition to the rolling of the wheels, that determines the ı component of ωC .Examining the relative velocities of the gears with respect to each other and with respect to pointO will allow us to find ωC1 .

vF/O= ωC×rF/O

= (ωC1ı + ωC2

)×r1

= ωC1r1k

vF/O= ωC1r1

k (2)


vF/O= vE/O

+ vF/E

= [(ωA1ı + ωA2

)×(−r1 ı + r1 )] + [(ωG31

ı + ωG32 )×r1 ı ]

= (ωA1r1 + ωA2r1 − ωG32r1)k


k (3)


ωC1 = ωA1 + ωA2 − ωG32 (4)


vF/O= vJ/O

+ vF/J

= [(ωB1ı + ωB2

)×(r1 ı + r1 )] + [(ωG31

ı + ωG32 )×−r1 ı ]

= (ωB1r1 − ωB2r1 + ωG32r1)k


k (5)


ωC1 = ωB1 − ωB2 + ωG32 (6)


ωC1 =ωA1 + ωB1

2=

33.3 + 33.32

rad/s = 33.3 rad/s (7)

Thus,ωC = ωC1

ı + ωC2 = 33.3ı rad/s

We now haveωA = 33.3ı rad/s (8)

ωB = 33.3ı rad/s (9)

ωG3= 33.3ı rad/s + φ (10)

104

where φ is the rotational speed of Gear G3 with respect to the carrier C.We can now apply the formula relating the velocity of two points on a rigid body, the body inquestion being gear G3.

vE = ωA×rE/G

= 33.3ı rad/s×[(d− r1)ı + (r2 + r1) ]= 33.3(r2 + r1)

k rad/s

(11)

vJ = ωB×rJ/H

= 33.3ı rad/s×[−(d− r1)ı + (r2 + r1) ]= 33.3(r2 + r1)

k rad/s

(12)

We’ll now applyvJ = vE + ωG3

×rJ/E(13)

(11), (12), (13) ⇒

33.3(r2 + r1)k rad/s = 33.3(r2 + r1)

k rad/s + (33.3ı rad/s + φ )×(2r1 ı )

φ = 0 (14)

(10), (14) ⇒ ωG3= 33.3ı rad/s

105

8.4.12GOAL: Find the angular velocity of Gear G4.GIVEN: vA = 0, vB = −6

k m/s. r

1= 0.2 m, r

2= 0.3 m, r

3= 0.3 m, d = 0.5 m.

DRAW:




ı + ωA2

ωB = ωB1ı + ωB2

ωC = ωC1

ı + ωC2

ωG3= ωG31

ı + ωG32

ωG4= ωG41

ı + ωG42


vA = ωA×rA/G

vB = ωB×rB/H

vAk = (ωA1

ı + ωA2 )×r2 vB

k = (ωB1

ı + ωB2 )×r2

⇒ ωA1 = vA/r2 ⇒ ωB1 = vB/r2

The above yields:

ωA1 =0 m/s0.3 m

= 0

ωB1 =−6 m/s0.3 m

= −20 rad/s


vB = vA + ωB×rB/A

vBk = vA

k + (ωB1

ı + ωB2 )×2dı


2d=

6 m/s− 02(0.5 m)

= 6 rad/s

106

Thus we haveωC2 = ωB2 = ωA2 = 6 rad/s (1)

From the figure it is clear that gear G4 will have the same angular velocity as the carrier C as wellas an angular velocity with respect to C. Gears G3 and G4 roll against gears G1 and G2, and itis this rolling, in addition to the rolling of the wheels, that determines the ı component of ωC .Examining the relative velocities of the gears with respect to each other and with respect to pointO will allow us to find ωC1 .

vF/O= ωC×rF/O

= (ωC1ı + ωC2

)×r1

= ωC1r1k

vF/O= ωC1r1

k (2)


vF/O= vE/O

+ vF/E

= [(ωA1ı + ωA2

)×(−r1 ı + r1 )] + [(ωG31

ı + ωG32 )×r1 ı ]

= (ωA1r1 + ωA2r1 − ωG32r1)k


k (3)


ωC1 = ωA1 + ωA2 − ωG32 (4)


vF/O= vJ/O

+ vF/J

= [(ωB1ı + ωB2

)×(r1 ı + r1 )] + [(ωG31

ı + ωG32 )×−r1 ı ]

= (ωB1r1 − ωB2r1 + ωG32r1)k


k (5)


ωC1 = ωB1 − ωB2 + ωG32 (6)


ωC1 =ωA1 + ωB1

2=

0− 202

rad/s = −10 rad/s

Thus,ωC = ωC1

ı + ωC2 = (−10ı + 6 ) rad/s

We now haveωA = 6 rad/s (7)

ωB = (−20ı + 6 ) rad/s (8)

ωG4= −10ı rad/s + (6 rad/s + φ) (9)

107

where φ is the rotational speed of Gear G4 with respect to the carrier C.We can now apply the formula relating the velocity of two points on a rigid body, the body inquestion being gear G4.

vK = ωA×rK/G

= 6 rad/s×[(d− r1)ı + (r2 − r1) ]= −6(d− r1)

k rad/s

(10)

vL = ωB×rL/H

= (−20ı + 6 ) rad/s×[−(d− r1)ı + (r2 − r1) ]= [−20(r2 − r1) + 6(d− r1)]

k rad/s

(11)

We’ll now applyvL = vK + ωG4

×rL/K(12)

(10), (11), (12) ⇒

[−20(r2 − r1) + 6(d− r1)]k rad/s = −6(d− r1)

k rad/s + [−10 rad/sı + (6 rad/s + φ) ]×(2r1 ı )

−20(0.1 m/s) + 6(0.3 m/s) = −6(0.3 m/s)− (0.4 m)(6 rad/s + φ)

4 rad/s = −0.4φ

φ = −10 rad/s (13)

(9), (13) ⇒ ωG4= (−10ı − 4 ) rad/s

108

8.4.13GOAL: Determine the highest velocity point on a rolling ice cream cone.GIVEN: Cone dimensions. Time to make a full circle on the floor is 2 s.DRAW:

FORMULATE EQUATIONS: We’ll use the equations for finding velocities on rotating bodies.SOLVE: Example 8.4 shows how to determine the angular velocities of a rotating disk on theend of a bent shaft. The current homework problem can be thought of as precisely the sameproblem. The motion of OA will be the same as that of Example 8.4’s shaft. From observationwe can deduce that the maximal speed will be found at the point B, the topmost point of thecone at the pictured instant. From geometry we have β = sin−1(0.25) = 14.48. |AB| = 1 in and|OA| =

√42 − 12 in = 3.873 in

The cone takes 2 s to complete one full rotation around the floor and thus we have

ωOA

=2π2 s

= π rad/s

The correspondence between our problem and that of Example 8.4 isωcone → ω

Wω

AO→ ω

S

3.873 in → L1

1 in → L2

Substituting these values into the expression for ωW

gives us

ωcone = (π rad/s)(

sinβ − 3.873 cosβ + sinβ1

)c1 + (π rad/s) cosβc3

= (−11.78c1 + 3.042c3) rad/s

vB = ωcone×rB/O

= (−11.78c1 + 3.042c3) rad/s×(3.873c1 + c3) in

vB = 23.56c2 m/s

109

8.4.14GOAL: Show that the angular velocity of a rolling wheel is oriented horizontally.GIVEN: Wheel’s angular velocity in terms of unit vectors attached to the wheel.DRAW:

b1

b2

b3

c1 cosβ 0 sinβc2 0 1 0c3 − sinβ 0 cosβ

FORMULATE EQUATIONS: We’ll use the angular velocity found in the example:

ωW = ωS

[sinβ −

L1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

SOLVE:Re-expressing the c1,

c3 unit vectors in terms ofb1,

b3 gives us

ωW = ωS

[sinβ − L

1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

= ωS

[sinβ − L

1L

2cosβ − sinβ

]c1 + ω

Scosβc3

=[−L1L

2cosβc1 + cosβc3

]ω

S

= L cosβL

2

[−L1L

c1 +L

2L

c3

]ω

S

= −L cosβL

2[cosβc1 − sinβc3]ωS

= −ω

Stanβ

b1

110

8.4.15GOAL: Find the acceleration of point D in the illustrated mechanism.GIVEN: System geometry.DRAW:

b1

b2

b3


ASSUME: The wheel rolls without slip.FORMULATE EQUATIONS: D is attached to the bent shaft and thus we can view it as apoint on a rotating rigid body. We’ll use

vD = ωS×rD/O

where S indicates the shaft and then differentiate to find the acceleration.SOLVE:

vD = ωS

b3×(L

1cosβ

b1 + L

1sinβ

b3) = ω

SL

1cosβ

b2

aD = αSL

1cosβ

b2 + ω

SL

1cosβ d

dt

b2

= αSL

1cosβ

b2 + ω

SL

1cosβ(−ω

S

b1)

= L1cosβ(α

S

b2 − ω2

S

b1)

= L1cosβ[α

Sc2 − ω2

S(cosβc1 − sinβc3)])

aD = L1cosβ(−ω2

Scosβc1 + α

Sc2 + ω2

Ssinβc3)

111

8.4.16GOAL: Explain what the physical root is for the angular acceleration components of the disk inExample 8.4.GIVEN: Angular velocity and acceleration of the disk.DRAW:

b1

b2

b3


FORMULATE EQUATIONS: The problem statement involvesb1 and

b3. We’ll start with the

angular velocity found in the example and re-express it in terms of these unit vectors. Then we’lldifferentiate to find the angular acceleration.SOLVE:ωW is given by

ωW = ωS

[sinβ −

L1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

Re-expressing the c1,c3 unit vectors in terms of

b1,

b3 gives us

ωW = ωS

[sinβ − L

1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

= ωS

[sinβ − L

1L

2cosβ − sinβ

]c1 + ω

Scosβc3

=[−L1L

2cosβc1 + cosβc3

]ω

S

= L cosβL

2

[−L1L

c1 +L

2L

c3

]ω

S

= −L cosβL


= −ω

Stanβ

b1

We can differentiate by viewing this as a vector in a rotating reference frame, the frame defined bythe bent shaft. Let S indicate a rotating frame of reference that rotates at the same rate as theshaft.

112

αW =d

dt

∣∣∣∣∣N

ωW =d

dt

∣∣∣∣∣S

ωW + ωS×ωW

=d

dt

∣∣∣∣∣S

(−

ωS

tanβb1

)+ ω

S

b3×

(−

ωS

tanβb1

)

= −α

Stanβ

b1 −

ω2S

tanβb2

Both terms depend upon the angular speed and acceleration of the shaft because the wheel’srotation is determined (through a rolling constraint) on the velocity of its center (which is drivenby one end of the rotating shaft).The first term, −

αS

tanβb1, arises when differentiating the angular velocity with respect to the rotat-

ing frame. Physically it corresponds to the “usual” way we’d expect to see an angular accelerationarise, namely by the shaft’s rotation rate increasing (or decreasing) so as to create a non-zeroacceleration. If the shaft were to rotate at a constant rate this term would disappear.

The second term,ω2

Stanβ

b2, comes about because the angular velocity (which points in the −

b1

direction) is swept around in a circle as the shaft rotates. The rotation is positive in theb3

direction (counter-clockwise when looking down at the system along the −b3 direction) and this

rotation causes the angular velocity vector to rotate as well, its tip moving in the −b2 direction.

113

8.4.17GOAL: Find the acceleration of point C in the illustrated mechanism. L

1= 10 cm, β = 15.

ωshaft = 3b3 rad/s.

GIVEN: System geometry.DRAW:

b1

b2

b3


ASSUME: The wheel rolls without slip.FORMULATE EQUATIONS: We’ll start with the angular velocity of the wheel, as found inthe example. After re-expressing this in terms of

b1,

b2,

b3 we’ll differentiate to find αW . Then, by

viewing the body as rotating about the fixed point O, we’ll determine the acceleration using ourrigid body acceleration formula

aC = αW ×rC/O+ ωW ×

(ωW ×rC/O

)(1)

SOLVE:ωW = ω

S

[sinβ −

L1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

Re-expressing the c1,c3 unit vectors in terms of

b1,

b3 gives us

ωW = ωS

[sinβ − L

1cosβ + L

2sinβ

L2

]c1 + ω

Scosβc3

= ωS

[sinβ − L

1L

2cosβ − sinβ

]c1 + ω

Scosβc3

=[−L1L

2cosβc1 + cosβc3

]ω

S

= L cosβL

2

[−L1L

c1 +L

2L

c3

]ω

S

= −L cosβL


= −ω

Stanβ

b1

αW = −α

S

tanβb1 −

ωS

tanβd

dt

b1 = − 1

tanβ

(α

S

b1 + ω2

S

b2

)Using these expressions in (1) gives us

114

aC = − 1tanβ

(α

S

b1 + ω2

S

b2

)×L

b1 −ω

S

tanβb1×

(−

ωS

tanβb1×L

b1

)Interestingly, the second term drops out (

b1×

b1 = 0), leaving us with

aC =Lω2

S

tanβb3

From geometry we have

L = 10.35 cm

aC =(0.1035 m)(3 rad/s)2

tan 15b3 = 3.48

b3 m/s2

115

8.4.18GOAL: Find the velocity and acceleration of point A.GIVEN: System geometry. ωP = 2

b3 rad/s, αP = 4

b3 rad/s, ωgun/plate

= 3b1 rad/s.

DRAW:

FORMULATE EQUATIONS:vA = ωgun×rA/O

(1)

aA = αgun×rA/O+ ωgun×

(ωgun×rA/O

)(2)

SOLVE: The angular velocity of the gun is given byωgun = (2

b3 + 3

b1) rad/s (3)

(1), (3) ⇒ vA = (2b3 + 3

b1) rad/s×(2 m)(cos 30

b2 + sin 30

b3)

vA = (−3.464b1 − 3

b2 + 5.196

b3) m/s

αgun = 4b3 rad/s2 + (2

b3 rad/s)×(2

b3 + 3

b1) rad/s

αgun = (4b3 + 6

b2) rad/s2 (4)

(2)–(4) ⇒aA = (6

b2 + 4

b3) rad/s2×(2 m)(cos 30

b2 + sin 30

b3)

+(2b3 + 3

b1) rad/s×(−3.464

b1 − 3

b2 + 5.196

b3)

aA = (5.07b1 − 22.5

b2 − 9.0

b3) m/s2

116

8.4.19GOAL: Find the rate β at which the water cannon is pivoting up at a given instant (β = 10).GIVEN: Base rotates about Z-axis at 4 rad/s, the magnitude of the tip B’s velocity is 6 m/s atthe instant when β = 10.DRAW: The figure below depicts the canon with the original X,Y, Z axes and a set of unit vectors

bi that are fixed to the rotating base.

ASSUME: We assume that the given rotation rate of the base is in the positivek direction. It

could just as easily be assumed that the rotation is in the negativek direction, without affecting

the magnitude of the tip’s velocity or the calculated value of β.FORMULATE EQUATIONS: We’ll use the equation for velocity on a rotating body.SOLVE: The rotation rate of the base is 4

k rad/s, and that of the barrel with respect to the base

is βb1. The total angular velocity of the barrel is thus:

ω = 4k + β

b1 = 4

b3 + β

b1

ask =

b3.

The velocity of the tip B is:

vB

= ω×rB/O

=(4

b3 + β

b1

)×(1.2 cos 10

b2 + 1.2 sin 10

b3

)= −4(1.2) cos 10

b1 − 1.2β sin 10

b2 + 1.2β cos 10

b3 m/s

The magnitude of the velocity is (vB· v

B)1/2 and is given as 6 m/s:

∣∣∣ vB

∣∣∣ =√(

−4.8 cos 10 m/s)2

+(1.2β cos 10 m/s

)2+(1.2β sin 10 m/s

)2

6 m/s =√(

−4.8 cos 10 m/s)2

+(1.2β m/s

)2

β = 3.08 rad/s

117

8.4.20GOAL: Determine the total angular acceleration of a model airplane propeller using two meth-ods: a) differentiation of the components of the angular velocity vector, and b) the formula fordifferentiating a vector in a rotating frame:

d

dt

∣∣∣∣N

p =d

dt

∣∣∣∣S

p + ω×p

GIVEN: The airplane is attached to a tether and circles its attachment point O with a currentangular velocity of ω

3rad/s and an angular acceleration of α

3rad/s2. The propeller turns with

angular velocity ω2

rad/s and angular acceleration α2

rad/s2, both with respect to the model.

DRAW: Theb1,

b2,

b3 frame is attached to the plane.

FORMULATE EQUATIONS: We’ll use the equation for angular velocity and angular acceler-ation of a rotating body.SOLVE: The angular velocity of the propeller is the sum of the angular velocity of the plane andthe angular velocity of the propeller with respect to the plane:

ω = ω3

b3 + ω

2

b2

a) Differentiating:

α = ω = ω3

b3 + ω

3

b 3 + ω

2

b2 + ω

2

b 2

= α3

b3 + ω

3(0) + α

2

b2 + ω

2(−ω

3

b1)

α = −ω2ω

3

b1 + α

2

b2 + α

3

b3

b) Using formula for differentiation in the rotating frame B (withbi attached):

α =d

dt

∣∣∣∣N

ω =d

dt

∣∣∣∣B

ω + ωB×ω

=d

dt

∣∣∣∣B

(ω

3

b3 + ω

2

b2

)+(ω

3

b3

)×(ω

3

b3 + ω

2

b2

)= α

3

b3 + α

2

b2 − ω

2ω

3

b1

α = −ω2ω

3

b1 + α

2

b2 + α

3

b3

118

8.4.21GOAL: Find the velocity and acceleration of point B.GIVEN: System geometry. ωOA ≡ ω1 = 0.8

b3 rad/s, αOA ≡ α1 = 2

b3 rad/s2, ωAB/torso

≡ ω2 =

−1.6b1 rad/s, αAB/torso

≡ α2 = −0.5b1 rad/s2,

DRAW:

FORMULATE EQUATIONS: We’ll use the general rigid body equationsvB = vA + ω×rB/A

(1)

aB = aA + α×rB/A+ ω×

(ω×rB/A

)(2)

SOLVE:(1) ⇒ vA = (0.8

b3) rad/s×(0.24

b1 m) = 0.192

b2 m/s (3)

(2) ⇒ aA = 2b3 rad/s2×(0.24

b1 m) + 0.8

b3 rad/s×(0.192

b2 m/s)

= (0.48b2 − 0.1536

b1) m/s2

(4)

(1), (3) ⇒ vB = 0.192b2 m/s + (0.8

b3 − 1.6

b1) rad/s×(−0.24

b3 m)

= (0.192− 0.384)b2 m/s

vB = −0.192b2 m/s

αAB = α1 + α2 + ω1×(ω1 + ω2)= (2

b3 − 0.5

b1) rad/s2 + (0.8

b3 rad/s)×(0.8

b3 rad/s− 1.6

b1 rad/s)

= (−0.5b1 − 1.28

b2 + 2.0

b3) rad/s2

Using (1),(2) and (4) gives usaB = aA + (−0.5

b1 − 1.28

b2 + 2.0

b3) rad/s2×(−0.24

b3 m)

+(0.8b3 − 1.6

b1)×

[(0.8

b3 − 1.6

b1)×(−0.24

b3 m)

]= (0.48

b2 − 0.1536

b1) m/s2

+(−0.12b2 + 0.3072

b1) m/s2

+(0.8b3 − 1.6

b1) rad/s×(−0.384

b2 m/s)

aB = (0.461b1 + 0.36

b2 + 0.614

b3) m/s2

119

8.4.22GOAL: Determine the velocity of a point T on the tip of a fan blade as a function of time.GIVEN: Dimensions of fan and blade, tilt angle of fan, spin rate of blades, and the side to sideoscillation.DRAW: Let

bi be the set of unit vectors attached to the rotating base of the fan; let unit vectors

ci be the set attached to the tilt arm OA; and letdi be a set attached to the blade itself.

