CEE 271: Applied Mechanics II, Dynamics Lecture 33: Ch.19

CEE 271: Applied Mechanics II, Dynamics– Lecture 33: Ch.19, Sec.1–3 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Date: __________________

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LINEAR AND ANGULAR MOMENTUM,PRINCIPLE OF IMPULSE AND MOMENTUM

Today’s objectives: Studentswill be able to

1 Develop formulations forthe linear and angularmomentum of a body.

2 Apply the principle of linearand angular impulse andmomentum.

In-class activities:• Reading Quiz• Applications• Linear and Angular

Momentum• Principle of Impulse and

Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 The angular momentum of a rotating two-dimensional rigidbody about its center of mass G is .(a) mvG(b) IGvG(c) mω(d) IGω

ANS: (d)

2 If a rigid body rotates about a fixed axis passing through itscenter of mass, the body’s linear momentum is .(a) a constant(b) zero(c) mvG(d) IGω

ANS: (b)

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APPLICATIONS

• The swing bridge opens andcloses by turning using amotor located under thecenter of the deck at A thatapplies a torque M to thebridge.

• If the bridge was supported at its end B, would the sametorque open the bridge in the same time, or would it openslower or faster?

• What are the benefits of making the bridge with thevariable depth (thickness) substructure as shown?

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APPLICATIONS(continued)

• As the pendulum of the Charpy tester swings downward,its angular momentum and linear momentum bothincrease. By calculating its momenta in the verticalposition, we can calculate the impulse the pendulum exertswhen it hits the test specimen.

• As the pendulum rotates about point O, what is its angularmomentum about point O? ANS: H = IOω

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APPLICATIONS(continued)

• The space shuttle has several engines that exert thrust onthe shuttle when they are fired. By firing different engines,the pilot can control the motion and direction of the shuttle.

• If only engine A is fired, about which axis does the shuttletend to rotate?ANS: +x-axis

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LINEAR AND ANGULAR MOMENTUM (Section 19.1)

• The linear momentum of a rigid body is defined as

L = mvG

• This equation states that the linear momentum vector Lhas a magnitude equal to (mvG) and a direction defined byvG.

• The angular momentum of a rigid body isdefined as

HG = IGω

• Remember that the direction of HG isperpendicular to the plane of rotation.

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LINEAR AND ANGULAR MOMENTUM (continued)

Translation only

• When a rigid body undergoesrectilinear or curvilineartranslation, its angularmomentum is zero because

ω = 0

• Therefore, L = mvG and HG = 0

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LINEAR AND ANGULAR MOMENTUM (continued)

Rotation about a fixed axis.

• When a rigid body is rotatingabout a fixed axis passingthrough point O, the body’s linearmomentum and angularmomentum about G are:

L = mvG

andHG = IGω

• It is sometimes convenient to compute the angularmomentum of the body about the center of rotation O.

HO = (rG ×mvG) + IGω = IOω

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LINEAR AND ANGULAR MOMENTUM (continued)

General plane motion.

• When a rigid body is subjected togeneral plane motion, both the linearmomentum and the angular momentumcomputed about G are required.

L = mvG

HG = IGω

• The angular momentum about point Ais then (in magnitude)

HA = IGω + mvG(d)

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PRINCIPLE OF IMPULSE AND MOMENTUM(Section 19.2)

• Linear impulse-linear momentum equation:

L1 + Σ

∫ t2

t1

F dt = L2

(mvG)1 + Σ

∫ t2

t1

F dt = (mvG)2

• Angular impulse-angular momentum equation:

(HG)1 + Σ

∫ t2

t1

MGdt = (HG)2

IGω1 + Σ

∫ t2

t1

MGdt = IGω2

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PRINCIPLE OF IMPULSE AND MOMENTUM(continued)

• The previous relations can be represented graphically bydrawing the impulse-momentum diagram.

• To summarize, if motion is occurring in the x− y plane, thelinear impulse-linear momentum relation can be applied tothe x and y directions and the angular momentum-angularimpulse relation is applied about a z-axis passing throughany point (i.e., G). Therefore, the principle yields threescalar equations describing the planar motion of the body.

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PROCEDURE FOR ANALYSIS

1 Establish the x, y, z inertial frame of reference.2 Draw the impulse-momentum diagrams for the body.3 Compute IG, as necessary.4 Apply the equations of impulse and momentum (one vector

and one scalar or the three scalar equations).5 If more than three unknowns are involved, kinematic

equations relating the velocity of the mass center G andthe angular velocity ω should be used to furnish additionalequations.

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EXAMPLE

• Given: The 300 kg wheel has aradius of gyration about its masscenter O of kO = 0.4 meter. Thewheel is subjected to a couplemoment of M = 300 N · m.

