Question 2–11 Solution to Question 2–11

26 Chapter 2. Kinematics

Question 2–11

A rod of length L with a wheel of radius R attached to one of its ends is rotating

about the vertical axis OA with a constant angular velocity Ω relative to a fixed

reference frame as shown in Fig. P2-11. The wheel is vertical and rolls without

slip along a fixed horizontal surface. Determine the angular velocity and angular

acceleration of the wheel as viewed by an observer in a fixed reference frame.

A

B

LO

R

Ω

Figure P2-11

Solution to Question 2–11

Let F be a reference frame fixed to the ground. Then choose the following

coordinate system fixed in reference frame F :

Origin at OEx = Along OB when t = 0

Ez = Along Ω

Ey = Ez × Ex

Next, let A be a reference frame fixed to the direction of OB. Then choose the

following coordinate system fixed in reference frame A:

Origin at Oex = Along OBez = Along Ω

ey = ez × ex

Finally, let D be the reference frame of the wheel. Then choose the following

coordinate system fixed in reference frame D:

Origin at Bux = Along OBuy = In the plane of the wheel

uz = In the plane of the wheel = ux × uy

27

Now the angular velocity of the arm OB as viewed by an observer fixed to the

ground, denoted FωA, is given as

FωA = Ω = Ωez (2.136)

Next, the position of point B is given as

rB = Lex (2.137)

Computing the rate of change of rB in reference frame F , we obtain the velocity

of point B in reference frame F as

FvB =FdrB

dt=AdrB

dt+ FωA × rB (2.138)

Now we have

AdrB

dt= 0 (2.139)

FωA × rB = Ωez × Lex = LΩey (2.140)

Consequently,FvB = LΩey (2.141)

Next, suppose we let Q be the point of contact of the wheel with the ground.

Then, because the wheel rolls without slip along the ground, we know that

FvDQ = 0 (2.142)

Then, using Eq. (2–517) on page 107, we can obtain a second expression for FvBas

FvB = FvDQ + FωD × (rB − rQ) (2.143)

Now we know that FωD is given from the angular velocity addition theorem as

FωD = FωA +AωD (2.144)

We already have FωA from earlier. Then, because the wheel rotates about the

ex-direction (≡ ux-direction) and ex = ux is fixed in reference frameA, we have

AωD =ωux (2.145)

where ω is to be determined. Adding Eqs. (2.136) and (2.145), we obtain

FωD = Ωez +ωux (2.146)

Also, rB − rQ is given as

rB − rQ = REz = Rez (2.147)


Therefore,FvB = (Ωez +ωex)× Rez = −Rωey (2.148)

Setting the expressions for FvB from Eqs. (2.148) and (2.148) equal, we obtain

LΩ = −Rω (2.149)

from which we obtain ω as

ω = −LRΩ (2.150)

The angular velocity of the wheel as viewed by an observer fixed to the ground

is then given as

DωF = Ωez −L

RΩex (2.151)

The angular acceleration of the wheel in reference frame F is then given as

FαD =Fddt

(

FωD)

=Fddt

(

FωA)

+Fddt

(

AωD)

(2.152)

Now, because FωA = Ωez = ΩEz and Ez is fixed in reference frame F , we have

Fddt

(

FωA)

= Ωez = 0 (2.153)

because Ω is constant. Next, because AωD = −(L/R)Ωex and ex is fixed in ref-

erence frameA, we can apply the transport theorem to AωD between reference

frames A and F as

Fddt

(

AωD)

=Addt

(

AωD)

+ FωA ×AωD (2.154)

Now we have

Addt

(

AωD)

= −LRΩex = 0 (2.155)

FωA ×AωD = Ωez ×(

−LRΩex

)

= −LRΩ

2ey (2.156)

where we have again used the fact that Ω is constant. Therefore,

Fddt

(

AωD)

= −LRΩ

2ey (2.157)

Consequently, the angular acceleration of the disk as viewed by an observer

fixed to the ground is given as

FαD = −LRΩ

2ey (2.158)