The transformations frombi to ci and from ci to

di are:

b1

b2

b3

c1c2

c3c1 cosβ sinβ 0

d1 1 0 0

c2 − sinβ cosβ 0d2 0 cosψ sinψ

c3 0 0 1d3 0 − sinψ cosψ

ASSUME: We assume that the blades are spinning in the positive c1 direction.FORMULATE EQUATIONS: The velocity of a point T at the blade tip may be written as:

vT

= vA

+ vT/A

= vA

+ ωAT×r

T/A

+ vrel (1)

SOLVE: Since the rotating frame used in the construction of (1) is that of the blade itself, vrel = 0.We have:

vT

= vA

+ ωAT×r

T/A

(2)

= ωOA×r

A/O

+ ωAT×r

T/A

(3)

Let the length of arm OA be L1

= 8 in. and the length of a blade AT be L2

= 4 in. The angularvelocity of the blade is sum of the angular velocity of the arm and the angular velocity of the bladewith respect to the arm. From (3):

vT

= θb2×L1

c1 +(θ

b2 + ψc1

)×L

2

d2

= θ (sinβc1 + cosβc2)×L1c1 +

+[θ (sinβc1 + cosβc2) + ψc1

]×L

2(cosψc2 + sinψc3)

= −θL1cosβc3 +

(θ sinβ + ψ

)L

2

(cosψ c3 − sinψ c2

)+ θL

2cosβ sinψ c1

120

Given θ = θ0sin (ωt), taking the time derivative yields θ = θ

0ω cos (ωt). Letting ψ

0= 0 such that

ψ = ψt = 26t gives:vT

= 4 θ0ω cos (ωt) cosβ sin (26t) c1 −

(4 θ

0ω cos (ωt) sinβ + 104

)sin (26t) c2+

+[(

4 θ0ω cos (ωt) sinβ + 104

)cos (26t)− 8 θ

0ω cos (ωt) cosβ

]c3 in/s

121

8.4.23GOAL: Find the acceleration of collar D.GIVEN: Angular velocity of link OB, angular velocity of link BC with respect to OB, and velocityat which collar D slides along link BC.DRAW:

Unit vectorsbi are fixed to link OB, while unit vectors ci are fixed to link BC. At the illustrated

instant, both sets of unit vectors and the X,Y, Z axes are aligned.FORMULATE EQUATIONS: We’ll use the equation for motion on a three-dimensional body:

aD

= aB

+ α×rD/B

+ ω×(

ω×rD/B

)+ arel + 2ω×vrel (1)

SOLVE: The vectors quantities appearing in Equation (1) are:a

B= ω

OB×(

ωOB×r

B/O

)= θ

b1×

(θ

b1×−0.5

b3

)= 0.5 θ2

b3

ω = ωBC

= θb1 + φ

b3

α = ωBC

= θb1 + θ

db1dt + φ

b3 + φ

db3dt = 0 + 0 + 0− φθ

b2 = −φθ

b2vrel = 5 c2 m/sarel = 0rD/B

= 0.2 c2 m

The individual terms in (1), with θ = 4 rad/s and φ = 5 rad/s, are now:

122

aB

= (0.5 m) (4 rad/s)2b3 = 8

b3 m/s2

α×rD/B

= −φθb2×0.2 c2 = 0

ω×(

ω×rD/B

)=(θ

b1 + φ

b3

)×[(θ

b1 + φ

b3

)×0.2 c2

]= −0.2

(θ2 + φ2

)b2

= − (0.2 m)((4 rad/s)2 + (5 rad/s)2

)b2 = −8.2

b2 m/s2

2ω×vrel = 2(θ

b1 + φ

b3

)×5 c2 = 10θ

b3 − 10φ

b1

= (10 m/s) (4 rad/s)b3 − (10 m/s) (5 rad/s)

b1 =

(40

b3 − 50

b1

)m/s2

Substituting into (1):

aD

= 8b3 + 0− 8.2

b2 + 0 + 40

b3 − 50

b1 m/s2

aD

=(−50

b1 − 8.2

b2 + 48

b3

)m/s2

123

8.5 Moments and Products of Inertia for a Three-DimensionalBody

8.6 Parallel Axis Expressions for Inertias

124

8.6.1GOAL:Determine the mass moments of inertia and products of inertia along the x, y, z axes forthe the illustrated body. Then calculate the inertias about axes parallel to x, y, z that go throughthe body’s mass center. Express results in terms of the body’s mass m.GIVEN: The dimensions of the body are illustrated in the figure; we assume a linear density ρ.DRAW:

FORMULATE EQUATIONS:First, we note the relationship between the mass and linear density of the body:

ρ =m

L1

+ L2

(1)

Since moments and products of inertia are additive for composite bodies, we will treat the illustratedbody as two narrow rod segments and add the computed inertias of each segment to arrive at theresultant value for the entire body.SOLVE:

Ixx =∫

Body

(y2 + z2

)dm =

L1∫

0

y2ρdy +

L2∫

0

z2ρdz = ρy3

3

∣∣∣∣∣L

1

0

+ ρz3

3

∣∣∣∣∣L

2

0

=ρ

3

(L3

1+ L3

2

)= m

3

(L3

1+L3

2L

1+L

2

)

Iyy =∫

Body

(x2 + z2

)dm =

L1∫

0

(0 + 0) dm +

L2∫

0

z2ρdz = ρz3

3

∣∣∣∣∣L

2

0

= m3

(L3

2L

1+L

2

)

Izz =∫

Body

(x2 + y2

)dm =

L1∫

0

y2ρdy +

L2∫

0

(0 + 0) dm = ρy3

3

∣∣∣∣∣L

1

0

= m3

(L3

1L

1+L

2

)

The segment of length L1

has two planes of symmetry, the x−y plane and the y−z plane. Likewise,the segment of length L

2has two planes of symmetry, the x−z plane and the y−z plane. Since

each segment has two planes of symmetry all of the products of inertia are zero. (Also, note that

125

since each segment is one dimensional, the product of dimensions appearing in the integral termsmust be zero.)

Ixy = Iyx = Ixz = Izx = Iyz = Izy = 0

Denote the segment of length L1

as mass m1

and the segment length L2

as mass m2. The center

of mass of the body is located at:

rG

=m

1rG−1 +m

2rG−2

m1

+m2

=m

(L

1L

1+L

2

)(−L

12

)+m

(L

2L

1+L

2

)(L

22

k

)m

= −12

L21

L1

+ L2

+12

L22

L1

+ L2

k

Let x′, y′, z′ be a set of coordinate axes through the center of mass G that are parallel to thex, y, z axes, respectively. Using the parallel axis expressions for moments and products of inertiawe obtain:

Ix′x′

= Ixx −m(r22

+ r23

)=

m

3

(L3

1+ L3

2

L1

+ L2

)−m

(−12

L21

L1

+ L2

)2

+

(12

L22

L1

+ L2

)2

= m(L

1+L

2)2

[112

(L4

1+ L4

2

)+ 1

3

(L

1L3

2+ L

2L3

1

)]Iy′y′

= Iyy −m(r21

+ r23

)=

m

3

(L3

2

L1

+ L2

)−m

0 +

(12

L22

L1

+ L2

)2

=mL3

2(L

1+L

2)2

(112L2

+ 13L1

)Iz′z′

= Izz −m(r21

+ r22

)=

m

3

(L3

1

L1

+ L2

)−m

0 +

(−1

2

L21

L1

+ L2

)2

=mL3

1(L

1+L

2)2

(112L1

+ 13L2

)Iy′z′

= Iz′y′

= Iyz −mr2r3

= 0−m

(−1

2

L21

L1

+ L2

)(12

L22

L1

+ L2

)

= m4

(L

1L

2L

1+L

2

)2

The center of mass of the body lies on the plane x = 0, and thus the y′, z′ plane is also a plane ofsymmetry. The coordinate axis normal to this plane is the x′ axis. Therefore all products of inertiainvolving the x′ axis are zero:

Ix′y′

= Iy′x′

= Ix′z′

= Iz′x′

= 0

126

Alternatively, note that since the center of mass lies on the plane x = 0, r1

= 0.

127

8.6.2GOAL:Determine the mass moments of inertia and products of inertia along the x, y, z axes forthe the illustrated body. Express results in terms of the body’s mass m.GIVEN: The dimensions of the body are illustrated in the figure. We assume a linear density ρ.DRAW:

FORMULATE EQUATIONS:First, we note the relationship between the mass and linear density of the body:

m = 3aρ

Because moments and products of inertia are additive for composite bodies, we will treat theillustrated body as three narrow rod segments and add the computed inertias of each segment toarrive at the resultant value for the entire body.SOLVE:

Ixx = (ρa)a2

12+ 2(ρa)

(a

2

)2

=7ρa3

12

Iyy = 2(ρa)a2

3=

2ρa3

3

Izz =(ρa)a2

12+ 2

[(ρa)a2

12+ (ρa)

((a

2

)2

+(a

2

)2)]

=5ρa3

4

Ixy = 2(ρa)(a

2

)(a

2

)=ρa3

2, Iyz = Izx = 0

Expressing these in terms of the overall mass m gives us

Ixx = 7ma2

36 , Iyy = 2ma2

9 , Izz = 5ma2

12 , Ixy = ma2

6 , Iyz = Izx = 0

128

8.6.3GOAL: Determine the rotational inertias of the illustrated flat plate.GIVEN: Body’s orientation and dimensions.DRAW:

FORMULATE EQUATIONS: From Appendix B we have

Ix′x′

=ma2

12, I

y′y′=mb2

12, I

z′z′=m(a2 + b2)

12

Ix′y′

= Iy′z′

= Iz′x′

= 0

SOLVE:To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r

1= b/2,

r2

= a/2 and r3

= 0.

Ixx = Ix′x′

+m(r22

+ r23) =

ma2

3

Iyy = Iy′y′

+m(r23

+ r21) =

mb2

3

Izz = Iz′z′

+m(r21

+ r22)

= m

(a2 + b2

12

)+m

[(b2)2

+(a2)2]

= m(a2 + b2)3

Ixy = Ix′y′

+mr1r2

= m

(ab

4

)Iyz = Izx = 0

Ixx = ma2

3 , Iyy = mb2

3 , Ixx = m(a2 + b2)3 , Ixy = mab

4 , Iyz = Izx = 0 ,

129

8.6.4GOAL: Determine the moments and products of inertia along the x, y, z axes for the the illustratedtriangular body. Then calculate the inertias about axes parallel to x, y, z that go through the body’smass center. Express results in terms of the areal density ρ.GIVEN: The dimensions of the body are illustrated in the figure.DRAW:

SOLVE: The borders of the triangular region are defined by the lines: x = 0, y = 0, and x =a(1− y/b). The moments and products of inertia are computed by integrating over this region:

Ixx =∫

Body

(y2 + z2

)dm =

b∫0

a(1−y/b)∫0

(y2 + 0

)ρ dx dy

= ρ

b∫0

a

(1− y

b

)y2 dy

= ρa

(b3

3− b4

4b

)= 1

12ρab3

Iyy may be found similarly, by reversing the order of integration in x and y and using integrationlimits of 0 and b(1−x/a) on y. By the symmetry of the problem we should find that Iyy = 1

12ρa3b.

However, let’s proceed by maintaining the order above and performing the integration:

Iyy =∫

Body

(x2 + z2

)dm =

b∫0

a(1−y/b)∫0

(x2 + 0

)ρ dx dy

= ρ

b∫0

a3(1− y

b

)33

dy

= −ρa3b

12

(1− y

b

)4∣∣∣∣∣b

0

= 112ρa

3b

For Izz :

Izz =∫

Body

(x2 + y2

)dm =

b∫0

a(1−y/b)∫0

(x2 + y2

)ρ dx dy

130

Notice that this is just the sum of Ixx and Iyy (this is true for all planar figures). Thus,

Izz = Ixx + Iyy = 112 ρ

(ab3 + a3b

)Since the x−y plane is a plane of symmetry for the body, all products of inertia involving thenormal z-axis are zero:

Ixz = Izx = Iyz = Izy = 0

The remaining products of inertia are:

Ixy = Iyx =∫

Body

xy dm =b∫

0

a(1−y/b)∫0

xy ρ dx dy

= ρ

b∫0

a2(1− y

b

)22

y dy

= ρa2

2

(b2

2− 2b2

3+b2

4

)= 1

24ρa2b2

To determine the moments and products of inertia about the center of mass, we first need todetermine its location.

rG

=1m

∫Body

r dm =2ρab

b∫0

a(1−y/b)∫0

(xı + y ) ρ dx dy

=2ab

b∫0

(a2

2

(1− y

b

)2ı + a

(1− y

b

)y

)dy

=2ab

−a2b

6

(1− y

b

)3

ı + a

(y2

2− y3

3b

)

b

0

=13

(a ı + b

)Using our parallel axis expressions for inertias, with r

1= a/3, r

2= b/3, and r

3= 0 we have:

Ix′x′

= Ixx −m(r22

+ r23

)=

112ρab3 − ρab

2

(b2

9

)= 1

36 ρab3

Iy′y′

= Iyy −m(r21

+ r23

)=

112ρa3b − ρab

2

(a2

9

)= 1

36 ρa3b

Iz′z′

= Izz −m(r21

+ r22

)= I

x′x′+ I

y′y′= 1

36 ρ(ab3 + a3b

)Iy′x′

= Ix′y′

= Ixy − mr1r2

=124ρa2b2 − ρab

2

(a

3

)(b

3

)= − 1

72 ρa2b2

Due to the fact that r3

= 0, or the fact that the z′-axis is normal to the plane of symmetry throughthe mass center, all other products of inertia are zero.

Ix′z′

= Iz′x′

= Iy′z′

= Iz′y′

= 0

131

8.6.5GOAL: Determine the rotational inertias of the illustrated flat plate.GIVEN: Body’s orientation and dimensions.DRAW:

FORMULATE EQUATIONS: We’ll find the solution by taking the difference of the rotationalinertias corresponding to a plate with dimensions a

2, b

2and one corresponding to the “missing

chunk” having dimensions a1, b

1. Each plate will have the same density ρ. From Appendix B we

have, for a plate with dimensions a, b:

Ix′x′

=mb2

12, I

y′y′=ma2

12, I

z′z′=m(a2 + b2)

12

Ix′y′

= Iy′z′

= Iz′x′

= 0

SOLVE: We’ll start with a complete plate (a2, b

2) and put a “2” to the upper left of the I as an

identifier. To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) withr1

= a2/2, r

2= b

2/2 and r

3= 0.

2Ixx = 2Ix′x′

+m2(r2

2+ r2

3) =

m2b22

3

2Iyy = 2Iy′y′

+m2(r2

3+ r2

1) =

m2a2

2

32Izz = 2I

z′z′+m

2(r2

1+ r2

2)

= m2

(a2

2+ b2

212

)+m

2

[(b22

)2

+(a

22)2]

=m

2(a2

2+ b2

2)

3

2Ixy = 2Ix′y′

+m2r1r2

= m2

(a

2b2

4

)2Iyz = 2Izx = 0

Next we’ll consider the a plate corresponding to the “missing chunk” (a1, b

1) and put a “1” to the

upper left of the I as an identifier. To find the mass moments of inertia and products of inertia weuse (8.24)-(8.29) with r

1= a

1/2, r

2= b

1/2 and r

3= 0.

1Ixx = 1Ix′x′

+m1(r2

2+ r2

3) =

m1b21

3

1Iyy = 1Iy′y′

+m1(r2

3+ r2

1) =

m1a2

1

3

132

1Izz = 1Iz′z′

+m1(r2

1+ r2

2)

= m1

(a2

1+ b2

112

)+m

1

[(b12

)2

+(a

12)2]

=m

1(a2

1+ b2

1)

3

1Ixy = 1Ix′y′

+m1r1r2

= m1

(a

1b1

4

)1Iyz = 1Izx = 0

We can now take the difference of the two sets of rotational inertias, using m1

= ρa1b1

andm

2= ρb

1b2

to obtain

Ixx =ρa

2b32

3−ρa

1b31

3

Iyy =ρb

2a3

2

3−ρb

1a3

1

3

Izz =ρ(a3

2b2

+ a2b32)

3−ρ(a3

1b1

+ a1b31)

3

Ixy =ρa2

2b22

4−ρa2

1b21

4Finally, expressing these in terms of the system mass m = ρ(a

2b2− a

1b1) gives us

Ixx =m(a

2b32− a

1b31)

3(a2b2− a

1b1) Iyy =

m(b2a3

2− b

1a3

1)

3(a2b2− a

1b1) Izz =

m[a

2b2(a2

2+ b2

2)− a

1b1(a2

1+ b2

1)]

3(a2b2− a

1b1)

Ixy =m(a

2b2

+ a1b1)

4 Iyz = Izx = 0

133

8.6.6GOAL: Determine the mass moments of inertia and products of inertia along the x, y, z axes forthe illustrated body. Express answers in terms of the body’s areal density ρ.GIVEN: Dimension of the body are given in the figure.DRAW:

FORMULATE EQUATIONS: To determine the inertia values of the composite body, we willmodel it as a triangular region with a quarter-circular region removed. Since inertias are additive,we may obtain the overall inertia values by subtracting those of the quarter-circular region fromthose of the triangular region.

I =

I triangle −

Iq.circle

SOLVE: The triangular region is bounded by the lines x = 0, y = 0, and x = b− y. The momentsand products of inertia are determined by integrating over this area:

I4xx =∫

Body

(y2 + z2

)dm =

b∫0

b−y∫0

(y2 + 0

)ρ dx dy

= ρ

b∫0

(b− y) y2 dy

= ρ

(b4

3− b4

4

)

=112ρb4

By symmetry,

I4yy = I4xx =112ρb4

Computing I4zz , noting that z2 = 0 over the region,

I4zz =∫

Body

(x2 + y2

)dm = I4yy + I4xx

=16ρb4

Since the x−y plane is a plane of symmetry for the body, all products of inertia involving thenormal z-axis are zero:

I4xz = I4zx = I4yz = I4zy = 0

134

Computing I4xy = I4yx for the triangular region,

I4xy = I4yx =∫

Body

xy dm =b∫

0

b−y∫0

xyρ dx dy

= ρ

b∫0

(b− y)2

2y dy

= ρ

(b4

4− b4

3+b4

8

)

=124ρb4

The quarter-circular area may be defined using polar r, θ coordinates, with x = r cos θ, y = r sin θ,and r2 = x2 + y2. The bounds on the region are 0 ≤ r ≤ a and 0 ≤ θ ≤ π/2. In r, θ coordinates,the differential area may be written as dA = dr(rdθ), thus the differential mass unit becomesdm = ρ r dr dθ.

Ixx =∫

Body

(y2 + z2

)dm =

π2∫

0

a∫0

(r sin θ)2 ρ r dr dθ

= ρ

π2∫

0

a4

4sin2θ dθ

= ρa4

8(θ − sin θ cos θ)

∣∣∣∣π2

0

=116ρπa4

By symmetry,

Iyy = Ixx =116ρπa4

Now for Izz ,

Izz =∫

Body

(x2 + y2

)dm =

π2∫

0

a∫0

r2ρ r dr dθ

= ρ

π2∫

0

a4

4dθ

=18ρπa4

The x−y plane is a plane of symmetry and thus all moments of inertia involving the normal z-axisare zero:


135

The remaining products of inertia are:

Ixy = Iyx =∫

Body

xy dm =

π2∫

0

a∫0

r2 sin θ cos θρ r dr dθ

= ρa4

4

π2∫

0

sin θ cos θ dθ

= ρa4

8sin2θ

∣∣∣∣π2

0

=18ρa4

The moments and products of inertia of the composite body are obtained by subtracting the inertiasof the quarter-circular region from the respective inertias of the triangular region:

Ixx = I4xx − Ixx = 112ρb

4 − 116ρπa

4

Iyy = I4yy − Iyy = 112ρb

4 − 116ρπa

4

Izz = I4zz − Izz = 16ρb

4 − 18ρπa

4

Ixy = I4xy − Ixy = 124ρb

4 − 18ρa

4

Iyx = Ixy = 124ρb

4 − 18ρa

4

By symmetry, all products of inertia involving the z-axis are zero:


136

8.6.7GOAL: Determine the rotational inertias of the illustrated body.GIVEN: Body’s orientation. a = 0.1 m, b = 0.25 m, c = 0.15 m, d = 0.1 m, e = 0.1 m, f = 0.2 m,g = 0.15 m. Body’s density is ρ and mass is m.DRAW:

FORMULATE EQUATIONS: We’ll solve this problem by breaking it up into two rectangularbodies, one with dimensions a, b, c and the other with dimensions c, e, g. From Appendix B wehave:

Ix′x′

= m

(c2 + b2

12

), I

y′y′= m

(a2 + c2

12

), I

z′z′= m

(a2 + b2

12

)To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with appropriatevalues of r

1, r

2and r

3.

SOLVE: Let’s first consider the body with dimensions a, b, c. We’ll put a “1” to the upper left ofthe I as an identifier.For this body r

1= a/2, r

2= b/2 and r

3= −c/2.

1Ixx = 1Ix′x′

+m1(r2

2+ r2

3)

= m1

(c2 + b2

12

)+m

1

(c2 + b2

4

)= m

1

(c2 + b2

3

)1Iyy = 1I

y′y′+m

1(r2

3+ r2

1)

= m1

(a2 + c2

12

)+m

1

(a2 + c2

4

)= m

1

(a2 + c2

3

)1Izz = 1I

z′z′+m

1(r2

1+ r2

2)

= m1

(a2 + b2

12

)+m

1

(a2 + b2

4

)= m

1

(a2 + b2

3

)1Ixy = 1I

x′y′+m

1r1r2

= m1

(ab

4

)1Iyz = 1I

y′z′+m

1r2r3

= −m1

(cb

4

)1Izx = 1I

z′x′+m

1r3r1

= −m1

(ac

4

)Now we’ll consider the body with dimensions c, e, g. We’ll put a “2” to the upper left of the I as

137

an identifier.For this body r

1= a+ e/2, r

2= d+ g/2 and r

3= −c/2.

2Ixx = 2Ix′x′

+m2(r2

2+ r2

3)

= m2

(c2 + g2

12

)+m

2

[(c2)2

+(d+ g

2

)2]2Iyy = 2I

y′y′+m

2(r2

3+ r2

1)

= m2

(e2 + c2

12

)+m

2

[(c2)2

+(a+ e

2

)2]2Izz = 2I

z′z′+m

2(r2

1+ r2

2)

= m2

(e2 + g2

12

)+m

2

[(d+ g

2)2

+(a+ e

2

)2]2Ixy = 2I

x′y′+m

2r1r2

= m2

(a+

e

2

)(d+

g

2

)2Iyz = 2I

y′z′+m

2r2r3

= m2

(d+

g

2

)(− c

2

)2Izx = 2I

z′x′+m

2r3r1

= m2

(− c

2

)(a+

e

2

)The individual masses are given by

m1

= 3.75×10−3ρ, m2

= 2.25×10−3ρ

Adding the rotational inertias together and evaluating for the given dimensional values gives us

Ixx = 1Ixx + 2Ixx = (1.96×10−4 m5)ρ

Iyy = 1Iyy + 2Iyy = (1.1×10−4 m5)ρ

Izz = 1Izz + 2Izz = (2.16×10−4 m5)ρ

Ixy = 1Ixy + 2Ixy = (8.25×10−5 m5)ρ

Iyz = 1Iyz + 2Iyz = (−6.469×10−5 m5)ρ

Izx = 1Izx + 2Izx = (−3.94×10−5 m5)ρ

Finally, evaluating in terms of the object’s mass, m = (abc+ ceg)ρ = (6.0×10−3 m3)ρ, gives us ourfinal results:

Ixx = (3.27×10−2 m2)m, Iyy = (1.83×10−2 m2)m, Izz = (3.60×10−2 m2)m

Ixy = (1.375×10−2 m2)m, Iyz = (−1.078×10−2 m2)m, Izx = (−6.56×10−3 m2)m

138

8.6.8GOAL: Determine the mass moments of inertia and products of inertia for the illustrated washeralong the x, y, z axes (which go through the body’s mass center). Express the results in terms of ρ.GIVEN: The outer radius of the washer is ro = 0.03 m, the inner radius is r

i= 0.02 m, and the

thickness is h = 0.006 m.DRAW:

SOLVE: We may compute the inertias in cylindrical r, θ, z coordinates, with x = r cos θ, y = r sin θ,and x2 + y2 = r2. First, we’ll compute Izz .