• Find: The angular velocity after 6 seconds if it starts fromrest and no slipping occurs.

• Plan: Time as a parameter should make you think Impulseand Momentum! Since the body rolls without slipping, pointA is the center of rotation. Therefore, applying the angularimpulse and momentum relationships along withkinematics should solve the problem.

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EXAMPLE (Solution)

• Impulse-momentum diagram:

+ =

Wt

FtIGω2

mvG2

Nt

Mt

mvG1

IGω1

• Impulse and Momentum using kinematics: (vG)2 = rω2:

(HA)1 + Σ

∫ t2

t1

MAdt = (HA)2

0 + Mt = m(vG)2r + IGω2

= mr2ω2 + m(kO)2ω2

= m[r2 + (kO)2

]ω2

ω2 (t = 6 s) =300(6)

300(0.62 + 0.42)= 11.5 rad/s

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CONCEPT QUIZ

1 If a slab is rotating about its center of massG, its angular momentum about anyarbitrary point P is its angularmomentum computed about G (i.e., IGω).

(a) larger than(b) less than(c) the same as(d) None of the above

ANS: (c)2 The linear momentum of the slab in question 1 is .

(a) constant(b) zero(c) increasing linearly with time(d) decreasing linearly with time

ANS: (b)16 / 36

GROUP PROBLEM SOLVING

• Given: A gear set with:mA = 10 kgmB = 50 kgkA = 0.08 meterkB = 0.15 meterM = 10 N · m

• Find: The angular velocity of gear B after 5 seconds if thegears start turning from rest.

• Plan: Time is a parameter, thus Impulse and Momentumis recommended. First, relate the angular velocities of thetwo gears using kinematics. Then apply angular impulseand momentum to both gears.

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GROUP PROBLEM SOLVING(Solution)

Impulse-momentum diagrams: Note that the initial momentumis zero for both gears.

Ayt

Gear A:

Gear B:

AxtMt

IAωAWAt

F t

=

=Bxt

y

x

rA

rB

IBωB

WBt

Byt

F t

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GROUP PROBLEM SOLVING (continued)

• Kinematics: rAωA = rBωB

• Angular impulse andmomentum relation:

• For gear A: Mt− (Ft)rA = IAωA

• For gear B: (Ft)rB = IBωB so that (Ft) = (IBωB)/rB• Combining the two equations yields:

Mt = IAωA + (rA/rB)IBωB

• Substituting from kinematics for ωA = (rB/rA)ωB, yields

Mt = ωB[(rB/rA)IA + (rA/rB)IB] (1)

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GROUP PROBLEM SOLVING (continued)

IA = mA(kA)2 = 10(0.08)2 = 0.064kg · m2

IB = mB(kB)2 = 50(0.15)2 = 1.125kg · m2

• Using Eq. (1),

Mt = ωB[(rB/rA)IA + (rA/rB)IB]

10(5) = ωB[(0.2/0.1)0.064 + (0.1/0.2)1.125]

50 = 0.6905ωB (2)

• Therefore, ωB = 72.4 rad/sand ωA = (rB/rA)ωB = (0.2/0.1)72.4 = 144 rad/s

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ATTENTION QUIZ

1. If a slender bar rotates about end A, its angularmomentum with respect to A is?

������������������

������������������A

ω

l

G

(a) 112ml2ω

(b) 16ml2ω

(c) 13ml2ω

(d) ml2ω

ANS: (c)

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ATTENTION QUIZ (Continued)

2. As in the principle of work and energy, if a force does nowork, it does not need to be shown on the impulse andmomentum diagram/equation.(a) False(b) True(c) Depends on the case(d) No clue!

ANS: (a)

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CONSERVATION OF MOMENTUM

Today’s objectives: Studentswill be able to

1 Understand the conditionsfor conservation of linearand angular momentum.

2 Use the condition ofconservation of linear/angular momentum.

In-class activities:• Reading Quiz• Applications• Conservation of Linear and

Angular Momentum• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 If there are no external impulses acting on a body .(a) only linear momentum is conserved(b) only angular momentum is conserved(c) both linear momentum and angular momentum are

conserved(d) neither linear momentum nor angular momentum are

conservedANS: (c)

2 If a rigid body rotates about a fixed axis passing through itscenter of mass, the body’s linear momentum is .(a) a constant(b) zero(c) mvG(d) IGω

ANS: (b)

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APPLICATIONS

• A skater spends a lot of time either spinning on the ice orrotating through the air. To spin fast, or for a long time, theskater must develop a large amount of angular momentum.

• If the skater’s angular momentum is constant, can the skatervary her rotational speed? How?

• The skater spins faster when the arms are drawn in andslower when the arms are extended. Why?

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APPLICATIONS(continued)

• Conservation of angular momentum allows cats to land ontheir feet and divers to flip, twist, spiral and turn. It alsohelps teachers make their heads spin!