Izz =∫

Body

(x2 + y2

)dm =

h2∫

−h2

2π∫0

ro∫ri

r2ρ r dr dθ dz

= ρ

h2∫

−h2

2π∫0

(r4o4 −

r4

i4

)dθ dz

= 2πρh(

r4o4 −

r4

i4

)= 2πρ (0.006 m)

((0.03 m)4

4 − (0.02 m)4

4

)Izz =

(6.13×10−9 m5

)ρ

By symmetry, Ixx = Iyy , so we will only need to calculate Ixx :

Ixx =∫

Body

(y2 + z2

)dm =

ro∫ri

2π∫0

h2∫

−h2

(r2sin2θ + z2

)ρ r dz dθ dr

= ρh

ro∫ri

2π∫0

(r3sin2θ + r

h2

12

)dθ dr

= ρh

ro∫ri

([r3

2(θ − sin θ cos θ)

]2π

0

+ 2πrh2

12

)dr

= πρh

ro∫ri

(r3 + r

h2

6

)dr

=14πρh

(r4o − r4

i

)+

112πρh3

(r2o − r2

i

)=

14πρ (0.006 m)

((0.03 m)4 − (0.02 m)4

)+

+112πρ (0.006 m)3

((0.03 m)2 − (0.02 m)2

)139

Iyy = Ixx =(3.09×10−9 m5

)ρ

Note that when perfoming the above integrations we could have substituted the lower limit riwith

0, to find the moments of inertia of a solid disk. Subtracting the inertia values of a solid disk ofradius r

ifrom those of a solid disk of radius ro yields identical results.

Since all three planes, x−y, x−z, and y−z, are planes of symmetry for the body, all of the productsof inertia are zero:

Ixy = Iyx = Ixz = Izx = Iyz = Izy = 0

140

8.6.9GOAL: Determine the mass moments and products of inertia for the illustrated triangular prismGIVEN: Body’s orientation and dimensions.DRAW:

FORMULATE EQUATIONS: From Appendix B we have the rotational inertias about thebody’s mass center G of a flat right-triangle:

Ix′x′

=mb2

18, I

y′y′=ma2

18, I

z′z′=m(a2 + b2)

18

Ix′y′

=mab

36, I

y′z′= I

z′x′= 0

To find the mass moments of inertia and products of inertia we’ll use (8.24)-(8.29) with r1

= a3 ,

r2

= b3 and r

3= c

2 after determining the rotational inertias of a solid prism from those of the flatright-triangle.SOLVE:The mass of a thin slice having thickness dz is ρab dz2 .

Ixx =c∫0

[ρab3

36 + ρab2

(z2 +

(b3)2)]

dz = ρab2

c∫0

[b2

6 + z2

]dz

= ρab2

(b2z6 + z3

3

) ∣∣∣c0

Ixx = ρabc

(b2

12 + c2

6

)= m

(b2

6 + c2

3

)From symmetry we have

Iyy = ρabc

(b2

6 + c2

12

)= m

(b2

3 + c2

6

)

Izz = m

(a2

18+b2

18

)+m

(a2

9+b2

9

)= m

(a2

6+b2

6

)

Izz = m

(a2

6 + b2

6

)= ρabc

(a2

12 + b2

12

)

Ixy = Ix′y′

+m

(a

3

)(b

3

)= −mab

36+mab

9=mab

12

Ixy = mab12 = ρa2b2c

24

141

Iyz = m

(b

3

)(c

2

)

Iyz = mbc6 = ρab2c2

12

From symmetry we have

Izx = mac6 = ρa2bc2

12

142

8.6.10GOAL: Determine the mass moments and products of inertia for the illustrated cylindrical body.GIVEN: Body’s orientation and dimensions.DRAW:

FORMULATE EQUATIONS: From Appendix B we have the rotational inertias about thebody’s mass center G:

Iy′y′

=mr2

2, I

x′x′=m(3r2 + h2)

12

Ix′y′

= Iy′z′

= Iz′x′

= 0

SOLVE:To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r

1= 0,

r2

= h2 and r

3= r.

Iyy = Iy′y′

+mr2 = 3mr22

Ixx = Ix′x′

+m

(r2 +

(h2)2)

= mh2

3 + 5mr24

From symmetry we see that Ixy = 0 .The mass of the body is given by

m = ρπr2h

giving us

Ixx = ρπr2h3

3 + 5ρπr4h4 , Iyy = 3ρπr4h

2

Ixy = 0

143

8.6.11GOAL: Determine Ixx , Ixy , and Izz , for the illustrated body. Express the results in terms of boththe areal density ρ and the body’s mass m. Denote the mass of each half-circular piece by m

hcand

the mass of the rectangular piece by mr .GIVEN: The body’s dimensions are given in the figure.DRAW:

FORMULATE EQUATIONS: We’ll use the rotational inertia formulas for a half-circle and flatrectangular body from Appendix B and then apply the parallel axis theorem.SOLVE: We’ll first look at Body A (the left half-circle) then B (the right half-circle) and finallythe rectangular piece C that joins the two half-circles.Body A:

AIx′′x′′

=12m

hc

(b

2

)2

=18m

hcb2

d is the distance from the y axis to the body’s mass center.Using the parallel axis theorem gives us

AIx′x′

= AIx′′x′′

−mhcd2

and finally

AIxx = AIx′x′

+mhc

(d2 +

(b

2

)2)

=18m

hcb2 +m

hc

(b

2

)2

=38m

hcb2

AIzz = AIz′z′

+mhc

(b

2

)2

=14m

hc

(b

2

)2

+mhc

(b

2

)2

=516m

hcb2

Because Body A lies in the y,z plane Ixy = 0.Body B :We’ll again use x′, y′, z′ to refer to axes through that body’s mass center. From observation wehave the same value for BIxx as for Body A:

BIxx =38m

hcb2

144

BIzz = BIz′z′

+mhc

[(b2)2

+ a2

]

= 116mhc

b2 +mhc

[(b2)2

+ a2

]

= 516mhc

b2 +mhca2

BIxy = mhca

(b

2

)=m

hcab

2

Body C:CIxx = mrb

2

CIzz =13mr(a

2 + b2)

CIxy = mr

(a

2

)(b

2

)=mrab

4

The masses of the individual pieces are given bym

hc= (0.1 m)2π/2 = (1.57×10−2 m2)ρ

mr = (0.4 m)(0.1 m)ρ = (8.00×10−2 m2)ρ

Ixx = AIxx + BIxx + CIxx

= (1.57×10−2 m2)ρ(2)(38)

(0.2 m)2 + (8.00×10−2 m2)ρ(0.2 m)2

=(3.67×10−3 m5

)ρ

Izz = AIzz + BIzz + CIzz

= (1.57×10−2 m2)ρ[(2)(

516)

(0.2 m)2 + (0.4 m)2]+ 1

3(8.00×10−2 m2)ρ[(0.4 m)2 + (0.2 m)2

]=(8.24×10−3 m5

)ρ

Ixy = AIxy + BIxy + CIxy

= (1.57×10−2 m2)ρ (0.2m)(0.4m)2 + (8.00×10−2 m2)ρ (0.2m)(0.4m)

4

=(2.23×10−3 m5

)ρ

The total mass of the body is given by

mt = mr + 2mhc

= (0.111 m2)ρ

Ixx = 3.30×10−2mt Izz = 7.40×10−2mt Ixy = 2.00×10−2mt

145

8.6.12GOAL: Determine Ixx , Iyy , Izz , and Ixy for the illustrated body. Express the results in terms ofboth the areal density ρ and the body’s mass mc . Denote the mass of each half circular piece bym = 1

2mcGIVEN: The body’s dimensions are given in the figure.DRAW:

SOLVE: To determine the moments and products of inertia for the composite body, we will requiresome inertia values for a half-circular body. Consider a circle with normal axis z′ through its masscenter. Since the moment of inertia of a full circle with mass mc about an axis through its masscenter and in the plane of the circle is 1

4mcr2, that of a half circle through the same point is:

Iparallel

=12

(14mcr

2)

=12

(14

(2m) r2)

=14mr2

The moment of inertia of a full circle about an axis through its mass center and normal to theplane of the circle is 1

2mcr2, thus for a half circle through the same point:

Inormal

=12

(12mcr

2)

=12

(12

(2m) r2)

=12mr2

By symmetry, Ixx = Izz for the composite body. Thus we will only calculate Izz . Using the parallelaxis expressions of inertia for half circle A we obtain:

IAzz = IA

z′z′+m

(r2 + d2

)= IA

z′′z′′+mr2

=12mr2 +mr2

=32mr2

For half circle B:

IBzz = IB

z′z′+mr2

=14mr2 +mr2

=54mr2

146

Thus, for the composite body,

Izz = IAzz + IB

zz =114mr2

Ixx = Izz = 118 mcr

2 = 118 ρπr

4

The moment of inertia about the y-axis is the same for both half circles. Iyy is therefore quitestraightforward to compute,

Iyy = IAyy + IB

yy =14mr2 +

14mr2 =

12mr2

Iyy = 14mcr

2 = 14ρπr

4

The product of inertia, IBxy , will be zero since the y−z plane is a plane of symmetry for half circle

B (with normal x-axis). The product of inertia is then Ixy = IAxy . One method of obtaining IA

xy

is to integrate over the half circle A using integration parameters x, y and the bounds x = 0 andx2 + (y − r)2 = r2. (⇒ x = 0 and x2 = 2yr − y2)

IAxy =

∫Body

xy dm =2r∫

0

√2yr−y2∫

0

xyρ dx dy

= ρ

2r∫0

2yr − y2

2y dy

= ρ

(8r4

3− 2r4

)

=23ρr4

Thus,

Ixy = IAxy = 2

3ρr4 = 2

3πmcr2

147

8.6.13GOAL: Determine the Ixx , Iyy , Izz and Ixy for the illustrated body.GIVEN: Body’s orientation and shape. a = 0.1 m, b = 0.02 m, c = 0.07 m, d = 0.06 m. Thedensity of the large block is 1000 kg /m3 and the density of the small block is 800 kg /m3.DRAW:

FORMULATE EQUATIONS: We’ll solve this problem by breaking it up into two rectangularbodies, one with dimensions a, a, b and the other with dimensions e, e, d, where e = a − c and isintroduced for convenience. e = 0.03 m. From Appendix B we have:

Ix′x′

= m

(a2 + b2

12

), I

y′y′= m

(a2 + b2

12

), I

z′z′= m

(a2 + a2

12

)for the large block and

Ix′x′

= m

(d2 + e2

12

), I

y′y′= m

(d2 + e2

12

), I

z′z′= m

(e2 + e2

12

)for the small block.To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with appropriatevalues of r

1, r

2and r

3.

SOLVE: Let’s first consider the body with dimensions a, a, b. We’ll put a “1” to the upper left ofthe I as an identifier.For this body r

1= a/2, r

2= a/2 and r

3= b/2.

1Ixx = 1Ix′x′

+m1(r2

2+ r2

3)

= m1

(a2 + b2

12

)+m

1

(a2 + b2

4

)= m

1

(a2 + b2

3

)1Iyy = 1I

y′y′+m

1(r2

3+ r2

1)

= m1

(a2 + b2

12

)+m

1

(a2 + b2

4

)= m

1

(a2 + b2

3

)1Izz = 1I

z′z′+m

1(r2

1+ r2

2)

= m1

(a2 + a2

12

)+m

1

(a2 + a2

4

)= m

1

(a2 + a2

3

)1Ixy = 1I

x′y′+m

1r1r2

= m1

(a2

4

)Now we’ll consider the body with dimensions d, e, e. We’ll put a “2” to the upper left of the I as

148

an identifier.For this body r

1= e/2, r

2= c+ e/2 and r

3= b+ d/2.

2Ixx = 2Ix′x′

+m2(r2

2+ r2

3)

= m2

(d2 + e2

12

)+m

2

[(c+ e

2)2

+(b+ d

2)2]

2Iyy = 2Iy′y′

+m2(r2

3+ r2

1)

= m2

(d2 + e2

12

)+m

2

[(e2)2

+(b+ d

2)2]

2Izz = 2Iz′z′

+m2(r2

1+ r2

2)

= m2

(e2 + e2

12

)+m

2

[(e2)2

+(c+ e

2)2]

2Ixy = 2Ix′y′

+m2r1r2

= m2

(e

2

)(c+

e

2

)The individual masses are given by

m1

= (1000 kg /m3)(0.1 m)(0.1 m)(0.02 m) = 0.2 kg

m2

= (800 kg /m3)(0.03 m)(0.03 m)(0.06 m) = 0.0432 kg

Adding the rotational inertias together and evaluating for the given dimensional values gives us

Ixx = 1Ixx + 2Ixx = (1.6×10−2 m2)(0.2 kg) + (1.01×10−2 m2)(0.0432 kg)

Iyy = 1Iyy + 2Iyy = (1.6×10−2 m2)(0.2 kg) + (3.1×10−3 m2)(0.0432 kg)

Izz = 1Izz + 2Izz = (6.6×10−3 m2)(0.2 kg) + (7.6×10−3 m2)(0.0432 kg)

Ixy = 1Ixy + 2Ixy = (2.5×10−3 m2)(0.2 kg) + (1.275×10−3 m2)(0.0432 kg)

Ixx = 3.77×10−3 kg·m2 Iyy = 3.47×10−3 kg·m2

Izz = 1.66×10−3 kg·m2 Ixy = 5.55×10−4 kg·m2

149

8.7 Angular Momentum

150

8.7.1GOAL: Determine the angular momentum of the illustrated body about the fixed point O.GIVEN: The body rotates about the Y -axis with angular speed ω, and has areal density ρ.DRAW:

FORMULATE EQUATIONS: Let x, y, z be a set of body-fixed axes with origin O. At theinstant illustrated, these axes are aligned with the ground-fixed X,Y, Z axes. The body rotatesabout the Y -axis with angular speed ω; thus the z and Z axes remain aligned and the angularvelocity may be expressed as: ω = ω . The angular momentum of the body about point O is then:

HO =

[IXX

ω1− I

XYω

2− I

XZω

3

]ı +[

IYY

ω2− I

YZω

3− I

YXω

1

] +[

IZZω

3− I

ZXω

1− I

ZYω

2

]k

HO = −IXY

ω ı + IY Y

ω − IZYω

k

Once we have determined the relevant inertia values about the point O, we may substitute andsolve for the angular momentum.SOLVE: We only need to determine I

XY, I

Y Yand I

Y Z. I

Y Yis easily found as the mass moment

of inertia about the Y axis is simply that of two thin rods, with masses ρab, ρac and lengths b, c,respectively, rotated about their ends:

IY Y

=(ρac)c2

3+

(ρab)b2

3=ρa(b3 + c3)

3Using the appropriate equations from (8.21)-(8.29) lets us determine the products of inertia. Eachrectangle has no product of inertia when evaluated about its mass center and thus the only finiteproducts of inertia come about because we’re shifting our attention from each plate’s mass centerto the point O.

IXY

= ρ(ab)a

2

(−b2

)= −ρa

2b2

4

IY Z

= ρ(ac)a

2c

2=ρa2c2

4

HO =(ab2

4ı + b3 + c3

3 − ac2

4k

)ρaω

151

8.7.2GOAL: Determine the angular momentum of the illustrated rectangular body about the fixedpoint O.GIVEN: The body rotates about the Z-axis with angular speed ω, and has mass m.DRAW:

FORMULATE EQUATIONS: Let x, y, z be a set of body-fixed axes with origin O. At theinstant illustrated, these axes are aligned with the ground-fixed X,Y, Z axes. The body rotatesabout the Z-axis with angular speed ω; thus the z and Z axes remain aligned and the angularvelocity may be expressed as: ω = ω

k . The angular momentum of the body about point O is then:

HO =

[Ixxω1

− Ixyω2− Ixzω3

]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

HO = −Ixzωı − Iyzω

+ Izzωk (1)

Once we have determined the relevant inertia values about the point O, we may substitute andsolve for the angular momentum.SOLVE: The products of inertia about parallel axes through the center of mass G are all zero, dueto the symmetry of the body. The products of inertia about the point O may be computed fromthe parallel axis expressions for inertia.

Ixz : Ixz = Ix′z′

+mr1r3

= 0 +m

(a

2

)(− b

2

)= −1

4mab

Iyz : Iyz = Iy′z′

+mr2r3

= 0 +m

(c

2

)(− b

2

)= −1

4mcb

The moment of inertia about the z′-axis through the mass center is: Iz′z′

= 112m

(a2 + c2

). Using

the parallel axis expressions for inertia again, we obtain:

Izz :Izz = I

z′z′+ m

(r21

+ r22

)=

112m(a2 + c2

)+m

[(a

2

)2

+(c

2

)2]

=13m(a2 + c2

)Substituting into (1),

HO = mω(14 ab

ı + 14 cb

+ 13(a2 + c2

)k)

152

8.7.3GOAL: Determine the angular momentum of the illustrated ring.GIVEN: The outer radius of the washer is r

2, the inner radius is r

1m, the mass is m and the

angular velocity is ωk .

DRAW:

FORMULATE EQUATIONS: From Appendix B we have the following inertias, taken aboutthe principal axes x′, y′, z′ of a solid cylinder, oriented as the ring is for this problem.

Ix′x′

= Iz′z′

=m(3r2 + h2)

12, I

y′y′=mr2

2All products of inertia are zero.SOLVE: For our problem we’ll find the moments of inertia for two solid cylinders, one of radiusr1

and one of radius r2. By subtracting we’ll obtain the inertias for the ring. The mass of the ring

is given by

m = πhρ(R22−R2

1)

Outer radius cylinder:

oIx′x′

= oIz′z′

=πR2

2hρ(3R2

2+ h2)

12

oIy′y′

=πR2

2hρR2

2

2=πhρR4

2

2Inner radius cylinder:

iIx′x′

= iIz′z′

=πR2

1hρ(3R2

1+ h2)

12

iIy′y′

=πR2

1hρR2

1

2=πhρR4

1

2We now subtract the inertias of the smaller cylinder from the inertias of the larger one.

Ix′x′

= Iz′z′

=π hρ

[3(R4

2−R4

1

)+ h2

(R2

2−R2

1

)]12

=πhρ

(R2

2−R2

1

) [h2 + 3

(R2

1+R2

2

)]12

=m[h2 + 3

(R2

1+R2

2

)]12

153

Iy′y′

= IY Y

=π hρ

(R2

2−R2

1

) (R2

1+R2

2

)2

=m(R2

1+R2

2

)2

Using the parallel axis theorem to find IXX

and IZZ

we obtain

IXX

= IZZ

=m[h2 + 3

(R2

1+R2

2

)]12

+m

(h

2

)2

=mh2

3+m(R2

1+R2

2)

4

The products of inertia about X, Y , Z are zero from symmetry.The general expression for angular momentum (using ω = ω

k ) is therefore

HO =

[mh2

3 +m(R2

1+R2

2)

4

]ω

k

154

8.7.4GOAL: Calculate the angular momentum of the system about point O.GIVEN: The system consists of four thin bars of length a, and has an angular velocity ω =(10 + 18

k)

rad/s.DRAW:

FORMULATE EQUATIONS: The framework is symmetric in the x−y, x−z, and y−z planes,thus all products of inertia about the mass center G are zero. The body-fixed x, y, z axes aretherefore a set of principle axes for the system, and the angular momentum may be expressed as:

HG = Ixxω1

b1 + Iyyω2

b2 + Izzω3

b3 (1)

To determine the angular momentum about the point O, we’ll use the expression,

HO =

HG + rG/O×mv

G(2)

SOLVE: We need to find the moments of inertia Ixx , Iyy , and Izz about the mass center. Assumethe bars have linear density ρ, and consider a differential mass unit dm = ρ dL, where dL is adifferential unit of length along a bar. For a unit change in length dL along the bar, there is acorresponding change

√2 dz along the z-axis. dm can thus be expressed as:

dm = ρ dL = ρ√

2 dz

Since the bars are one-dimensional, a single variable of integration (chosen to be z) is all that’sneeded to calculate the inertias. We’ll integrate over one of the bars, bar A in the diagram, andthen use our results to find the moments of inertia of the entire system. Note that the limits on zare 0 and a√

2. For IA

yy :

IA

yy =∫

BodyA

(x2 + z2

)dm =

a√2∫

0

(0 + z2

)ρ√

2 dz =16ρa3

By symmetry, we expect IA

zz = IA

yy . Although we could perform this integration over z with thesubstitution y = a√

2− z along bar A, it will be easier to just apply the same arguments as above

to the y-axis and integrate over y:

IA

zz =∫

BodyA

(x2 + y2

)dm =

a√2∫

0

(0 + y2

)ρ√

2 dy =16ρa3

155

As x2 = 0 in the above equations, for IA

xx we have:

IA

xx =∫

BodyA

(y2 + z2

)dm = IA

yy + IA

zz =13ρa3

The moments of inertia of all four bars about the center of mass of the framework will be the same,due to the symmetry of the system. Thus,

Ixx = IA

xx + IB

xx + IC

xx + ID

xx = 4IA

xx =43ρa3

Similarly,

Iyy = Izz =23ρa3

The total mass of the system is M = ρ(4a), so rewriting these expressions in terms of M we obtain:

Ixx =13Ma2 Iyy = Izz =

16Ma2 (3)

There are, of course, a number of ways to determine the above moments of inertia. A secondmethod worth mentioning involves the use of symmetry arguments and the parallel axis expressions.Through the symmetry of the framework, we know that Iyy = Izz , and since x2 = 0, we haveIxx = Iyy + Izz = 2Iyy = 2Izz . Thus, we only need to compute one of the inertia values, Ixx . Thismay be computed from the known moment of inertia of a thin bar about a transverse axis, 1

12mL2,

the fact that there are 4 bars, and the parallal axis expressions:

Ixx = 4

(Ix′x′

+m(y2 + z2

))= 4

(112

(M

4

)a2 +

M

4

(a

2

)2)

=13Ma2

where a2 is the distance from G to the center of any bar.

We are now ready to compute the angular momentum of the system. Note that at the giveninstant, the body-fixed

b1,

b2,

b3 unit vectors are aligned with the ground-fixed ı , ,

k unit vectors.

Returning to equation (1), with substitutions from (3),

HG = Ixxω1

b1 + Iyyω2

b2 + Izzω3

b3

=13Ma2 (0) ı +

16Ma2 (10) +

16Ma2 (18)

k

=53Ma2 + 3Ma2

k

Now substituting into equation (2), with vG

= ω×rG/O

HO =

HG + rG/O×mv

G

=53Ma2 + 3Ma2

k +a√2

k ×M

[(10 + 18

k)× a√

2k

]=

203Ma2 + 3Ma2

k

HO = Ma2(

203

+ 3k)

156

8.7.5GOAL: Calculate the angular momentum of the object about point O.GIVEN: The object has an angular velocity ω = (−6 + 10 ) rad/s.DRAW:

FORMULATE EQUATIONS: The angular momentum of the body about point O is given by:

HO =

[Ixxω1


]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

(1)

Once we have determined the relevant inertia values about the point O, we can substitute and solvefor the angular momentum.SOLVE: We’ll break the body into two pieces: AB and OC and identify the horizontal bar AB as“H” and the vertical bar OC as “V .”

mH

=2m3, m

V=m

3

HIXX

= HIZZ

=m

HL2

3=

2mL2

9HI

Y Y= HI

XY= HI

Y Z= HI

ZX= 0

V IXX

= V IY Y

=m

HL2

3=mL2

9V I

ZZ= V I

XY= V I

Y Z= V I

ZX= 0

Our overall rotational inertias are given by the sum of the two components:

IXX

=2mL2

9+mL2

9=mL2

3

IY Y

=mL2

9, I

ZZ=

2mL2

9

IXY

= IY Z

= IZX

= 0

Using these values in (1), along with ω1

= −6 rad/s, ω2

= 10 rad/s, gives us

HO = Ixxω1ı + Iyyω2

+ Izzω3

k

= mL2

3 (−6 rad/s)ı + mL2

9 (10 rad/s)

HO = mL2(−2ı + 10

9)

rad/s

157

8.7.6GOAL: Calculate the angular momentum of the disk D about G and O.GIVEN: The object has an angular velocity ω = (76ı − 24 ) rad/s. a = 0.3 m, r = 0.25 m,m = 10 kg.DRAW:

Note that the ω1

and ω2

of the original figure don’t correspond to angular rotation rates about theı , axes, respectively.

FORMULATE EQUATIONS: The angular momentum expressions for the body when rotatingabout the points O and G are given by:

HG =

[Ixxω1


]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

(1)

HO =

[Ixxω1


]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

(2)

where ωD = ω1

ı + ω2

+ ω3

k = (76ı − 24 ) rad/s.

Once we have determined the relevant inertia values about the point O, we can substitute and solvefor the angular momentum.SOLVE:

Iy′y′

= Iz′z′

=mr2

4, I

x′x′=mr2

2(3)

All products of inertia are zero.We can use the parallel axis theorem to find the corresponding rotational inertias about O:

IY Y

= IZZ

=mr2

4+ma2, I

XX=mr2

2(4)

Due to symmetry the products of inertia are still zero.

(1), (3) ⇒

HG = mr2

2 ω1

ı + mr2

4 ω2

= (10 kg)(0.25 m)22 (76ı − 12 ) rad/s

= (23.75ı − 3.75 ) kg·m2/ s

158

(2), (4) ⇒

HO = mr2

2 ω1

ı +m

(r2

4 + a2

)ω

2

= (10 kg)[(

(0.25 m)22

)76ı − 24

((0.25 m)2

4 + (0.3 m)2)]

rad/s

= (23.75ı − 25.35 ) kg·m2/ s

159

8.7.7GOAL: Calculate the angular momentum of the system about O.GIVEN: The framework has a mass of 0.005 kg and each arm has a length of h = 0.12 m. Thewheels have radii r = 0.007 m and negligible thickness. The framework rotates about the Y axisat 2.2 rad/s and the wheels rotate without slip. Each wheel has a mass of 0.001 kg.DRAW:

FORMULATE EQUATIONS:

HO =

[Ixxω1


]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

(1)

Once we have determined the relevant inertia values about the point O, we can substitute and solvefor the angular momentum.SOLVE:Although the system contains several pieces, its symmetry allows us to simplify the analysis substan-tially. Each individual wheel will generate an angular momentum contribution due to its rotationabout the axis along the corresponding arm (the X axis for wheel A in the figure, for instance).Because an identical contribution will be generated by each wheel, their sum will go to zero. Thuswe’re left with the same angular momentum as we’d have for the 3-arm assemble and attachedwheels rotating about the Y axis as a single rigid body.Each arm of the 3-arm body has mass moment of inertia about the Y axis of

IY Y

=m

3−a

3

(h2

3

)=m

3−ah2

9

For each wheel we have

IY Y

=mwr

2

4+mwh

2

The complete rotational inertia is therefore

IY Y

= 3

[(0.005 kg)(0.12 m)2

9+ (0.001 kg)

((0.007 m)2

4+ (0.12 m)2

)]= 6.72×10−5 kg·m2

Thus we have

HO =(6.72×10−5 kg·m2

)(2.2 rad/s) = 1.48×10−4 kg·m2/ s

160

8.7.8GOAL: Determine the angular momentum of the illustrated rectangular body about the fixedpoint O.GIVEN: The body rotates about the Z-axis with angular speed ω

3, and has a linear density ρ.

DRAW:

FORMULATE EQUATIONS: Let x, y, z be a set of body-fixed axes with associated unit vectorsb1,

b2,

b3. The general expression for the angular momentum of the body about point O is:

HO =

[Ixxω1


]ı +[

Iyyω2− Iyzω3

− Iyxω1

] +[

Izzω3− Izxω1

− Izyω2

]k

Letting ω1

= ω2

= 0 gives us

HO = (−Ixz ı +−Iyz + Izzk )ω

3

Once we have determined the relevant inertia values about the point O, we can substitute and solvefor the angular momentum.SOLVE: Denote the mass of the segment OA by m

1and the mass of the segment AB by m

2.

Izz =m

1L2

1

3+m

2L2

1

Ixz = m2L

1

(−L2

2

)= −

m2L

1L

2

2

Iyz = 0

HO =

[m

2L

1L

2

2ı +

(m

1L2

1

3+m

2L2

1

)k

]ω

3

Re-expressing m1

and m2

in terms of ρ, L1, L

2gives us

HO = ρL1ω

3

[L2

22

ı +

(L2

13 + L

1L

2

)k

]

161

8.7.9GOAL: Find the angular momentum of the illustrated structure about the point O.GIVEN: The angular velocity of the structure is ω = ω

b1 and its areal density is ρ.

DRAW:

FORMULATE EQUATIONS: The equation of interest is:

HO =[Ixxω1


]b1 +[

Iyyω2− Iyzω3

− Iyxω1

]b2 + (1)[

Izzω3− Izxω1

− Izyω2

]b3

SOLVE: Since ω = ωb1, equation (1) gives us:

HO = Ixxωb1 − Iyxω

b2 − Izxω

b3

We need to calculate Ixx , Iyx , and Izx .The moment of inertia of a thin, rectangular plate about an axis perpendicular to its surface andthrough the center of mass is 1

12m(l2 +w2), where l is the length of the plate and w its width. Themoment of inertia about a transverse axis lying in the plane of the plate, through the center ofmass and parallel to an edge of length w is 1

12ml2. Using the parallel axis expressions, the moment

of inertia about the x axis for each plate is:

IB

xx = IB

x′′x′′+mB

(r22

+ r23

)=

112mB

(b2 + c2

)+mB

[(b

2

)2

+(− c

2

)2]

=13mB

(b2 + c2

)

IA

xx = IA

x′x′+mA

(r22

+ r23

)=

112mAc

2 +mA

(− c

2

)2

=13mAc

2

For the composite structure,

Ixx = IB

xx + IA

xx =13ρbc

(c2 + b2

)+

13ρacc2 =

13ρ(ac3 + b3c+ bc3

)162

The plane of symmetry for plate A is the x−z plane, so IA

yx = 0. The plane of symmetry for plate Bis the y−z plane, so IB

yx = IB

zx = 0. The only remaining moment of inertia needed for substitutioninto equation (1) is IA

zx . Notice that all products of inertia about the center of mass of each plateare zero, thus by the parallel axis expressions we have:

Izx = IA

zx = IA

z′x′+mAr3r1 = 0 +mA

(− c

2

)(a

2

)= −1

4ρa2c2

Substition into (1) yields

HO = 13 ρ

(ac3 + b3c+ bc3

)ω

b1 + 1

4 ρa2c2ω

b3

163

8.8 Equations of Motion for a Three-Dimensional Body

164

8.8.1GOAL: Determine the bearing forces for a rotating shaft.GIVEN: System configuration. Each lumped mass is 2 kg. a = 14 cm and b = 11 cm. ω =100 rad/s.DRAW:

FORMULATE EQUATIONS: We’ll use the a force balanceF = maG

and three moment balances about the shaft’s mass center∑M

1= H

1+ (ω

2H

3− ω

3H

2)∑

M2

= H2

+ (ω3H

1− ω

1H

3)∑

M3

= H3

+ (ω1H

2− ω

2H

1)

where

HG = H1

b1 +H

2

b2 +H

3

b3

SOLVE: From (8.32) we have:

HG = −Ix′z′ωb1 − Iy′z′ω

b2 + Iz′z′ω

b3

Ix′z′ = m1

(3a2

)(0) +m

2

(a

2

)(−b) +m

3

(−a

2

)(0) +m

4

(−3a2

)(b)

Ix′z′ = −ab2

(m2

+ 3m4) = −2mab

Iy′z′ = m1

(3a2

)(−b) +m

2

(a

2

)(0) +m

3

(−a

2

)(b) +m

4

(−3a2

)(0)

Iy′z′ = −ab2

(3m1

+m3) = −2mab

Iz′z′ = b2(m1

+m2

+m3

+m4) = 4mb2

HG = 2mabωb1 + 2mabω

b2 + 4mb2ω

b3

165

H1 = 2mabωH2 = 2mabωH3 = 4mb2ω

M1 = 0− 2mabω2 = (R4−R

2)5a2

(1)

M2 = 0 + 2mabω2 = (R1−R

3)5a2

(2)

M3 = 0 (3)R

1+R

3= 0 (4)

R2

+R4

= 0 (5)

(5) → (1) ⇒ 5aR4

= −2mabω2 ⇒ R4

= −2mbω2

5(6)

(6) → (5) ⇒ R2

=2mbω2

5

(4) → (2) ⇒ 5aR1

= 2mabω2 ⇒ R1

=2mbω2

5(7)

(7) → (4) ⇒ R3

= −2mbω2

5Using the given values:

R1

= 2(2 kg)(0.11 m)(100 rad/s)25 = 880 N

R3

= −880 N R2

= 880 N R4

= −880 N

166

8.8.2GOAL: Find the moment the shaft must exert on the disk for it to rotate with constant angularspeed ω.GIVEN: The radius of the disk is R and its areal density is ρ. The disk is inclined an angle φfrom the plane normal to the axis of the shaft, as shown.DRAW: Let x, y, z be a set of body-fixed axes attached to the disk, with the origin at its centerof mass. The z-axis is chosen to be normal to the plane of the disk.

FORMULATE EQUATIONS: The angular momentum equation of motion has the form:

∑

MG

=

HG

One method of solution is therefore to form

HG

and take its derivative. A second method of solutionwould be to use Euler’s equations, since the x, y, z axes are a set of principle axes for the disk. Wewill use both methods and check that the solutions are the same.SOLVE: The moments of inertia of the disk about the x, y, z axes are Ixx = Iyy = 1

4mR2, and

Izz = 12mR

2. The angular velocity of the disk is:

ω = ωk = ω

(cosφ

b3 + sinφ

b1

)

Since the angular velocity is constant, the angular acceleration α is zero.By symmetry, all products of inertia in the x, y, z coordinate system are zero (and thus x, y, z areprinciple axes). The angular momentum then becomes:

HG

= Ixxω1

b1 + Iyyω2

b2 + Izzω3

b3

=14mR2ω sinφ

b1 + 0 +

12mR2ω cosφ

b3 (1)

167

Taking the derivative:

∑

MG

=

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG

+ ω×

HG

= 0 + ω(cosφ

b3 + sinφ

b1

)×(

14mR2ω sinφ

b1 +

12mR2ω cosφ

b3

)= −1

4mR2ω2 sinφ cosφ

b2

Note that the unit vectorb2 is always in the direction perpendicular to the axis of the shaft. With

m = ρπR2, we have:

M = − 14 ρπR

4ω2 sinφ cosφb2

Since α and ω2

are zero, the only non-zero Euler’s equation is

M2

= Iyyα2+ ω

1ω

3

(Ixx − Izz

)= 0 + (ω sinφ) (ω cosφ)

(14mR2 − 1

2mR2

)

M2

= − 14 ρπR

4ω2 sinφ cosφ

which yields an identical result as above.

168

8.8.3GOAL: Determine the bearing forces at A.GIVEN: System configuration. The shaft has a mass m

1and the half-disk has a mass m

2.

DRAW:

The primed axes are attached to the disk and rotate along with it, as do the unit vectorsb1,

b2,

b3. The radius of the half-disk is given by r = d/2.

FORMULATE EQUATIONS: We’ll use the a force balanceF = maG


1= H

1+ (ω

2H

3− ω

3H

2)∑

M2

= H2

+ (ω3H

1− ω

1H

3)∑

M3

= H3

+ (ω1H

2− ω

2H

1)

where

HG = H1

b1 +H

2

b2 +H

3

b3

SOLVE: From (8.32) we have:

HG = −Iz′x′ωb1 − Iy′z′ω

b2 + Iz′z′ω

b3

Iz′x′

= Iy′z′

= 0

Izz =m

2r2

4

HG =m

2r2ω

4b3

H1

= 0, H2

= 0, H3

=m

2r2ω

4We can now consider our force and moment balances.

169

M1

= 0 = (R4−R

2)3d2

(1)

M2

= 0 = (R1−R

3)3d2

(2)

R2

+R4− (m

1+m

2)g cos θ = −m

2rω2 = −

4m2rω2

3π(3)

R1

+R3− (m

1+m

2)g sin θ = 0 (4)

(2), (4) ⇒ 2R1

= g(m1

+m2) sin θ

R1

= R3

=g(m

1+m

2) sin θ

2

(1), (3) ⇒ 2R2

= g(m1

+m2) cos θ −

4m2

(d

2

)ω2

3π

R2

= R4

=g(m

1+m

2) cos θ

2 − m2dω2

3π

170

8.8.4GOALS:(0.1) Find the full equations of motion for the illustrated system under the condition ψ = 0.(0.2) Find the equilibrium conditions for which ψ = ψ

0(constant), θ = θ = 0 and θ = θ

06= 0.

GIVEN: The rod is attached by a horizontal pivot to a vertical shaft, as shown in the figure. Thelinear density of the rod is ρ.DRAW: Let x, y, z be a body-fixed coordinate system attached to the rod, with correspondingunit vectors c1,

c2,c3. Let

b1,

b2,

b3 be a set of unit vectors that rotate with the vertical shaft,

and, at the instant depicted, are aligned with the ground-fixed X,Y, Z axes.

The coordinate transformation matrix betweenbi and ci is:

b1

b2

b3

c1 1 0 0c2 0 cos θ − sin θc3 0 sin θ cos θ

GOVERNING EQUATIONS: The full equations of motion will consist of three force equationsand three moment equations. Newton’s law provides the force equations, and since x, y, z areprinciple axes for the rod (two planes of symmetry, x−z and y−z), Euler’s equations will providethe moment equations:

F = maG (1)

∑M

1= Ixxα1

+ ω2ω

3

(Izz − Iyy

)(2)∑

M2

= Iyyα2+ ω

1ω

3(Ixx − Izz ) (3)∑

M3

= Izzα3+ ω

1ω

2

(Iyy − Ixx

)(4)

where

MO = M1

c1 +M2

c2 +M3

c3 is the net moment acting on the rod about the point O.SOLVE: We’ll begin with the left side of equations (1-4). The forces acting on the rod are theforce of gravity and the reaction force between the rod and shaft at point O. Denote the reaction

171

force acting on the rod as having components Ri

in the ci frame. The net force acting on the rodis then:

F = R

1c1 +R

2c2 +R

3c3 −mg

b3

= R1

c1 +R2

c2 +R3

c3 − ρLg (− sin θc2 + cos θc3) (5)

The rod pivots freely about the c1 direction, so we may write the reaction torque at O as havingonly two components, T

2and T

3, in the c2 and c3 directions. The net moment

MO acting on therod is then:

MO = T2

c2 + T3

c3 + rG/O×(−mg

b3

)= T

2c2 + T

3c3 +

(−L

2c3

)×−ρLg

(− sin θc2 + cos θc3

)= T

2c2 + T

3c3 +

12ρgL2 sin θc1 (6)

That takes care of the left side of our equations. We will now address the kinematic and inertialterms appearing on the right side of equations (1-4). We begin with the angular velocity of therod, which is:

ω = ψb3 − θ

b1 = ψ (cos θ c3 − sin θ c2)− θc1 (7)

Taking a time derivative, the angular acceleration is

α = ψb3 + ψ

db3

dt− θ

b1 − θ

db1

dt

= (0)b3 + ψ(0)− θ

b1 − θψ

b2

= −θc1 − θψ cos θc2 − θψ sin θc3 (8)

Since O is fixed in space, the velocity and acceleration of the rod’s center of mass may be computedas

vG = ω×rG/O

=(ψ cos θ c3 − ψ sin θ c2 − θc1

)×−L

2c3

=12ψL sin θ c1 −

12θLc2 (9)

aG = α×rG/O+ ω×vG

=(−θc1 − θψ cos θc2 − θψ sin θc3

)×−L

2c3 +

+(ψ cos θ c3 − ψ sin θ c2 − θc1

)×(12ψL sin θ c1 −

12θL c2

)= ψθL cos θ c1 +

12L(ψ2 sin θ cos θ − θ

)c2 +

12L(ψ2sin2θ + θ2

)c3 (10)

The moments of inertia of the thin rod about the x, y, z axes are Ixx = Iyy = 13mL

2 = 13ρL

3 andIzz = 0. Substitution of (6),(7),(8) into our moment balance equations (2-4), with m = ρL, nowyields:

172

(2) →

[∑M

1=

12mgL sin θ

]=

13ρL3

(−θ)

+(−ψ sin θ

) (ψ cos θ

)(0− 1

3ρL3

)12ρgL2 sin θ =

13ρL3

(−θ + ψ2 sin θ cos θ

)

⇒ θ + 3g2L sin θ − ψ2 sin θ cos θ = 0 (11)

This is the equation of motion for θ, integration of which will yield θ(t).

(3) →[∑

M2

= T2

]=

13ρL3

(−θψ cos θ

)+(−θ) (ψ cos θ

)(13ρL3 − 0

)

⇒ T2

= −23ρL

3θψ cos θ (12)

(4) →[∑

M3

= T3

]= 0

(−θψ sin θ

)+(−θ) (−ψ sin θ

)(13ρL3 − 1

3ρL3

)

⇒ T3

= 0 (13)

Note that the constraint torque T3

has been included in our analysis since rotation of the rod aboutc3 is constrained (unlike the rotation about c1, for example). However, our asssumption that the

rod is one-dimensional requires that this constraint torque is always zero.Let F

i=

F · ci and a

i= a · ci. A force balance (1) then takes the form (with m = ρL):

F = maG ⇒ F

i= ρLa

i(i = 1, 2, 3)

(F = maG) · c1 ⇒ R

1= ρψθL2 cos θ (14)

(F = maG) · c2 ⇒ R

2+ ρLg sin θ = 1

2ρL2(ψ2 sin θ cos θ − θ

)(15)

(F = maG) · c3 ⇒ R

3− ρLg cos θ = 1

2ρL2(ψ2sin2θ + θ2

)(16)

You might notice that force equation (15) and motion equation (11) share some common terms.Although unnecessary for our purposes, this leads to an interesting simplification for R

2, in that

both the second and first order time-derivatives may be eliminated:

(11) → (15) ⇒ R2

= −14ρLg sin θ

(0.2) Substitution of the given equilibrium conditions into the equation of motion (11) yields:

0 +3g2L

sin θ0− ψ

02 sin θ

0cos θ

0= 0

⇒ 3g2L

− ψ02 cos θ

0= 0

⇒ θ0

= cos−1

(3g

2Lψ 20

)

173

which is the equilibrium value of θ for a given ψ0.

In order for the above inverse cosine function to exist, the operand must lie in the range [-1,1].Thus,

−1 ≤ 3g2Lψ 2

0

≤ 1

Since all quantities appearing in the operand are positive, this is equivalent to:

ψ 20

≥ 3g2L

If ψ0

is too small, no such equilibria (θ06= 0) exists. Also note that the same argument implies

cos θ ≥ 0, assuring −π/2 ≤ θ0≤ π/2.

174

8.8.5GOAL: Find the rotational speed that allows θ to remain constant at 45. Is this the same resultas if all the mass were concentrated at C?GIVEN: System configuration. L = 0.5 m, m = 2kg.DRAW:

FORMULATE EQUATIONS:We’ll use the a force balance

F = maG


1= H

1+ (ω

2H

3− ω

3H

2)∑

M2

= H2

+ (ω3H

1− ω

1H

3)∑

M3

= H3

+ (ω1H

2− ω

2H

1)

where

HG = H1

b1 +H

2

b2 +H

3

b3

SOLVE:(a) Using (8.33) and ω = ω

k gives us

HO = −Izxωb1 − Iyzω

b2 + Izzω

b3

Izz =mL2

12+mL2 sin2 θ = mL2

(112

+ sin2 θ

)Iyz = mL2 cos θ sin θIzx = 0

H1

= 0, H2

= −mL2 cos θ sin θω, H3

= mL2(

112 + sin2 θ

).