• Conservation of angular momentum makeswater circle the drain faster as it gets closer to the drain.

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CONSERVATION OF LINEAR MOMENTUM (19.3)

• Recall that the linear impulse and momentum relationshipis

L1 +∑∫ t2

t1

F dt = L2

where L1 = (mvG)1 and L2 = (mvG)2.• If the sum of all the linear impulses acting on the rigid body

(or a system of rigid bodies) is zero, all the impulse termsare zero. Thus, the linear momentum for a rigid body (orsystem) is constant, or conserved. So L1 = L2.

• This equation is referred to as theconservation of linear momentum. The conservation oflinear momentum equation can be used if the linearimpulses are small or non-impulsive.

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CONSERVATION OF ANGULAR MOMENTUM

• Similarly, if the sum of all the angular impulses due toexternal forces acting on the rigid body (or a system ofrigid bodies) is zero, all the impulse terms are zero. Thus,angular momentum is conserved

HG1 +∑∫ t2

t1

MGdt = HG2

where HG1 = IGω1 and HG2 = IGω2.• The resulting equation is referred to as the conservation of

angular momentum or HG1 = HG2.• If the initial condition of the rigid body (or system) is known,

conservation of momentum is often used to determine thefinal linear or angular velocity of a body just after an eventoccurs.

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PROCEDURE FOR ANALYSIS

1 Establish the x, y, z inertial frame of reference and drawFBDs.

2 Write the conservation of linear momentum equation.3 Write the conservation of angular momentum equation

about a fixed point or at the mass center G.4 Solve the conservation of linear or angular momentum

equations in the appropriate directions.5 If the motion is complicated, use of the kinematic

equations relating the velocity of the mass center G andthe angular velocity ω may be necessary.

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EXAMPLE

• Given: A 10 kg wheel(IG = 0.156 kg m2) rolls withoutslipping and does not rebound.

• Find: The minimum velocity, vG,the wheel must have to just rollover the obstruction at A.

• Plan: Since no slipping or rebounding occurs, the wheelpivots about point A. The force at A is much greater thanthe weight, and since the time of impact is very short, theweight can be considered non-impulsive. The reaction forceat A is a problem as we don‚Äôt know either its direction ormagnitude. This force can be eliminated by applying theconservation of angular momentum equation about A.

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EXAMPLE (Solution)

• Impulse-momentum diagram:

• Conservation of angular momentum

HA1 = HA2

r′mvG1 + IGω1 = rmvG2 + IGω2

(0.2 − 0.03)(10)vG1 + 0.156ω1 = 0.2(10)vG2 + 0.156ω2

• Kinematics: Since there is no slip, ω = vG/r = 5vG.Substituting and solving the momentum equation yields

vG2 = 0.892vG1

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EXAMPLE

To complete the solution, conservation of energy can be used.Since it cannot be used for theimpact (why?), it is applied justafter the impact. In order to rollover the bump, the wheel mustgo to position 3 from 2. WhenvG2 is a minimum, vG3 is zero.Why?

Energy conservation equation: T2 + V2 = T3 + V3

12(10)vG,22 + 1

2(0.156)ω22 + 0 = 0 + 98.1(0.03)

Substituting ω2 = 5vG2 and vG2 = 0.892vG1 and solving yields

vG1 = 0.729 m/s

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GROUP PROBLEM SOLVING

• Given: Two children(mA = mB = 30 kg) sit at theedge of the merry-go-round,which has a mass of 180 kg anda radius of gyration of kz = 0.6 m.

• Find: The angular velocity of the merry-go-round if A jumpsoff horizontally in the +t direction with a speed of 2 m/s,measured relative to the merry-go-round.

• Plan: Draw an impulse-momentum diagram. Theconservation of angular momentum can be used to find theangular velocity.

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GROUP PROBLEM SOLVING(Solution)

MB

B

A

vA/M

AM

Figure: Impulse-momentumdiagram

• Apply the conservation of angularmomentum equation:∑

H1 =∑

H2

180(0.6)2(2) + 2 × [(30)2(0.75)2]

= 180(0.6)2ω + (30)ω(0.75)2

+(30)(0.75ω + 2)(0.75)

Now we solve

197.1 = 98.55ω + 45

ω = 1.54 rad/s

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ATTENTION QUIZ

1. Using conservation of linear and angular momentumrequires that(a) all linear impulses sum to zero(b) all angular impulses sum to zero(c) both linear and angular impulses sum to zero(d) None of the above

ANS: (c)

2. The angular momentum of a body about a point A that isthe fixed axis of rotation but not the mass center (G) is(a) IAω(b) IGω(c) rG(mvG) + IGω(d) Both (a) and (c)

ANS: (d)

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Note

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