H1

= 0, H2

= 0 and H3

= 0.The first of our moment balance equations yields∑

M1

= −ωH2

= ω2mL2 cos θ sin θ = mgL sin θ

ω2 =g

L cos θ=

9.81 m/s2

(0.5 m)(1/√

2)= 27.75 rad/s2

175

ω = 5.27 rad/s(b) If all the mass were at C then we’d have:

Izz = mL2 sin2 θ

Iyz = mL2 cos θ sin θIzx = 0

Iyz is the same as in (a). Therefore the results for this case will not differ from those of (a).

176

8.8.6GOAL: Describe the effect of lateral perturbations on the front wheel of a bicycle, when one isriding with no hands.DRAW:

ASSUME: Initially, the bike travels in a perfectly straight line and is perfectly upright.FORMULATE EQUATIONS: The sum of the moments about the mass center of the wheelmust equal the time rate of change of its angular momentum.∑

MG =d

dt

HG

SOLVE: Let ı be the forward vector, the right-to-left vector, andk the vertical up direction.

Initially, the front wheel has an angular momentum about its mass center with only one componentin the direction,

HG = H2

. Assume that instead of riding with no hands, the handlebars areheld fixed by the rider. If the bike leans to the left (ω · ı 6= 0), the angular momentum vectorwill tilt towards the ground, in the −

k direction. Thus, the net moment on the wheel must alsohave component in the −

k direction (provided by the rider’s grip on the handlebars). We conclude,therefore, that with no hands the wheel will have a tendency to turn in the positive

k direction,

or to the left.A similar argument shows that when the rider leans right, the wheel will turn to the right as well.

Leaning causes the bike to turn in the direction of the lean.

We may obtain the same result with an examination of a moment balance. Letb1,

b2,

b3 be a set

of body-fixed unit vectors initially aligned with ı , k . If the bike is traveling forward and leaning

to the right then its angular velocity may be represented as ω = ω1

b1 + ω

2

b2 and the angular

momentum as

HG = H1

b1 + H

2

b2. Our third moment balance equation, corresponding to the

b3

component, is ∑M

3= Izzα3

+ ω1ω

2(Iyy − Ixx)

For a wheel, Iyy > Ixx . With no hands, and assuming a negligible resistance to turning at thecontact point between the wheel and ground, the net moment

∑M

3will be zero and thus

α3

= −ω

1ω

2(Iyy − Ixx)Izz

< 0

Thus the wheel turns to the right.

177

The resulting motion of the wheel will tend to stabilize the bicycle. When the wheel turns, thecentripetal acceleration of the bicycle requires a frictional force between the wheels and groundtoward the center of curvature of the rider’s path. The moment this force exerts on the bicycle willtend to counter the moment due to gravity.

178

8.8.7GOAL: Find the total force

F and moment

MG acting on the dynamical potato.GIVEN: Linear velocity and acceleration; angular velocity and acceleration; mass of potato; andall moments and products of inertia.DRAW:

GOVERNING EQUATIONS:F = ma (1)

∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG (2)

SOLVE: Determing the net force acting on the dynamical potato is a simple matter.

(1) →

F =

(25 kg

)(4ı m/s2

)F = 100ı N

To apply (2), we require the angular momentum about G, which may be computed from theexpression

[Ixxω1


]b1 +

HG =[Iyyω2

− Iyzω3− Iyxω1

]b2 + (3)[

Izzω3− Izxω1

− Izyω2

]b3

At the given instant,

[(0.05)(8)− (0.02)(−12)− (0.04)(3)] ı +

HG = [(0.03)(−12)− (0.02)(8)− (0.07)(3)] +

[(0.10)(3)− (0.04)(8)− (0.07)(−12)]k

= (0.52 ı − 0.73 + 0.82k ) kg·m2/s

179

Now, with the substitution of (3) into (2),

∑

MG =d

dt

∣∣∣∣S

HG + ω×

HG

=[Ixxα1

− Ixyα2− Ixzα3

]b1 +[

Iyyα2− Iyzα3

− Iyxα1

]b2 +[

Izzα3− Izxα1

− Izyα2

]b3 + ω×

HG

= [(0.05)(85)− (0.02)(−100)− (0.04)(30)] ı +[(0.03)(−100)− (0.02)(85)− (0.07)(30)] +

[(0.10)(30)− (0.04)(85)− (0.07)(−100)]k +

(8 ı − 12 + 3k )×(0.52 ı − 0.73 + 0.82

k )

MG =(−2.60 ı − 11.8 + 7.00

k)

N·m

Note that the linear velocity does not appear in any of our computations.

180

8.8.8GOAL: Determine the angular acceleration of the bar and the support force FO at the instant thebar is released. Consider two cases: a) θ = 0 and θ = 30.GIVEN: Mass and dimensions of bar, initial positions before release.DRAW:

FORMULATE EQUATIONS: The equations of motion for a rigid body areF = ma (1)

∑

MO =

HO

∣∣∣∣N

=d

dt

∣∣∣∣S

HO + ω×

HO (2)

Since we are only concerned with the instant of release, ω = 0, and equation (2) becomes:∑

MO =[Ixxα1

− Ixyα2− Ixzα3

]b1 +[

Iyyα2− Iyzα3

− Iyxα1

]b2 + (3)[

Izzα3− Izxα1

− Izyα2

]b3

SOLVE: The moment of inertia for a slim rod about a transverse axis through its mass centeris 1

12mL2. With a little help from the superposition principle and the parallel axis theorem, the

inertias of our rigid body about point O are:

Ixx =13

(37m

)b2 =

17mb2 = 0.0386 kg·m2

Iyy =13

(47m

)a2 +

37ma2 =

1321ma2 = 0.297 kg·m2

Izz = Ixx + Iyy =37m

[a2+

(b

2

)2]=

1321ma2 +

17mb2 = 0.336 kg·m2

Ixy =37ma

(b

2

)=

314mab = 0.0771 kg·m2

Ixz = 0Iyz = 0

181

where m = 3 kg, a = 0.40 m, and b = 0.30 m.The center of mass of the bar is located at

xG =47

(a

2b1

)+

37

(a

b1 +

b

2b2

)=

57a

b1 +

314b

b2 = 0.286

b1 + 0.0643

b2

(a) When θ = 0, the space-fixed ı , ,k and body-fixed

bi unit vectors are aligned. The only

moment about O is that due to gravity, which at the given instant is∑MO = xG×(−mg

k ) =5mga

7 − 3mgb

14ı

Using this result in (3) yields three equations

−3mgb14

= Ixxα1− Ixyα2

5mga7

= Iyyα2− Ixyα1

0 = Izzα3

which may be solved for the angular accelerations

α1

= 15.7 rad/s2; α2

= 32.4 rad/s2; α3

= 0

α = (15.7ı + 32.4 ) rad/s2

The acceleration of the mass center is

a = α×xG = (15.7ı + 32.4 )×(0.286 ı + 0.0643 ) = −8.24k rad/s2

Equation (1) may now be solved for the reaction force,

FO −mg

k = ma ⇒

FO = (3 kg)(−8.24k rad/s2) + (3 kg)(9.81 m/s2)

k

FO = 4.71 N

k

(b) When θ = 30, the gravity vector becomes −g sin 30b2 − g cos 30

b3. The moment about O is

now: ∑MO =

(57a

b1 +

314b

b2

)×(−mg sin 30

b2 −mg cos 30

b3)

= −5amg14

b3 +

5√

3amg14

b2 −

3√

3bmg28

b1

Using this result in (3) and solving for the angular acceleration components yields

α1

= 13.6 rad/s2; α2

= 28.0 rad/s2; α3

= −12.5 rad/s2

α = (13.6b1 + 28.0

b2 − 12.5

b3) rad/s2 = (13.6ı + 30.5 + 3.17

k ) rad/s2


a = α×xG = (13.6b1 + 28.0

b2 − 12.5

b3)×(0.286

b1 + 0.0643

b2)

= (0.805b1 − 3.58

b2 − 7.14

b3) rad/s2

182

and the forceFO is

FO = ma +mg(sin 30

b2 + cos 30

b3)

= (3 kg)(0.805b1 − 3.58

b2 − 7.14

b3) + (3 kg)(9.81 m/s2)(sin 30

b2 + cos 30

b3)

FO = (2.42

b1 + 3.98

b2 + 4.08

b3) N = (2.42ı + 1.41 + 5.52

k ) N

183

8.8.9GOAL: Determine the angular acceleration α of the disk and the reaction force

FO at the given

instant.GIVEN: We’re given the mass and dimensions of the disk as well as its initial position and angularvelocity.DRAW:

GOVERNING EQUATIONS: The equations of motion for a rigid body areF = maG (1)∑

MO =

HO

∣∣∣∣N

=d

dt

∣∣∣∣S

HO + ω×

HO (2)

SOLVE: The center of mass of the disk is located at

rG/O= −Rı − 4R

3π =

(−ı − 4

3π

)m (3)

The mass moments of inertia of the disk about point O are

Ixx =14mR2 =

14(12 kg)(1.0 m)2 = 3 kg·m2

Iyy =14mR2 +mR2 =

54mR2 =

54(12 kg)(1.0 m)2 = 15 kg·m2

Izz =12mR2 +mR2 =

32mR2 =

32(12 kg)(1.0 m)2 = 18 kg·m2

Ixy = m (−R)(−4R

3π

)=

43πmR2 =

43π

(12 kg)(1.0 m)2 =16π

kg·m2

Ixz = Iyz = 0

To apply (2) we require the angular momentum of the disk about O, which may be computed from[Ixxω1


]b1 +

HO =[Iyyω2


]b2 + (4)[

Izzω3− Izxω1

− Izyω2

]b3

184

At the given instant,

HO = [0− (16/π)(2)− 0] ı + [(15)(2)− 0− 0] + [(18)(5)− 0− 0]k

= (−32/π ı + 30 + 90k ) kg·m2/s (5)

Substituting (4) and (5) into (2), with∑

MO = rG×m(−gk ), and then taking the dot products

with ı , ,k yields three equations:

(2) · ı ⇒ 16g/π = [3α1− (16/π)α

2] + (2)(90)− (5)(30)

(2) · ⇒ −12g = [15α2− (16/π)α

1] + (5)(−32/π)

(2) ·k ⇒ 0 = 18α3− (2)(−32/π)

Solving the above for α1, α

2, and α

3,

α1

= −2.14 rad/s2 α2

= −5.18 rad/s2 α3

= −1.13 rad/s2

α = (−2.14ı − 5.18 − 1.13k ) rad/s2


aG = α×rG = (−0.480ı + 1.13 − 4.27k ) m/s2

Equation (1) may now be solved for the forceFO ,

FO −mg

k = maG

⇒ FO = (12 kg)(9.81 m/s2)

k + (12 kg)(−0.480ı + 1.13 − 4.27

k ) m/s2

FO = (−5.76ı + 13.6 + 66.5

k ) N

185

8.8.10GOAL: Find the acceleration aG of the mass center, the angular acceleration α, and the acceler-ation aP of the point P at the instant given.GIVEN: Weight and dimensions of body, its angular velocity; and the force

FQ applied at point

Q.DRAW:

FORMULATE EQUATIONS: The equations of motion for a rigid body areF = maG (1)∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG (2)

SOLVE: From (1), the acceleration of the mass center is

aG =F

m=

[(20ı − 8 + 10k )− 40 ] lbs(

40 lbs32.2 ft/s2

) = (16.1ı − 38.64 + 8.05k ) ft/s2 (3)

The rigid body can be viewed as a composite body with three sections: a lower-left cube, an upper-right cube, and a rectangular parallelpiped in the middle. The moments of inertia for the body arefound by superimposing the inertias of all three sections, and applying the parallel-axis expressionsas needed. Let x′, y′, z′ be a new set of coordinates axes parallel to x, y, z, and with an origin atthe mass center. By symmetry, I

x′z′= I

y′z′= 0.

Ix′x′

= 2[16

(m

4

)h2 +

(m

4

)(h

2

)2]+[16

(m

2

)h2]

=724mh2

Iy′y′

= 2[16

(m

4

)h2 +

(m

4

)(3h2

)2]+[

112

(m

2

)(h2 + (2h)2

)]=

1712mh2

Iz′z′

= 2[16

(m

4

)h2 +

(m

4

)(3h2

)2

+(h

2

)2]+[

112

(m

2

)(h2 + (2h)2

)]=

3724mh2

Ix′y′

=m

4

(32h

)(h

2

)+m

4

(−3

2h

)(−h

2

)=

38mh2

The moment exerted on the body about its mass center is∑

MG = rQ/G×FQ = (−2hı − 0.5h

k )×(20ı − 8 + 10

k ) = −8

3ı +

203

+323

k ft·lbs

186

To apply (2), we also need the angular momentum

HG ,[Ix′x′

ω1− I

x′y′ω

2− I

x′z′ω

3

]b1 +

HG =[Iy′y′

ω2− I

y′z′ω

3− I

y′x′ω

1

]b2 + (4)[

Iz′z′

ω3− I

z′x′ω

1− I

z′y′ω

2

]b3

=(

40 lbs32.2 ft/s2

)(8 in

12 in/ft

)2([ 724

(−5)− 38(4)]

ı +[1712

(4)− 38(−5)

] +

[3724

(−1)]

k

)= (−1.63ı + 4.16 − 0.851

k ) slug·ft2/s (5)

Equation (2) may be rewritten as∑

MG =(Ix′x′

α1− I

x′y′α

2

)ı +

(Iy′y′

α2− I

y′x′α

1

) +

(Iz′z′

α3

)k + ω×

HG (6)

Substitution of (4), (5), and∑

MG into (6) yields three equations that may be solved for α,

α = α1

ı + α2

+ α3

k = (−9.10ı + 9.47 + 29.3

k ) rad/s2

The acceleration of point P will be

aP = aG + α×rP/G+ ω×

(ω×rP/G

)= (16.1ı − 38.64 + 8.05

k ) +

(−9.10ı + 9.47 + 29.3k )×

(43

ı +23

+13

k

)+

(−5ı + 4 − k )×

[(−5ı + 4 −

k )×(

43

ı +23

+13

k

)]aP = (−34.6ı − 41.9 − 20.3

k ) ft/s2

187

8.8.11GOAL: Find the angular acceleration α of the body, the acceleration aG of the mass center, andthe velocity and acceleration vQ and aQ of the point mass Q, all at time t = 0.GIVEN: Geometry and mass of system components; the initial linear and angular velocities of thesystem; the force applied to the system and its point of application.DRAW:

ASSUME: The thickness of the cylindrical shell is negligible.FORMULATE EQUATIONS: The equations of motion for a rigid body are

F = maG (1)

∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG (2)

SOLVE: From (1), the acceleration of the mass center is

aG =F

M + 2m=

(−60ı + 40 − 50k ) N

5 kg + 2(3 kg)= (−5.45ı + 3.64 − 4.55

k ) m/s2 (3)

To compute the moments of inertia for the system we may superimpose the individual inertiavalues for the two point masses and the cylindrical shell. Note that the origin of the given x, y, z

188

coordinate system is the current center of mass. By symmetry, Ixz = Iyz = 0.

Ixx = MR2 + 2m(R+

L

4

)2

= (5 kg)(0.05 m)2 + 2(3 kg) (0.05 m + 0.20 m)2 =

= 0.3875 kg·m2

Iyy =12MR2 +

112ML2 + 2m

(L

4

)2

=

=12(5 kg)(0.05 m)2 +

112

(5 kg)(0.80 m)2 + 2(3 kg)(0.20 m)2 = 0.5129 kg·m2

Izz =12MR2 +

112ML2 + 2m

[(L

4

)2

+(R+

L

4

)2]

=12(5 kg)(0.05 m)2 +

+112

(5 kg)(0.80 m)2 + 2(3 kg)[(0.20 m)2 + (0.25 m)2

]= 0.8879 kg·m2

Ixy = m

(−L

4

)(R+

L

4

)+m

(L

4

)(−R− L

4

)= −2(3 kg)(0.20 m)(0.25 m) =

= −0.30 kg·m2

The moment exerted on the body about its mass center is∑

MG = r×F = (−0.20ı + 0.25 )×(−60ı + 40 − 50

k ) = −12.5ı − 10 + 7

k N·m

To apply (2), we also require the angular momentum

HG ,[Ixxω1


]b1 +

HG =[Iyyω2


]b2 + (4)[

Izzω3− Izxω1

− Izyω2

]b3

= [(0.3875)(5)− (−0.3)(2)]ı + [(0.5129)(2)− (−0.3)(5)] + [(0.8879)(−3)]k

= (2.54ı + 2.53 − 2.66k ) kg·m2/s (5)

Equation (2) may be rewritten as∑

MG =(Ixxα1

− Ixyα2

)ı +

(Iyyα2

− Iyxα1

) +

(Izzα3

)k + ω×

HG (6)

Substitution of (4), (5), and∑

MG into (6) yields three equations that may be solved for α,

α = α1

ı + α2

+ α3

k = (−26.2ı − 15.3 − 0.624

k ) rad/s2

The velocity and acceleration of the point mass Q will be

vQ = vG + ω×rQ/G= 0 + (5ı + 2 − 3

k )×(0.20ı − 0.25 )

vQ = (−0.75ı − 0.60 − 1.65k ) m/s2

aQ = aG + α×rQ/G+ ω×

(ω×rQ/G

)= (−5.45ı + 3.64 − 4.55

k ) + (−26.2ı − 15.3 − 0.624

k )×(0.20ı − 0.25 ) +

+(5ı + 2 − 3k )×

[(5ı + 2 − 3

k )×(0.20ı − 0.25 )

]

189

aQ = (−10.7ı + 14.0 + 3.57k ) m/s2

190

8.8.12GOAL: Determine the angular acceleration α of the T-bar, and the linear accelerations aG and aP

of the center of mass and point P , respectively, at the moment a given force F is applied.GIVEN: We’re given the mass m and dimension h for the body, and the applied force F with itslocation. The body is initially at rest.DRAW:

ASSUME: The thickness of the T-bar is negligible and may be modeled as a slender rod.FORMULATE EQUATIONS: The center of mass of the T-bar is located at

xG =1m

[23m(0) +

13m

(−h

2

)]= −h

6 = −0.05 m (1)

Let x′, y′, z′ be a set of axes that pass through the mass center and are parallel with the given x, y,z. The products of inertia about the x′, y′, z′ axes will all equal zero, due to the symmetry of thebody (and thus Euler’s equations are generally applicable). In addition, since the body is initiallyat rest, its angular velocity ω = 0 at the moment of interest. These conditions lead to a simplifiedversion of the angular equations of motion.

F = maG (2)

∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG = Ix′x′

α1

ı + Iy′y′

α2

+ Iz′z′

α3

k (3)

SOLVE: The moments of inertia about the x′, y′, z′ axes are

Ix′x′

=(

23m

)(h

6

)2

+112

(m

3

)h2 +

(m

3

)(h

3

)2

=112mh2 = 0.0225 kg·m2

Iy′y′

=112

(23m

)(2h)2 =

29mh2 = 0.06 kg·m2

Iz′z′

= Ix′x′

+ Iy′y′

=1136mh2 = 0.0825 kg·m2

The moment about the mass center is

∑

MG = r×F = (−0.15ı + 0.05 )×(−50ı + 40 + 20

k ) = (ı + 3 − 3.5

k ) N·m

191

Solving (3) for the angular acceleration components yields:

α1

=1 N·m

0.0225 kg·m2= 44.4 rad/s2

α2

=3 N·m

0.0600 kg·m2= 50 rad/s2

α3

=−3.5 N·m2

0.0825 kg·m2= −42.4 rad/s2

α = (44.4ı + 50 − 42.4k ) rad/s2

The center of mass acceleration is computed from (2),

aG =(−50ı + 40 + 20

k ) N

3 kg= (−16.7ı + 13.3 + 6.67

k ) m/s2

The acceleration of point P is then

aP = aG + α×rP/G+ ω×

(ω×rP/G

)= (−16.7ı + 13.3 + 6.67

k ) + (44.4ı + 50 − 42.4

k )×(0.3ı + 0.05 ) + 0

aP = (−14.5ı + 0.606 − 6.11k ) m/s2

192

8.8.13GOAL: Determine the ratio (ω

1− ω

2)/(L

1− L

2) that will result in a steady precession rate Ω

about the vertical, with shaft CD remaining in the horizontal plane.GIVEN: The mass and dimensions of the system components are assumed to be known. Theangular velocities of the disks are assumed to be ω

1and ω

2, in the directions indicated by the

diagram.DRAW: Let

b1,

b2,

b3 be a set of unit vectors fixed to the shaft CD.

b3 will thus always be the

vertical upward direction.

ASSUME: The mass of the shaft is negligible.FORMULATE EQUATIONS: We’ll choose to solve this problem by using the angular equationsof motion about the fixed point O. ∑

MO =

HO (1)

SOLVE: The moments of inertia of the two disks about points O will be very similar. Note theuse of the parallel axis expressions in Iyy and Izz below.

IA

xx =12mR2 IB

xx = 12mR

2

IA

yy = 14mR

2 +mL21

IB

yy = 14mR

2 +mL22

IA

zz = 14mR

2 +mL21

IB

zz = 14mR

2 +mL22

The products of inerta are all zero, by symmetry.If the shaft remains horizontal, the angular velocity of each disk will be

ω A = ω1

b1 + Ω

b3

ω B = −ω2

b1 + Ω

b3 (2)

Since we are using a set of principal axes, the angular momentum about O becomes

HO =

HA

O+

HB

O= IA

xxω1

b1 + IA

zzΩb3 − IB

xxω2

b1 + IB

zzΩb3 (3)

⇒

HO = IA

xxω1

b1 + IA

xxω1

b 1 + 0− IB

xxω2

b1 − IB

xxω2

b 1 + 0

=12mR2(ω

1− ω

2)

b1 +

12mR2(ω

1− ω

2)Ω

b2 (4)

193

The only moment about O is that due to gravity∑

MO = L1

b1×(−mg

b3) + (−L2

b1)×(−mg

b3) = mg(L1− L

2)

b2 (5)

Substitution of (4) and (5) into (1) yields

mg(L1− L

2)

b2 =

12mR2(ω

1− ω

2)

b1 +

12mR2(ω

1− ω

2)Ω

b2 (6)

Equating theb2 components gives the answer we’re looking for

ω1− ω

2

L1− L

2

=2gR2Ω

Note that since disks A and B spin freely about the shaft, ω1

= 0 and ω2

= 0, satisfying theb1

component of the above equation. Writing the equations of motion for each disk about its masscenter would quickly reveal that this is true. In fact, as an alternative solution we could haveemployed these equations instead, along with a moment balance for the massless shaft.

194

8.8.14GOAL: Examine the angular momentum and angular velocity of a discus after its release, andhow they change over time. Also examine the trajectory of the discus mass center and the distanceit is thrown.GIVEN: Mass and dimensions of discus; initial position, orientation, linear velocity and angularvelocity.DRAW:

ASSUME: The discus may be approximated as a circular disk with given mass and dimensions.SOLVE: Let

b1,

b2,

b3 be a set of body-fixed unit vectors with

b2 being the “spin” axis, or the

direction normal to the disk surface. Using our cicular disk approximation for the discus givesmoments of inertia equal to:

Ix′x′

=14m

(d

2

)2

+112mt2 =

14(2 kg)(0.11 m)2 +

112

(2 kg)(0.045 m)2 = 0.00639 kg·m2

Iy′y′

=12m

(d

2

)2

=12(2 kg)(0.11 m)2 = 0.0121 kg·m2

Iz′z′

= Ix′x′

= 0.00639 kg·m2

All products of inertia are zero, by symmetry.(a) The initial angular momentum of the discus is

HG = Ix′x′

ω1

b1 + I

y′y′ω

2

b2 + I

z′z′ω

3

b3

= [(0.00639)(−1.3)b1 + (0.0121)(20)

b2 + (0.00639)(0.8)

b3] kg·m2/s

HG = (−8.30×10−3 b1 + 0.242b2 + 5.11×10−3

b3) kg·m2/s

Since there are no moments acting on the discus about its mass center, the rate of change of angular

momentum is zero,∑

MG =

HG = 0. Thus

HG is constant.(b) From Euler’s equations, the motion of the discus is described by

Ix′x′

ω1

+ ω2ω

3(I

z′z′− I

y′y′) = 0

Iy′y′

ω2

+ ω3ω

1(I

x′x′− I

z′z′) = 0 (1)

Iz′z′

ω3

+ ω1ω

2(I

y′y′− I

x′x′) = 0

195

We can rearrange the above equations to yield (observe Ix′x′

− Iz′z′

= 0)

ω1

=Iy′y′

− Iz′z′

Ix′x′

ω2ω

3ω

2= 0 ω

3=

Ix′x′

− Iy′y′

Iz′z′

ω1ω

2(2)

and note that ω1

= Cω3

and ω1

= −Cω3, where C is a constant.

In general, if ω = ω1

b1 + ω

2

b2 + ω

3

b3, where

b1,

b2,

b3 are body-fixed unit vectors, the rate of

change of ω may be expressed as

ω = ω1

b1 + ω

2

b2 + ω

3

b3 (3)

For the present case,

ω =Iy′y′

− Iz′z′

Ix′x′

ω2ω

3

b1 +

Ix′x′

− Iy′y′

Iz′z′

ω1ω

2

b3

Examining (2) and (3), we see that there are three release conditions for which ω is constant.

• ω1

= ω3

= 0; Only spin imparted to discus, ω26= 0

• ω2

= 0; No spin imparted. Rotation is about a fixed transverse axis.• ω

1= ω

2= ω

3= 0; No rotation whatsoever, a special case of the above

(c) Neglecting air resistance, the discus will follow a typical parabolic trajectory for a projectileunder the influence of gravity. We can compute the distance of the throw by considering the verticaland horizontal components of the motion equations. The time that will elapse before the discusstrikes the ground may be computed from the vertical component,

t =voy

g+

√v2oy + 2gyo

g

=(25 m/s) sin 25

9.81 m/s2+

√(25 m/s)2(sin 25)2 + 2(9.81 m/s2)(1.5 m)

9.81 m/s2= 2.29 s

The throw distance is now computed from the horizontal component,

x = voxt = (25 m/s)(cos 25)(2.29 s) = 51.8 m

Our computation has come up short compared to an actual Olympic throw with the same releaseconditions. The reason for this is because we have neglected the lift provided by the 35 “attack”angle. While the frictional resistance of air will generally decrease a projectile’s speed, the liftprovided under these particular conditions is large enough to actually increase the maximum heightand flight time for the discus. This increase in flight time is apparently the more significant effect,allowing a properly-trained athlete to throw the discus farther than our prediction.

196

8.8.15GOAL: (a) Find expressions for the vertical components of the reaction forces at the sphere’scontact point C and at the support O; (b) Write an expression for the angle θ at which the contactforce goes to zero, and note any restrictions on m, R, and L for which this is impossible.GIVEN: The mass of the sphere, length of the rod, radius of the sphere, and angle θ are allassumed to be known. ω is constant.DRAW:

ı k

b1 cos θ 0 − sin θb2 0 1 0b3 sin θ 0 cos θ

ASSUME: The mass of the rod may be neglected. We assume that when the vertical reaction forcegoes to zero the sphere continues a “rolling” motion over the surface, although there is essentially nocontact. We will also assume that the horizontal component of the contact force in the ı directionis zero. This will be a valid assumption at the moment contact is lost, and without this assumptionthe ı -components of the reactions at C and O would be redundant. Of course, this assumptionwould also hold for a very smooth surface (we would just have to initiate “rolling” somehow).FORMULATE EQUATIONS: We will choose to write our moment equation about the centerof mass. The point O would work just as well, and we could even choose C, with a little morework. ∑

F = maG (1)∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG (2)

We are given the rotation rate ω about the vertical axis. Since the sphere is rolling its motion alsoconsists of a rotation about the shaft, which we will denote as a rate η. The angular velocity ofthe sphere is therefore ω = ω

k + η

b3. Since the point C is stationary, we may write a constraint

condition relating ω and η:

vC = ω×rC/O= (ω

k + η

b3)×(−L

b3 −Rk ) = −ωL sin θ

b2 + ηR sin θ

b2 = 0

⇒ ωL = ηR (3)

The locus of contact points on the sphere is a circle of radius R sin θ.

197

SOLVE: The moments of inertia for a sphere about its mass center are identical, and all productsof inertia about axes through G are zero.

Ixx = Iyy = Izz =25mR2

With ω = ωk + η

b3 = −ω sin θ

b1 + (ω cos θ + η)

b3, the angular momentum of the sphere becomes:

HG = −25mR2ω sin θ

b1 +

25mR2(ω cos θ + η)

b3 (4)

⇒

HG = −25mR2ω sin θ(ω cos θ)

b2 +

25mR2(ω cos θ + η)(ω sin θ)

b2

=25mR2ωη sin θ

b2 (5)

Let the reaction forces at O and C beFC = FCx

ı + FCy + FCz

k , and

FO = FOx

ı + FOy + FOz

k .

The moment at O will consist of a single component in theb1-direction,

MO = MO1

b1. The other

two components are zero because the system is pinned horizontally aboutb2, and the massless

shaft allows free rotation of the sphere aboutb3.∑

MG = (Lb3)×(FOx

ı + FOy + FOz

k ) + (−R

k )×(FCxı + FCy

+ FCz

k ) +M

O1

b1

= (RFCy cos θ − FOyL+MO1

)b1 + (−RFCx + FOxL cos θ − FOzL sin θ)

b2 +

+RFCy sin θb3 (6)

Equating components of (5) and (6) gives

0 = RFCy cos θ − FOyL+MO1

(7)25mR2ωη sin θ = −RFCx + FOxL cos θ − FOzL sin θ (8)

0 = RFCy sin θ (9)

The sphere’s center travels in a circle of radius L sin θ, with angular speed ω. It’s acceleration isaG = mLω2 sin θı and our force balance (1) yields the component equations

mLω2 sin θ = FCx + FOx (10)0 = FCy + FOy (11)

0 = FCz + FOz −mg (12)

From (9), (11), and (7) we see that FCy = 0, FOy = 0, and MO1

= 0. Interesting, but not what weneed. The relevant equations for our purposes are (8), (10), and (12). With our assumption thatFCx = 0 (as it must when the sphere loses contact, or on a very smooth surface), these equationsmay be solved for the vertical reactions at C and O,

(10),(3)→(8) ⇒25mRLω2 sin θ = mL2ω2 sin θ cos θ − FOzL sin θ

⇒ FOz = mω2(L cos θ − 2

5R

)(13)

(13)→(12) ⇒ FCz = mg −mω2(L cos θ − 2

5R

)(14)

198

(b) Letting FCz go to zero in equation (14) and solving for θ yields

cos θ =1L

(g

ω2+

25R

)

Since cos θ ≤ 1, we haveg

ω2+

25R < L

Thus no matter how large ω may be, it will be impossible for the vertical reaction to equal zerounless

25R < L

There is also a lower limit on ω, namely, ω >√g/L.

199

8.8.16GOAL: Determine the angular speed ω, and its direction, that is required if the angle θ betweenthe vertical and shafts AC and BD is to remain constant at θ = 30, for a given rotational rate Ω.GIVEN: Each disk has mass m = 5 kg and radius R = 0.08 m; shafts AC and BD have lengthL

1= 0.20 m with spring attachment points at a distance L

2= 0.15 m from points C and D;

the length of link CD is L3

= 0.30 m, which is the same as the unstretched length of the spring(k = 320 N/m); Ω = 3 rad/s in the direction shown.DRAW:

ı k


ASSUME: The mass of the spring and linkages is negligible.FORMULATE EQUATIONS: Since the system is symmetric, we can confine our work to onlyone disk and linkage. The linear and angular equations of motion for rigid bodies are∑

F = maG (1)

∑

MG =

HG

∣∣∣∣N

=d

dt

∣∣∣∣S

HG + ω×

HG (2)

200

SOLVE: The shaft is massless, so the only moments of inertia to consider are that of the disk.Since the disks are axisymmetric, we may use the shaft-fixed

b1,

b2,

b3 directions as “body-fixed”

unit vectors without loss of generality. All products of inertia are zero.

Ixx =14mR2 Iyy =

14mR2 Izz =

12mR2

With θ constant, the angular velocity of the disk isω = Ω

k + ω

b3 = −Ω sin θ

b1 + (Ω cos θ + ω)

b3 (3)

The angular momentum of the disk is then

HG = −IxxΩ sin θb1 + Izz (Ω cos θ + ω)

b3

= −14mR2Ω sin θ

b1 +

12mR2(Ω cos θ + ω)

b3 (4)

HG = −14mR2Ω sin θ(Ω cos θ

b2) +

12mR2(Ω cos θ + ω)(Ω sin θ

b2)

=14mR2Ω sin θ(Ω cos θ + 2ω) (5)

Let the reaction force at point D be Fxı +Fy

+Fzk . The link is pinned at D so that there is no

moment in theb2-direction; the negligible mass of the shaft and the fact that the disk spins freely

about is axis indicates that there can be no moment aboutb3 either. Thus the reaction moment is

just M1

b1.∑

MG =∑

(r×F) +M

1

b1

= −L1

b3×(Fx

ı + Fy + Fz

k ) +−(L

1− L

2)

b3×(−2kL

2sin θ)ı +M

1

b1

= (L1Fy +M

1)

b1 + [L

1Fz sin θ − L

1Fx cos θ + 2k(L

1− L

2)L

2sin θ cos θ]

b2 (6)

Substituting (5) and (6) into (2), and then equatingb1 and

b2 components gives

0 = L1Fy +M

1(7)

14mR2Ω sin θ(Ω cos θ + 2ω) = L

1Fz sin θ − L

1Fx cos θ + 2k(L

1− L

2)L

2sin θ cos θ (8)

The mass center travels a circular path of radius L3/2+L

1sin θ with constant speed. Its acceleration

isaG = Ω

k ×[Ω

k ×(L

3/2ı + L

1

b3)] = −Ω2(L

3/2 + L

1sin θ)ı

The component equations of our force balance (1) are

−mΩ2(L3/2 + L

1sin θ) = Fx − 2kL

2sin θ (9)

0 = Fy (10)0 = Fz −mg (11)

By (10) and (7), Fy = 0 and M1

= 0. Solving (9) and (11) for Fx and Fz and substituting into (8)yields:

14mR2Ω sin θ(Ω cos θ + 2ω) = L

1mg sin θ + 2k(L

1− L

2)L

2sin θ cos θ

−L1[2kL

2sin θ −mΩ2(L

3/2 + L

1sin θ)] cos θ (12)

Using the given parameter values in (12) allows us to solve for ω,

ω = 24.5 rad/s, in the positiveb3 direction

201

8.8.17GOAL: (a) Determine ω

1, ω

2, and the angular acceleration α at the instant a given force is applied

to the cylinder. (b) Find the reaction forces and moments at the support A at this instant.GIVEN: Mass and dimensions of system; initial rotation rates ω

1, ω

2; and the applied force F

and its location.DRAW:

Theb1,

b2,

b3 axes are attached to the shaft, with

b2 remaining vertical.

ASSUME: The mass of the 90-bend shaft is negligible.FORMULATE EQUATIONS: We employ the rigid body equations of motion. We will chooseto sum the applied moments about the fixed point O lying at the center of the 90-bend.∑

F = maG (1)

∑

MO =

HO

∣∣∣∣N

=d

dt

∣∣∣∣S

HO + ω×

HO (2)

SOLVE: The moments of inertia for the cylinder about its mass center are Ixx = 12mR

2 andIyy = Izz = 1

4mR2 + 1

12mt2. Since we have chosen the fixed point O we’ll use the parallel axis

expressions to obtain the inertia values about this point.

Ixx =12mR2

Iyy =14mR2 +

112mt2 +m

(L+

t

2

)2 14mR2 +

13mt2 +mL2 +mLt

Izz =14mR2 +

112mt2 +m

(L+

t

2

)2

=14mR2 +

13mt2 +mL2 +mLt

The angular velocity of the cylinder is

ω = ω1

b1 + ω

2

b2 (3)

The angular momentum of the system about O is

HO =12mR2ω

1

b1 +

(14mR2 +

13mt2 +mL2 +mLt

)ω

2

b2 (4)

202

Taking the derivative,

HO =12mR2ω

1

b1 +

12mR2ω

1(−ω

2

b3) +

(14mR2 +

13mt2 +mL2 +mLt

)ω

2

b2 (5)

The forces acting on the system are gravity −mgb2, the applied force

F = 40

b3 N, and the reaction

forces. Since the shaft is assumed to be massless, the net forces and moments acting on any portionof it must be zero (otherwise it would have infinite acceleration terms). It will be easier to visualizethe reactions if we initially consider only those existing at point O of the shaft. Doing so alsoremoves them from the moment equation. We’ll wait until part (b) below to consider those existingat point A.The reaction force at O may be written as

FO = F

O1

b1 + F

O2

b2 + F

O3

b3. The reaction moment at

O consists of only one component,

MO = MO3

b3. Since the cylinder spins freely about

b1 there can

be no reaction torque in this direction; and since the shaft revolves freely aboutb2 there can be no

reaction torque in this direction either. The sum of the moments about O is now∑

MO =(L+

t

2

)b1×(−mg)

b2 + (Lb1 +R

b2)×F

b3 +M

O3

b3

= RFb1 − LF

b2 +

[M

O3−mg

(L+

t

2

)]b3 (6)

Substitution of (5) and (6) into (2) and equating like components gives

12mR2ω

1= RF (7)(

14mR2 +

13mt2 +mL2 +mLt

)ω

2= −LF (8)

−12mR2ω

1ω

2= M

O3−mg

(L+

t

2

)(9)

The acceleration of the cylinder’s mass center is

aG = ω2

b2×

(L+

t

2

)b1 + ω

2

b2×

[ω

2

b2×

(L+

t

2

)b1

]=(L+

t

2

)(−ω

2

b3 − ω2

2

b1

)The component equations of the force balance (1) are now

FO1

= −mω22

(L+

t

2

)(10)

FO2−mg = 0 (11)

FO3

+ F = −mω2

(L+

t

2

)(12)

Equations (7)-(12) are our six rigid body equations of motion. Equations (7) and (8) may be solveddirectly for ω

1and ω

2:

ω1

=2FmR

=2(40 N)

(5 kg)(0.05 m)= 320 rad/s2 (13)

ω2

= − LF14mR2 +

13mt2 +mL2 +mLt

= − (0.15 m)(40 N)

(5 kg)[14(0.05 m)2 +

13(0.10 m)2 + (0.15 m)2 + (0.15 m)(0.10 m)

]

203

ω2

= −28.9 rad/s2 (14)

We can now solve for the angular acceleration α. Note ω1

= 120 rev/min = 4π rad/s, and ω2

=40 rev/min = 4

3π rad/s

α = ω = ω1

b1 + ω

2

b2 − ω

1ω

2

b3 = 320

b1 − 28.9

b2 − (4π)

(43π

)b3

α = (320b1 − 28.9

b2 − 52.6

b3) rad/s2 (15)

(b) Equations (9)-(12) may be solved for the force and moment reactions at O. To find the reactionsat A, we just have to balance the forces and moments acting on the AO segment of the shaft.Again, since the shaft is assumed massless these forces and moments must cancel. We’ll let theforce reaction at A be

FA = F

A1

b1 +F

A2

b2 +F

A3

b3, and the moment reaction

MA = MA1

b1 +M

A3

b3.

FA1

= FO1

= −mω22

(L+

t

2

)= −17.5 N

FA2

= FO2

= mg = 49.1 N

FA3

= FO3

= −F −mω2

(L+

t

2

)= −11.06 N

FA = (−17.5

b1 + 49.1

b2 − 11.06

b3) N

Summing the moments for massless segment AO about O, which must equal zero,∑

M = (−L)b2×(F

A1

b1 + F

A2

b2 + F

A3

b3) +M

A1

b1 +M

A3

b3 −M

O3

b3 = 0

⇒ MA1

= LFA3

= −1.659 N·m

MA3

= MO3− LF

A1= mg

(L+ t

2

)− 1

2mR2ω

1ω

2− LF

A1= 12.1 N·m

MA = (−1.659b1 + 12.1

b3) N·m

204

8.8.18GOAL: Find the tension in the string, and the reaction forces at each ball-and-socket joint, whenmass m3 has descended a distance s = 9

2πr2 from the given release position.GIVEN: m

1= 3 kg, m

2= 0.5 kg, m

3= 2 kg, r

1= 10 cm, r

2= 4 cm, d = 6 cm, and

L1

= L2

= L3

= 15 cm.DRAW: Let

b1,

b2,

b3 be a set of body-fixed unit vectors attached to the rotating shaft. Initially,

b1,

b2,

b3 and ı , ,

k coincide.

The rotational transformation from ı , ,k to

b1,

b2,

b3, with θ = s

r2, is

ı k


FORMULATE EQUATIONS: The system is viewed as two rigid bodies, the first being therod, disk, and pulley; the second being the hanging mass m

3. The governing equations are the

moment and force balances for the shaft, and a force balance for the hanging mass. There are eightunknowns: six reaction force components at A and B, the tension T in the string, and the motion

205

of the system. The moment and force balances will provide 7 equations, but we will see that twoforce components merely cancel, leaving 6 equations in 6 unknowns.It is convenient to write the moment balance about a fixed point along the shaft axis. We willchoose the point B, since this choice eliminates the reactions at B from the moment equations.Our governing equations are then:

shaft:∑

MB =

HB (1)∑ Fshaft = (m

1+m

2)aG−shaft (2)

hanging mass: T −m3g = m

3a (3)

As an alternative approach, the force equation for the shaft (2) could easily be replaced by a secondmoment balance, perhaps about A, to solve for the reactions at B.SOLVE: The angular velocity of the shaft may be written ω = ω = ω

b2. With only one rotational

component, the expression for the angular momentum of the shaft becomes:

HB = Iyyωb2 − Ixyω

b1 − Izyω

b3 (4)

The required moments of inertia may be computed using the known values for a disk, the parallelaxis theorem, and the additive property for composite bodies:

Iyy = I1yy + I2

yy =[I1

y′y′+m

1d2]

+[I2

y′′y′′+m

2(0)]

=12m

1r21

+m1d2 +

12m

2r22

Ixy = I1xy + I2

xy =[I1

x′y′+m

1(0)(2L)

]+[I2

x′′y′′+m

2(0)(L)

]= 0 + 0

Izy = I1zy + I2

zy =[I1

z′y′+m

1(−d)(2L)

]+[I2

z′′y′′+m

2(0)(L)

]= 0− 2m

1dL+ 0

The right side of equation (1) is now

HB = Iyyαb2 − Izyα

b3 − Izyω

2 b1

=(

12m

1r21

+m1d2 +

12m

2r22

)α

b2 + 2m

1dL

(α

b3 + ω2

b1

)(5)

Let the reactions at A and B be written asFA = FAx

ı +FAy +FAz

k and

FB = FBx

ı +FBy +FBz

k .

There are no constraint torques since the ball-and-socket joints rotate freely in all directions. Alsodenote L = L

1= L

2= L

3. The moment about B is then∑

MB = rA/B×FA + rG1/B

×(−m

1g)k + rG2/B

×(−m

2g)k + rC/B

×−Tk

=(−3L

)×(FAx

ı + FAy + FAz

k)+(−2L + d

b3

)×(−m

1g

k)+(

−L)×(−m

2g

k)+(−L + r

2ı)×(−T

k)

=(−3LFAz + 2Lm

1g + Lm

2g + TL

)ı +

(m

1gd sin s

r2+ Tr

2

) + 3LFAx

k (6)

Substitution of (5) and (6) into (1), and taking dot products with ı , ,k , yields the first three

equations of motion:

(1) · ı : −3LFAz + 2Lm1g + Lm

2g + TL = 2m

1dL(α sin s

r2+ ω2 cos s

r2

)(7)

(1) · : m1gd sin s

r2+ Tr

2=

(12m

1r21

+m1d2 +

12m

2r22

)α (8)

(1) ·k : 3LFAx = 2m1dL(α cos s

r2− ω2 sin s

r2

)(9)

206

The force balance for the shaft, as written in Eqn. (2), requires the acceleration of the center ofmass of the disk and pulley. However, since the pulley’s center of mass remains stationary, wecan balance the external forces acting on the system using the motion of the disk center alone.Equation (2) then becomes: ∑

Fshaft = m1

aG1

where

aG1 = α×r + ω×(ω×r ) = αb2×d

b3 + ω

b2×

(ω

b2×d

b3

)= m

1d(α

b1 − ω2

b3

)Using the above force balance, the three force equations for the shaft are then

ı : FAx + FBx = m1d(α cos s

r2− ω2 sin s

r2

)(10)

: FAy + FBy = 0 (11)k : FAz + FBz −

(m

1+m

2

)g − T = m

1d(−α sin s

r2− ω2 cos s

r2

)(12)

The acceleration of the hanging mass is a = ak = −αr

2

k . Using this result in (3) gives

T −m3g = −m

3r2α (13)

We now have all equations necessary for the solution. Specifically, these are equations (7)-(13).Equations (8) and (13) govern the motion of the system and may be solved for α:

α =m

1gd sin s

r2+m

3gr

2

12m1

r21

+m1d2 + 1

2m2r22

+m3r22

(14)

We also need to determine ω, which may be found by integrating our expression for α. To simplifythis task we will first apply the chain rule to obtain the relation α = sdw

ds = ωr2

dwds ⇒

αr2

ds = ωdω.Substitution into this relation, with dummy variables s′ and ω′, yields:

ω∫0

ω′ dω′ =1

12m1

r21

+m1d2 + 1

2m2r22

+m3r22

s∫0

(m

1gd

r2

sins′

r2

+m3g

)ds′

⇒ ω2 =m

1gd(1− cos s

r2

)+m

3gs

14m1

r21

+ 12m1

d2 + 14m2

r22

+ 12m3

r22

(15)

With the given mass and length parameters, equations (14) and (15) will give us the angular velocityand acceleration at the instant s = 9

2πr2 :

α = 86.8 rad/s2

ω = 29.6 rad/s

Using these values in equations (7),(9), and (13), with s = 92πr2 , we can solve for the tension T

and the reactions FAx and FAz :

T = 12.7 N; FAx = −105.0 N; FAz = 15.1 N

Inserting these values in equations (10) and (12) we find FBx and FBz :

FBx = −52.5 N; FBz = 16.3 N

207

The remaining equation of motion, (11), only tells us that

FAy = −FBy

That is, the y-components of the reaction forces at A and B must cancel. With no externallyapplied forces in this direction, and no preexisting stresses in the shaft along the y-axis, theseforces should remain zero.

208

8.8.19GOAL: Determine the precession rate and rotational kinetic energy of a football during flight.GIVEN: Weight and geometry of football; the spin rate is 15 rad/s about the long axis of thefootball and the long axis itself wobbles through a total angle of 15.DRAW:

FORMULATE EQUATIONS: Since there are no external moments acting on the football itsangular momentum is conserved. We can determine the precession rate by setting∑

MG =

HG = 0 (1)

The rotational kinetic enegy may be found using

KErot =12

ω ·

HG (2)

SOLVE: Modeling the football as a solid prolate spheroid, its has two equal semi-minor axesa = b = (19 in)/(2π) = 3.024 in. and a semi-major axis c = 5 in. First let’s get some unitconversions out of the way,

a = b =19 in

2π(12 in/ft)= 0.252 ft c =

5 in12 in/ft

= 0.416 ft

m =9 oz

(16 oz/lb)(32.2 ft/s2)= 0.0175 slug

The moments of inertia of an ellipsoid about principal axes through its mass center are

Ixx =15m(a2 + b2) =

25ma2 = 4.44×10−4 slug·ft2

Iyy = Izz =15m(a2 + c2) = 8.28×10−4 slug·ft2

The angular velocity of the football may be expressed as

ω = ω2

b1 + ω

1ı = (ω

2+ ω

1cos θ)

b1 − ω

1sin θ

b2

We may now write an expression for the angular momentum

HG ,

HG = Ixx(ω2

+ ω1cos θ)

b1 + Iyy (−ω

1sin θ)

b2

209

Taking the derivative with respect to time, and setting it equal to zero

⇒

HG = Ixx(ω2

+ ω1cos θ)

b 1 − Iyyω1

sin θb 2

= Ixx(ω2

+ ω1cos θ)ω

1sin θ

b3 − Iyyω

21sin θ cos θ

b3 = 0 (3)

Note that in the above we have assumed ω1

= 0 and ω2

= 0. Had these terms been included wewould have seen some

b1 and

b2 components in the above equation, but with no external moments

we would quickly find that they are zero. Solving (3) for ω1

yields

ω1

=

(Ixx

Iyy − Ixx

)ω

2

cos θ=

2a2ω2

(c2 − a2) cos θ=

2(3.024 in)2(15 rad/s)[(5 in)2 − (3.024 in)2] cos 15

ω1

= 17.9 rad/s

The rotational kinetic energy is computed from (2).

KErot =12

ω ·

HG

=12

[(ω

2+ ω

1cos θ)

b1 − ω

1sin θ

b2

]·[Ixx(ω

2+ ω

1cos θ)

b1 − Iyyω1

sin θb2

]=

12

[Ixx(ω

2+ ω

1cos θ)2 + Iyy (ω

1sin θ)2

]=

12

[(4.44×10−4 slug·ft2)[15 rad/s + (17.9 rad/s) cos 15]2 +

+(8.28×10−4 slug·ft2)[(17.9 rad/s) sin 15]2]

KErot = 0.240 ft·lb

210

8.8.20GOAL: Determine the direction of rotation that the student experiences after rotating the axle ofthe wheel as described.GIVEN: The wheel is given an intial spin ω about its axle in the direction indicated in the figure.There is no initial rotation of the swiveling platform. The student then rotates the wheel axle sohis right hand is directly above his left, with the axle of the wheel oriented vertically in space.DRAW:

SOLVE: Consider the mechanical “system” consisting of the student, wheel, and platform. Theonly external forces acting on the system are gravity and the reactions at the platform support.Since the platform swivels freely about the z-axis, no external moments may be applied to thesystem about this axis (or any point lying on the z-axis). The resultant of any external forcesacting on the platform must have a line of action that passes through the z-axis, or else be purelyvertical, parallel to this axis. Gravitational forces are assumed to be purely vertical as well.Since no external moments may be applied to our system about the z-axis, the angular momentumof the system about any point lying on this axis is conserved in the

k -direction. This is clear from

the angular momentum equation of motion:

∑

MO =

HO

⇒ 0 =

HO ·k

Initially, the wheel axle is oriented horizontally, and the student is at rest. The system’s initialangular momentum component in the

k -direction is therefore zero, and it must remain zero for all

211

time. As the student tilts the axle of the wheel, bringing his right hand above the left, the wheelitself begins to acquire an angular momentum component in the positive

k direction. To counter

this positive gain, the student and platform must begin rotating in the opposite direction, at anangular speed which maintains the system’s overall angular momentum component in this directionat zero. Thus,

Student rotates in -z direction, or Ω is negative

In the solution above, we have only addressed the vertical direction, and have made no mention ofthe horizontal components. So the question might be asked, “What about the change in angularmomentum in the horizontal direction? Initially the wheel has a large component in this direction,but in the final orientation it has none.” The answer is that while the horizontal angular momen-tum components do change, the platform support provides the external moments allowing for thischange, and the student’s muscles provide the torques required for him to remain upright.

212

8.8.21GOAL: (a) Determine the angle φ through which the student and platform rotate before comingto rest, after the described action is perfomed. (b) Find the initial and final kinetic energy of thesystem, as well as the work done by friction, and explain any differences.GIVEN: Mass and moments of inertia of the wheel; initial rotation rate ω of wheel and its direction;magnitude of the torsional friction; distance d between vertical axis and wheel center; approximatemoment of inertia I for student/platform combination about vertical z-axis.DRAW:

ASSUME: The described action is performed instantaneously. The student does not shift theposition of his mass center or that of the wheel. Neglect any friction in the wheel axle.FORMULATE EQUATIONS: The wheel itself spins freely about the body-fixed

b1 axis, and

thus its angular rotation rate ω is constant. This is evident from an examination of theb1-

component of Euler’s equation∑

Mdisk

1 = Ixxα1+ ω

2ω

3(Izz − Iyy )

⇒ 0 = Ixxω + ω2ω

3(0) ⇒ ω = 0

The angular velocity of the wheel after the rotation of its axle will therefore be

ωwheel = ωb1 = ω

k (1)

To find the angle through which the platform rotates we will have to determine the intial rotationrate Ω of the platform, and apply our angular equation of motion in the

k direction

∑

MP ·k =

HP ·k (2)

to find Ω, and then integrate to get φ.SOLVE: Since it is assumed that the student performs the 90 rotation of the wheel axis instan-taneously, the torsional friction does no work and the angular momentum of the system about thevertical z-axis is conserved during this action. Initially, the student is at rest and the wheel axle is

213

horizontal, so the angular momentum about thek direction is zero (

HP ·k = 0). We can use this

fact to determine the rotational rate Ω of the student immediately after the axle rotation:

HP ·k = IΩ +

HG,wheel ·k + (rG−G/P

×mvG−wheel ) ·k = 0

= IΩ + Ixxω +md2Ω = 0(3)

⇒ Ω0 = − Ixxω

I +md2= − (0.2 kg·m2)(18 rad/s)

0.9 kg·m2 + (2 kg)(0.55 m)2= −2.39 rad/s (4)

We now apply equation (2), where the only moment about point P in thek -direction is the torsional

friction Tf, in the direction opposite the rotation Ω,∑

MP ·k =

HP ·k

Tf

= (I +md2)Ω

⇒ Ω =T

f

I +md2= 0.133 rad/s2 (5)

Integrating, ∫Ω dΩ =

∫Ω dφ

⇒ 12(Ω2 − Ω2

0) = Ω(φ− φ0)

Set Ω = 0 to determine the angle through which the platform rotates before coming to rest.

φ− φ0 = −Ω2

0

2Ω= − (Ixxω)2

2Tf(I +md2)

= − (−2.39 rad/s)2

2(0.133 rad/s2)

Letting φ0 = 0,φ = −21.5 rad

(b) The kinetic energy of the system before the handlebars are rotated and after the platform comesto rest will be the same since the only motion in each case is the rotation ω of the wheel, whichhas constant amplitude.

KE initial = KEfinal =12

ω ·

HG =12ω

b1 · Ixxω

b1 =

12Ixxω

2

KE initial = KEfinal =12(0.2 kg·m2)(18 rad/s)2 = 32.4 J

The work done by the friction torque is

Wf

= Tf

k · φ

k = − (Ixxω)2

2(I +md2)

Wf

= (0.2 N·m)(−21.5 rad) = −4.31 J

The initial kinetic energy of the system is equal to the final kinetic energy, although there hasbeen negative work done by friction. The explanation for this is that the internal muscular forcesgenerated by the student when tilting the wheel added energy to the system. This exact amount ofenergy was then dissipated by friction to bring the platform to rest. In fact, we could have logicallycomputed the angle φ simply by determining the amount of rotation required to absorb the eneryadded by the student when he tilted the wheel axle.

214

8.8.22GOAL: Determine the ground reaction moment at point O.GIVEN: The given parameters are the mass and geometry of system. The system revolves aboutthe vertical with a constant angular velocity ω = 30

k rev/min = π

k rad/s.

DRAW: The body-fixedb1,

b2,

b3 unit vectors are attached to the rotating support beam. They

coincide with the absolute ı , ,k unit vectors at the instant depicted in the diagram.

ASSUME: Neglect the mass of the support beam; consider only the hemispherical shells.FORMULATE EQUATIONS: The angular momentum equation of motion for rigid bodies willallow us to compute the reaction moment at O.

∑

MO =

HO

∣∣∣∣N

=d

dt

∣∣∣∣S

HO + ω×

HO (1)

SOLVE: The angular momentum of the structure about point O is[Ixxω1


]b1 +

HO =[Iyyω2


]b2+ = −Ixzω3

b1 − Iyzω3

b2 + Izzω3

b3 (2)[

Izzω3− Izxω1

− Izyω2

]b3

where ω1

= ω2

= 0 and ω3

= π rad/s.Since the angular velocity is constant, α = 0, and (1) becomes∑

MO = ω×

HO = ω3

b3×(−Ixzω3

b1 − Iyzω3

b2 + Izzω3

b3)

= −Ixzω23

b2 + Iyzω

23

b1 (3)

215

We therefore only need to compute the two products of inertia Ixz and Iyz . Since the x-z planeis a plane of symmetry for the structure, Iyz = 0, leaving us with only Ixz to compute. The masscenter of the upper sphere is located at r

2

b1 + L

2

b3 while that of the lower sphere is located at

− r2

b1 + L

1

b3. Using the parallel axis expressions and the superposition principle for composite

bodies allows us to compute Ixz . By symmetry, Ix′z′

= Ix′′z′′

= 0 we

Iy′′y′′

+m

[L2

2+ Ixz = I

x′′z′′+mL

1

(−r

2

)+ I

x′z′+mL

2

r

2=

12mr(L

2− L

1)

The reaction moment at O may now be computed from (3):

∑

MO = −Ixzω23

b2 = −1

2mr(L

2− L

1)ω2

3

b2

= −12

(250 lb

32.2 ft/s2

)(4 ft)(24 ft− 16 ft)(π rad/s)2

b2

MO = −1.23×103 ft·lb

216

8.8.23GOAL: Determine the angular speed ω

1required for the support force between brace E and arm

BC to go to zero, for a given ω2.

GIVEN: Geometry and dimensions of system; the angular velocity ω2k of the vertical shaft AB.

DRAW: Letb1,

b2,

b3 be a set of unit vectors fixed to arm BC.

ı k

b1 cos θ 0 sin θb2 0 1 0b3 − sin θ 0 cos θ

ASSUME: Neglect the mass of arm BC.FORMULATE EQUATIONS: We will apply the angular equations of motion for a rigid bodyabout point B. We choose point B, instead of the mass center G, simply because it will eliminatethe reaction forces at B from our equations and avoid a consideration of the force equations.∑

MB =

HB

∣∣∣∣N

=d

dt

∣∣∣∣S

HB + ω×

HB (1)

SOLVE: The angular velocity of the disk is

ω = ω2

k + ω

1

b3 = ω

2sin θ

b1 + (ω

2cos θ + ω

1)

b3

To apply (1), we need to determine the angular momentum

HB . The moments of inertia of the diskabout the

bi axes through its mass center are Ixx = Iyy = 1

4mR2 and Izz = 1

2mR2. We utilize the

parallel axis expressions to determine the inertia values about point B:

Ixx =14mR2 +mL2 Iyy =

14mR2 +mL2 Izz =

12mR2

Ixy = Ixz = Iyz = 0 (by symmetry)

217

The angular momentum is now

HB = Ixxω2sin θ

b1 + Izz (ω

2cos θ + ω

1)

b3

=(

14mR2 +mL2

)ω

2sin θ

b1 +

(12mR2

)(ω

2cos θ + ω

1)

b3 (2)

The system is subject to the condition that ω1, ω

2, and θ are constant. The time derivative of

HB

is therefore

HB = ω2

k ×

HB = ω2(sin θ

b1 + cos θ

b3)×

HB

=(

14mR2 +mL2

)ω2

2sin θ cos θ

b2 −

(12mR2

)(ω

2cos θ + ω

1)ω

2sin θ

b2 (3)

With arm BC pinned aboutb2, the moment at B may be represented as M

1

b1 +M

3

b3 (although

intuitively we know these are both zero due to the steady motion of the system and the fact thatthe disk spins freely about its axis). When the contact force at the brace E goes to zero it willexert no moment on BC, and so the only external moment is that due to gravity. The sum of themoments about B is thus∑

MB = M1

b1 +M

3

b3 + (−L

b3)×[−mg(sin θb1 + cos θ

b3)]

= M1

b1 +M

3

b3 +mgL sin θ

b2 (4)

Applying (1), and equating theb2-components of (3) and (4) gives

mgL sin θ =(

14mR2 +mL2

)ω2

2sin θ cos θ −

(12mR2

)(ω

2cos θ + ω

1)ω

2sin θ (5)

Solving (5) for ω1

yields

ω1

=

(2L2

R2− 1

2

)ω

2cos θ − 2gL

R2ω2

=

(2(0.3 m)2

(0.1 m)2− 1

2

)(2 rad/s) cos 30 − 2(9.81 m/s2)(0.3 m)

(0.1 m)2(2 rad/s)

= −264 rad/s

ω1 = −264b3 rad/s

218

8.8.24GOAL: Determine the bending moment at the weld O in terms of m,ω, r, and g. Use two methods:(i) Equations of motion for the mass center G, and (ii) Equations of motion about the fixed pointO.GIVEN: The hemispherical shell rotates with constant angular velocity ω about the shaft axis, inthe direction indicated. The shell has mass m and radius r.DRAW: Let

b1,

b2,

b3 be a set of body-fixed unit vectors that, at the instant illustrated, are

aligned with the x, y, z axes, respectively.

GOVERNING EQUATIONS:

(i)

∑

MG =

HG (1)∑ F = maG (2)

(ii)∑

MO =

HO (3)

SOLVE: (i) The moments of inertia of the hemispherical shell about its mass center are Ixx = 23mr

2

and Iyy = Izz = 512mr

2. All products of inertia are zero. The angular velocity of the shell isω = ω

b3. The expression for angular momentum is thus:

HG = Izzωb3 =

512mr2ω

b3 (4)

Let the contraint forces and moments at the weld be written asF = F

1

b1 + F

2

b2 + F

3

b3 and

M = M1

b1 +M

2

b2 +M

3

b3. For a hemispherical shell, the mass center G is located a distance r

2along the x-axis, from either end of the shell. Substitution of the constraint forces and momentsinto equation (1) yields: ∑

MG =

HG

⇒ F2r

b1 +

(−F

1r +

12F

3r

)b2 −

12F

2r

b3 +M

1

b1 +M

2

b2 +M

3

b3 = 0 (5)

219

The acceleration of the mass center is aG = ω×(

ω× r2

b1

)= −1

2rω2 b1. The force balance, equation

(2), now yields:

F1

b1 + F

2

b2 + F

3

b3 −mg

b3 = −1

2mrω2

b1 (6)

Equation (6) may be solved for the constraint forces, F1

= −12mrω

2;F2

= 0;F3

= mg. Using theseresults in equation (5) gives us the moments:

M1

= 0; M2

= −12mr

(g + rω2

); M

3= 0

⇒

M = −12 mr

(g + rω2

) b2

(ii) If we center our work about the fixed point O we’ll see that we only need one equation to solvefor the bending moment, at a cost of slightly more complex inertia terms and a non-zero angularmomentum time-derivative. With only one angular velocity component, in the

b3 direction, the

angular momentum expression is

HO = IO

zzωb3 − IO

xzωb1 − IO

yzωb2 (7)

Using the parallel axis expressions, we find the moments and products of inertia appearing in theabove equation to be:

IO

zz = Izz +m(r21

+ r22

)=

512mr2 +m

(r2

4+ 0

)=

23mr2

IO

xz = Ixz +mr1r3

= 0 +m

(r

2

)(r)

=12mr2

IO

yz = Iyz +mr2r3

= 0 +m(0)(r)

= 0

And from equation (7) we now have:

HO =23mr2ω

b3 −

12mr2ω

b1

⇒

HO = −12mr2ω2

b2

Since the reaction forces at O exert no moment about O, equation (3) becomes;

M1

b1 +M

2

b2 +M

3

b3 +

12mgr

b2 = −1

2mr2ω2

b2

⇒

M = −12 mr

(g + rω2

) b2

220

8.8.25GOAL: (a) Determine the angular acceleration of the spherical shell when its angular velocity is

ω = 10k rad/s. (b) Find the reaction moment at the weld O at this same instant. Report in the

x, y, z coordinate frame.GIVEN: We are given the external motor torque that is driving the shaft, and the areal densityand radius of the shell.DRAW: Let

b1,

b2,

b3 be body-fixed axes aligned with the principle axes of the shell, while ı , ,

k

are body-fixed axes attached to the rotating shaft.

ı k

b1 sin θ 0 − cos θb2 0 1 0b3 cos θ 0 sin θ

ASSUME: The mass of the shaft is negligible.FORMULATE EQUATIONS: We will apply the rigid body equations of motion about pointO:

F = maG (1)

∑

MO =

HO

∣∣∣∣N

=d

dt

∣∣∣∣S

HO + ω×

HO (2)

SOLVE: The location of the mass center with respect to point O is

rG/O= rA/O

+ rG/A= r

k +

r

2b1 = r

(12− cos θ

)b1 + r sin θ

b3

The moments and products of inertia of the shell about its mass center are

Ix′x′

=23mr2 I

y′y′= I

z′z′=

512mr2 I

x′y′= I

x′z′= I

y′z′= 0

The moments and products of inertia about point O may be found using the parallel axis expres-

221

sions:

Ix′′x′′

=23mr2 +mr2 sin2 θ = 2ρπr4

(23

+ sin2 θ

)Iy′′y′′

=512mr2 +m

[r2(

12− cos θ

)2

+ r2 sin2 θ

]= 2ρπr4

(53− cos θ

)

Iz′′z′′

=512mr2 +mr2

(12− cos θ

)2

= 2ρπr4[

512

+(

12− cos θ

)2]

Ix′′z′′

= 0 +mr

(12− cos θ

)(r sin θ) = 2ρπr4 sin θ

(12− cos θ

)Ix′′y′′

= Iy′′z′′

= 0 + 0 = 0

Substituting the given geometric and mass properties gives

Ix′′x′′

= 7.03×10−4 kg·m2 Iy′′y′′

= 5.87×10−4 kg·m2

Iz′′z′′

= 3.18×10−4 kg·m2 Ix′′z′′

= −1.11×10−4 kg·m2

The angular velocity of the system may be expressed asω = ω

k = ω(− cos θ

b1 + sin θ

b3)

The angular momentum about point O is[Ix′′x′′

ω1− I

x′′y′′ω

2− I

x′′z′′ω

3

]b1 +

HO =[Iy′′y′′

ω2− I

y′′z′′ω

3− I

y′′x′′ω

1

]b2 +[

Iz′′z′′

ω3− I

z′′x′′ω

1− I

z′′y′′ω

2

]b3

=(−I

x′′x′′ω cos θ − I

x′′z′′ω sin θ

)b1 +

(Iz′′z′′

ω sin θ + Ix′′z′′

ω cos θ)

b3 (3)

Summing the moments about point O,∑

MO =(r

k +

r

2b1

)×(−mg

k ) +MO1

ı +MO2

+T

=12mgr sin θ

b2 +M

O1(sin θ

b1 + cos θ

b3) +M

O2

b2 − T cos θ

b1 + T sin θ

b3 (4)

where T = −0.05 N·m.Equation (2) may now be expressed in component form as∑

MO ·b1 = H

1+ (ω

2H

3− ω

3H

2)∑

MO ·b2 = H

2+ (ω

3H

1− ω

1H

3) (5)∑

MO ·b3 = H

3+ (ω

1H

2− ω

2H

1)

yielding the three equations

MO1

sin θ − T cos θ = −Ix′′x′′

ω cos θ − Ix′′z′′

ω sin θ + 0 (6)

MO1

cos θ + T sin θ = Iz′′z′′


ω cos θ + 0 (7)12mgr sin θ +M

O2= 0 + (ω sin θ)

(−I

x′′x′′ω cos θ − I

x′′z′′ω sin θ

)−

(−ω cos θ)(Iz′′z′′


ω cos θ)

= ω2(Iz′′z′′

− Ix′′x′′

)sin θ cos θ + ω2I

x′′z′′

(cos2 θ − sin2 θ

)(8)

222

Equations (6) and (7) may be solved for ω and MO1

, and with the substitutions θ = 40, T =−0.05 N·m and the computed inertia values we have

ω =T

Ix′′x′′

cos2 θ + Iz′′z′′

sin2 θ + 2Ix′′z′′

sin θ cos θ= −115 rad/s2

MO1

= ω[(I

z′′z′′− I

x′′x′′) sin θ cos θ + I

x′′z′′(cos2 θ − sin2 θ)

]= 0.0241 N·m

Equation (8) may be solved for MO2

,

MO2

= ω2(Iz′′z′′

− Ix′′x′′

)sin θ cos θ + ω2I

x′′z′′

(cos2 θ − sin2 θ

)− 1

2mgr sin θ =

= −2.81 N·m

α = ω = ωk = −115

k rad/s2

MO = (0.0241ı − 2.81 ) N·m

223

8.9 Energy of Three-Dimensional Bodies

224

8.9.1GOAL: Determine the kinetic energy of the illustrated system and compare how the energy changesfor finite vs infinitesimal width.GIVEN: Body’s orientation. ω = (20ı − 10 ) rad/s, a = 0.5 m, b = 0.4 m, h = 0.02 m or 0.DRAW:

FORMULATE EQUATIONS: From Appendix B we have:

Ix′x′

= m

(b2 + h2

12

), I

y′y′= m

(a2 + b2

12

), I

z′z′= m

(a2 + h2

12

)To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r

1= a/2,

r2

= h/2 and r3

= b/2.Ixx = I

x′x′+m(r2

2+ r2

3)

= m

(b2 + h2

12

)+m

(b2 + h2

4

)= m

(b2 + h2

3

)Iyy = I

y′y′+m(r2

3+ r2

1)

= m

(a2 + b2

12

)+m

(a2 + b2

4

)= m

(a2 + b2

3

)Izz = I

z′z′+m(r2

1+ r2

2)

= m

(a2 + h2

12

)+m

(a2 + h2

4

)= m

(a2 + h2

3

)

Ixy = Ix′y′

+mr1r2

= m

(ah

4

)

Iyz = Iy′z′

+mr2r3

= m

(bh

4

)

Izx = Iz′x′

+mr3r1

= m

(ab

4

)SOLVE:Having all the relevant rotational inertias and knowing that

KE =12

ω ·

HO

225

we can form the body’s kinetic energy using (8.33) with ω1

= 20 rad/s, ω2

= −10 rad/s, ω3

= 0:

KE =m

2[(20ı − 10 ) rad/s] ·

[(b2 + h2

3

)(20 rad/s)− ah

4 (−10 rad/s)]

ı +[(a2 + b2

3

)(−10 rad/s)− ah

4 (20 rad/s)]

+[−ab4 (20 rad/s)− bh

4 (−10 rad/s)]

k

KE =

m

2

(20 rad/s)

[20 rad/s

3

(b2 + h2

)+

10 rad/s4

ah

]− (10 rad/s)

[−10 rad/s

3

(a2 + b2

)− (5 rad/s)ah

]

KE = m

[a2(

503

)+ b2

(2503

)+ h2

(2003

)+ 50ah

]( rad/s)2

Using a = 0.5 m, b = 0.4 m and h = 0.02 m gives usKE = m

[(0.5)2

(503)

+ (0.4)2(2503)

+ (0.02)2(2003)

+ 50(0.5)(0.02)]

m/s2

= m(18.03 m/s2)

Neglecting h gives us

KE = m(17.5 m/s2)

Thus the reduction in kinetic energy is equal to(18.03− 17.5

18.03

)100 ⇒ 2.92%

226

8.9.2GOAL: Determine the kinetic energy of the illustrated system.GIVEN: Body’s orientation. vG = (1.5

b1 + 1.5

b2 + 10

b3) m/s, ω = (5

b1 + 15

b3) rad/s.

DRAW:


Ixx = m

(a2 + c2

12

), Iyy = m

(b2 + a2

12

), Izz = m

(b2 + c2

12

)The x,y,z axes are centered on the body’s mass center and parallel to the principal axes and thusthere are no products of inertia to consider.SOLVE:Having all the relevant rotational inertias and knowing that

KE =12

ω ·

HO +12

vG ·L

we can form the body’s kinetic energy. With no products of inertia the kinetic energy simplifies to

KE =12

(Ixxω21

+ Iyyω22

+ Izzω23) +

12m(v2

1+ v2

2+ v2

3)

KE = m2

[(a2 + c2

12

)ω2

1+(b2 + a2

12

)ω2

2+(b2 + c2

12

)ω2

3

]

+m2(v21

+ v22

+ v23

)= m

2

[(a2 + c2

12

)25 +

(b2 + c2

12

)225

]( rad/s)2

+m2 (2.25 + 2.25 + 100) (m/s)2

KE = m2

[(25a2 + 225b2 + 250c2

12 ( rad/s)2)

+ 104.5( m/s)2]

227

8.9.3GOAL: Determine the kinetic energy of the illustrated system.GIVEN: Body’s orientation. ω = (8ı + 6 ) rad/s, r = 0.2 m and h = 0.6 m.DRAW:

FORMULATE EQUATIONS:From Appendix B we have

Ix′x′

= Iy′y′

=

(3m(4r2 + h2)

80

), I

z′z′= I

ZZ=

(3mr2

10

)To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r

1= r

2= 0

and r3

= 3h4 .

IXX

= IY Y

= Ix′x′

+

(9mh2

16

)=

3m(r2 + 4h2)20

Ixy = Iyz = Izx = 0

SOLVE:Having all the relevant rotational inertias and knowing that

KE =12

ω ·

HO


= 8 rad/s, ω2

= 6 rad/s, ω3

= 0:

KE =12[(8ı + 6 ) rad/s] ·

3m(r2 + 4h2)

20 (8 rad/s)ı

3m(r2 + 4h2)20 (6 rad/s)

KE =3m2

[(8ı + 6 ) rad/s] ·

(0.2 m)2 + 4(0.6 m)2

20 (8 rad/s)ı

(0.2 m)2 + 4(0.6 m)220 (6 rad/s)

KE = m(7.104 + 3.996)(m/s)2 = 11.1m( m/s)2

228

8.9.4GOAL: Determine the kinetic energy of the illustrated system.GIVEN: Body’s orientation. ω = (20ı − 10 ) rad/s, a = 0.5 m, b = 0.4 m, h = 0.02 m or 0.DRAW:


Ixx = m

(b2

18

), Iyy = m

(a2

18

), Izz = m

(a2 + b2

18

), Ixy = −m

(ab

36

), Iyz = Izx = 0

To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r1

= a/3,r2

= b/2 and r3

= 0.IXX

= Ixx +mr22

= m

(b2

18

)+m

(b2

9

)= m

(b2

6

)IY Y

= Iyy +mr21

= m

(a2

18

)+m

(a2

9

)= m

(a2

6

)IZZ

= Izz +m(r21

+ r22)

= m

(a2 + b2

18

)+m

(a2 + b2

9

)= m

(a2 + b2

6

)

IXY

= Ixy +mr1r2

= −m(ab

36

)+m

(ab

9

)= m

(ab

12

)IY Z

= IZX

= 0


KE =12

ω ·

HO


= 6 rad/s, ω2

= 6 rad/s, ω3

= 0:

KE =m

2[(6ı + 6 ) rad/s] ·

[(b2

6

)(6 rad/s)− ab

12(6 rad/s)]

ı +[(a2

6

)(6 rad/s)− ab

12(6 rad/s)]

KE = 3m

(a2 + b2 − ab

)( rad/s)2

229

8.9.5GOAL: Determine the kinetic energy of the arm AB.GIVEN: Body’s orientation.DRAW:

FORMULATE EQUATIONS:Let L

1= |OA| and L

2= |AB|.

From the given information of the original problem we see that ωAB = ψb3 − θ

b2.

Using the general formula for the mass moment of inertia of a thin rod along with the parallel axistheorem gives us

IXX

= mL21, I

Y Y=mL2

2

3, I

ZZ=mL2

2

3+mL2

1

The only non-zero product of inertia is IXY

:

IXY

= ρ

L2∫

0

L1x dx =

ρL1L2

2

2=mL

1L

2

2


KE =12

ω ·

HO


= 0, ω2

= −θ, ω3

= ψ:

KE =12[(−θ

b2 + ψb3)] ·

mL1L

22

b1(θ)+

mL22

3 (−θ)b2+(

mL22

3 +mL21

)ψ

b3

KE = m

[L2

26 θ2 +

(L2

26 +

L21

2

)ψ2

]

230

8.9.6GOAL: Determine the kinetic energy of the rotating disk.GIVEN: Body’s orientation. r = 0.23 m, h = 0.01 m, a = 0.3 m, ω

1= 10 rad/s, ω

2= 70 rad/s,

m = 1.2 kg.DRAW:

FORMULATE EQUATIONS: Let the disk’s radius be denoted by r. From the given informa-tion in the original problem we see that ωAB = −ω

1 + ω

2ı .

We’ll determine the kinetic energy associated with the speed of the mass center (12mv

2G

) and thatassociated with rotation about the mass center.From Appendix B we have

Iyy = Izz =m(3r2 + h2

)12

, Ixx =mr2

2All products of inertia are zero.SOLVE:Having all the relevant rotational inertias and knowing that

KE =12mv2

G+

12

ωD ·

HG

we can form the body’s kinetic energy using (8.33) with ωD

= 70ı rad/s − 10 rad/s. Note thatfrom the definition in the original problem ω

1indicates rotational velocity about the (negative) Y

axis and ω2

indicates rotational velocity about the x axis.The component of kinetic energy due to motion of the mass center is given by

KE∣∣∣trans

=12mv2

G=

12m(aω

1)2 =

12

(1.2 kg)(0.3 m)2(10 rad/s)2 = 5.4 J

The rotational kinetic energy is found from

KE∣∣∣rot

=12[(ω

2ı − ω

1 )] ·

mr2

2 (ω2)ı +

m(3r2 + h2)12 (−ω

1)

KE = KE

∣∣∣trans

+KE∣∣∣rot

= 5.4 J +m

2

[r2ω2

2

2+

(3r2 + h2)ω21

12

]

KE = 5.4 J +1.2 kg

2

[(0.23 m)2(70 rad/s)2

2+

(3(0.23 m)2 + (0.01 m)2)(10 rad/s)2

12

]= 84.0 J

231

8.9.7GOAL: Determine the kinetic energy of the rotating system.GIVEN: Body’s orientation. r = 0.004 m, h = 0.1 m.DRAW:

FORMULATE EQUATIONS: From the given information in the original problem we see thatωA = ω

1

b1 + ω

2

b2 = (−50

b1 + 2

b2) rad/s.

We’ll consider three components of the kinetic energy of the system: the rotational energy of the3-arm body about O, the translational kinetic energy of the mass center of each wheel and therotational energy of each wheel about its mass center.Approximating the arms as thin rods lets us calculate the mass moment of inertia of the 3-armbody about the Y axis:

I3−a

= 3

(m

3−ah2

3

)= m

3−ah2

Approximating the wheels as thin disks we have, from Appendix B:

Izz = Iyy =mr2

4, Ixx =

mr2

2All products of inertia are zero.SOLVE:Having all the relevant rotational inertias and knowing that

KE =12I3−a

ω22

+12mv2

G+

12

ωw ·

HG

we can form the body’s kinetic energy using (8.33) with ωw = −50b1 rad/s + 2

b2 rad/s.

The kinetic energy due to rotation of the 3-arm body is given by

KE∣∣∣3−a

=12I3−a

ω22

=12

(0.005 kg)(0.1 m)2(2 rad/s)2 = 1.0×10−4 J

The component of kinetic energy due to motion of the mass center of the wheels is given by

KE∣∣∣trans

= 312mwh

2ω22

= 1.5(0.002 kg)(0.1 m)2(2 rad/s)2 = 1.2×10−4 J

Finally, the rotational kinetic energy of the wheels is found from

KE∣∣∣rot

= 312[(ω

1

b1 + ω

2

b2)] ·

mr2

2 ω1

b1+

mr2

4 ω2

b2

232

KE∣∣∣rot

= 3m2

[r2ω2

12 +

r2ω22

4

]

= 3(0.002) kg2

[(0.004 m)2(−50 rad/s)2

2 + (0.004 m)2(2 rad/s)24

]

= 6.00×10−5 J

Summing the three components gives us

KE = 1.0×10−4 J + 1.2×10−4 J + 6.00×10−5 J = 2.8×10−4 J

233

8.9.8GOAL: Determine the kinetic energy of a rolling cone.GIVEN: Body’s orientation and dimensions. m = 22 g.DRAW:

FORMULATE EQUATIONS: First we need to determine the cone’s angular velocity. Example8.4 shows how to determine the angular velocities of a rotating disk on the end of a bent shaft. Ourrolling cone can be thought of as precisely the same problem. The motion of OA will be the sameas that of Example 8.4’s shaft. From geometry we have β = sin−1(0.25) = 14.48. r = |AB| = 1 inand h = |OA| =

√42 − 12 in = 3.873 in

The cone takes 2 s to complete one full rotation around the floor and thus we have

ωOA

=2π2 s

= π rad/s

The correspondence between our problem and that of Example 8.4 isωcone → ω

Wω

AO→ ω

S

3.873 in → L1

1 in → L2

Substituting these values into the expression for ωW

gives us

ωcone = (π rad/s)(

sinβ − 3.873 cosβ + sinβ1

)c1 + (π rad/s) cosβc3

= (−11.78c1 + 3.042c3) rad/s

From Appendix B we have

(Note that these axes do not directly correspond to the ones in our problem). For this figure wehave

Ix′x′

= Iy′y′

=3m(4r2 + h2)

80, I

z′z′= I

ZZ=

3mr2

10For our problem m = 22 g, r = 1 in = 0.0254 m and h = 3.87 in = 0.0984 m.To find the mass moments of inertia and products of inertia we use (8.24)-(8.29) with r

2= r

3= 0

and r1

= 3h4 .

234

IY Y

= IZZ

= Iy′y′

+9mh2

16=

3m(r2 + 4h2)20

IXY

= IY Z

= IZX

= 0


KE =12

ω ·

HO


= 8 rad/s, ω2

= 6 rad/s, ω3

= 0:

KE =12[(−11.78c1 + 3.042c3) rad/s] ·

3mr210 (−11.78 rad/s)c1

3m(r2 + 4h2)20 (3.042 rad/s)c3

KE =3(0.022 kg)

2[(−11.78c1 + 3.042c3) rad/s] ·

(0.0254 m)2

10 (−11.78 rad/s)c1

(0.0254 m)2 + 4(0.0984 m)220 (3.042 rad/s)c3

KE = 8.96×10−4 J

235

8.9.9GOAL: Determine the angular momentum of the illustrated body about the fixed point O.GIVEN: The body rotates about the Y -axis with angular speed ω, and has areal density ρ.DRAW:

FORMULATE EQUATIONS: Let x, y, z be a set of body-fixed axes with origin O. At theinstant illustrated, these axes are aligned with the ground-fixed X,Y, Z axes. The body rotatesabout the Z-axis with angular speed ω; thus the z and Z axes remain aligned and the angularvelocity may be expressed as: ω = ω . The angular momentum of the body about point O isthen:

HO = IOω =

IXX −IXY −IXZ

−IY X IY Y −IY Z

−IZX −IZY IZZ

0ω0

= −IXY

ω ı + IY Y

ω − IY Zω

k (1)

Once we have determined the relevant inertia values about the point O, we may substitute andsolve for the angular momentum.SOLVE: We only need to determine I

XY, I

Y Yand I

Y Z. I

Y Yis easily found as the mass moment

of inertia about the Y axis is simply that of two thin rods, with masses ρab, ρac and lengths b, c,respectively, rotated about their ends:

IY Y

=(ρac)c2

3+

(ρab)b2

3=ρa(b3 + c3)

3Using the appropriate equations from (8.21)-(8.29) lets us determine the products of inertia. Eachrectangle has no product of inertia when evaluated about its mass center and thus the only finiteproducts of inertia come about because we’re shifting our attention from each plate’s mass centerto the point O.

IXY

= ρ(ab)a

2

(−b2

)= −ρa

2b2

4

IY Z

= ρ(ac)a

2c

2=ρa2c2

4Using these inertia values gives us

HO =

(ab2

4ı +

b3 + c3

3 − ac2

4

)ρaω

The kinetic energy is found from

KE =12

ω ·

HO =12ω ·

(b3 + c3

3

)ρaω =

(b3 + c3

6

)ρaω2

236

8.9.10GOAL: Determine the kinetic energy of the illustrated rectangular body about the fixed point O.GIVEN: The body rotates about the Z-axis with angular speed ω, and has mass m.DRAW:

FORMULATE EQUATIONS: Let x, y, z be a set of body-fixed axes with origin O. At theinstant illustrated, these axes are aligned with the ground-fixed X,Y, Z axes. The body rotatesabout the Z-axis with angular speed ω; thus the z and Z axes remain aligned and the angularvelocity may be expressed as: ω = ω

k . The angular momentum of the body about point O is

then:

HO = IOω =

Ixx −Ixy −Ixz

−Iyx Iyy −Iyz

−Izx −Izy Izz

0

0ω

= −Ixzωı − Iyzω

+ Izzωk (1)

Once we have determined the relevant inertia values about the point O, we may substitute andsolve for the angular momentum.SOLVE: The products of inertia about parallel axes through the center of mass G are all zero, dueto the symmetry of the body. The products of inertia about the point O may be computed fromthe parallel axis expressions for inertia.

Ixz : Ixz = Ix′z′

+mr1r3

= 0 +m

(a

2

)(− b

2

)= −1

4mab

Iyz : Iyz = Iy′z′

+mr2r3

= 0 +m

(c

2

)(− b

2

)= −1

4mcb

The moment of inertia about the z′-axis through the mass center is: Iz′z′

= 112m

(a2 + c2

). Using

the parallel axis expressions for inertia again, we obtain:

Izz :Izz = I

z′z′+ m

(r21

+ r22

)=

112m(a2 + c2

)+m

[(a

2

)2

+(c

2

)2]

=13m(a2 + c2

)Substituting into (1),

HO = mω

(14ab ı +

14cb +

13

(a2 + c2

)k

)The kinetic energy is found from

KE =12ω ·

HO =12ω

k · a

2 + c2

3mω

k = a2 + c2

6 mω2

237

8.9.11GOAL: Determine the kinetic energy of the illustrated system.GIVEN: Body’s orientation. ω = ω

k .

DRAW:

FORMULATE EQUATIONS:For our problem we’ll find the moments of inertia for two solid cylinders, one of radius r

1and one

of radius r2. By subtracting we’ll obtain the inertias for the ring. The mass of the ring is given by

m = πhρ(R22−R2

1)

Outer radius cylinder:

oIx′x′

= oIz′z′

=πR2

2hρ(3R2

2+ h2)

12

oIy′y′

=πR2

2hρR2

2

2=πhρR4

2

2Inner radius cylinder:

iIx′x′

= iIz′z′

=πR2

1hρ(3R2

1+ h2)

12

iIy′y′

=πR2

1hρR2

1

2=πhρR4

1

2We now subtract the inertias of the smaller cylinder from the inertias of the larger one.

Ix′x′

= Iz′z′

=π hρ

[3(R4

2−R4

1

)+ h2

(R2

2−R2

1

)]12

=m[h2 + 3

(R2

1+R2

2

)]12

Iy′y′

= IY Y

=π hρ

(R2

2−R2

1

) (R2

1+R2

2

)2

=m(R2

1+R2

2

)2

Using the parallel axis theorem to find IXX

and IZZ

we obtain

IXX

= IZZ

=m[h2 + 3

(R2

1+R2

2

)]12

+m

(h

2

)2

=mh2

3+m(R2

1+R2

2)

4

The products of inertia about X, Y , Z are zero from symmetry.The general expression for angular momentum (using ω = ω

k ) is therefore

HO =

[mh2

3+m(R2

1+R2

2)

4

]ω

k

SOLVE:Having the relevant rotational inertia and knowing that

238

KE =12

ω ·

HO

we can form the body’s kinetic energy using (8.33):

KE =12[ω

k ] ·

[mh2

3+m(R2

1+R2

2)

4

]ω

k

KE =

[mh2

6 +m(R2

1+R2

2)

8

]ω2

239

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