29

Question 2–13

A collar is constrained to slide along a track in the form of a logarithmic spiral

as shown in Fig. P2-13. The equation for the spiral is given as

r = r0e−aθ

where r0 and a are constants and θ is the angle measured from the horizontal

direction. Determine (a) expressions for the intrinsic basis vectors et , en, and

eb in terms any other basis of your choosing, (b) the curvature of the trajectory

as a function of the angle θ, and (c) the velocity and acceleration of the collar as

viewed by an observer fixed to the track.

r=r 0e−aθ

O

P

θ

Figure P2-13


(a) Intrinsic Basis

Let F be a reference frame fixed to the track. Then, choose the following coor-

dinate system fixed in reference frame F :

Origin at OEx To the Right

Ez Out of Page

Ey = Ez × Ex

Next, letA be a reference frame that rotates with the direction along Om. Then,

choose the following coordinate system fixed in reference frame A:

Origin at Oer Along OmEz Out of Page

eθ = Ez × er


The position of the particle is given in terms of the basis er ,eθ,ez as

r = rer = r0e−aθer (2.159)

Furthermore, the angular velocity of reference frame A in reference frame F is

given asFωA = θEz (2.160)

Applying the rate of change transport theorem between reference frame A and

reference frame F , the velocity of the particle in reference frame F is given as

Fv =Fdr

dt=Adr

dt+ FωA × r (2.161)

where

Adr

dt= rer = −ar0θe

−aθer (2.162)

FωA × r = θez × rer = θez × r0e−aθer = r0θe

−aθeθ (2.163)

Adding Eq. (2.162) and Eq. (2.163), we obtain the velocity of the particle in ref-

erence frame F asFv = −ar0θe

−aθer + r0θe−aθeθ (2.164)

Simplifying Eq. (2.161), we obtain Fv as

Fv = r0θe−aθ [−aer + eθ] (2.165)

The tangent vector in reference frame F is then given as

et =Fv

‖Fv‖ =Fv

‖Fv‖ (2.166)

where Fv is the speed of the particle in reference frame F . Now the speed of

the particle in reference frame F is given as

Fv = ‖Fv‖ = r0θe−aθ

√

1+ a2 (2.167)

Dividing Fv in Eq. (2.165) by Fv in Eq. (2.167), we obtain the tangent vector in

reference frame F as

et =−aer + eθ√

1+ a2(2.168)

Next, the principle unit normal vector is obtained as

en =Fdet/dt

‖Fdet/dt‖(2.169)

31

Now we have from the rate of change transport theorem that

Fdet

dt=Adet

dt+ FωA × et (2.170)

where

Fdet

dt= 0 (2.171)

FωA × et = θez ×−aer + eθ√

1+ a2= −θer + aeθ√

1+ a2(2.172)

Adding Eq. (2.171) and Eq. (2.172), we obtain

Fdet

dt= −θer + aeθ√

1+ a2(2.173)

Consequently,∥

∥

∥

∥

∥

Fdet

dt

∥

∥

∥

∥

∥

= θ (2.174)

Dividing Fdet/dt in Eq. (2.173) by ‖Fdet/dt‖ in Eq. (2.174), we obtain en as

en = −er + aeθ√

1+ a2(2.175)

Finally, the bi-normal vector is obtained as

eb = et × en =−aer + eθ√

1+ a2×−er + aeθ√

1+ a2= Ez (2.176)

(b) Curvature

The curvature of the trajectory in reference frame F is then obtained as

κ = ‖Fdet/dt‖

Fv(2.177)

Substituting ‖Fdet/dt‖ from Eq. (2.174) and Fv from Eq. (2.167), we obtain κas

κ = 1

r0e−aθ√

1+ a2(2.178)

(c) Velocity and Acceleration

The velocity of the particle in reference frameF can be expressed in the intrinsic

basis asFv = Fvet (2.179)


Using the expression for Fv from Eq. (2.167), we obtain

Fv = r0θe−aθ

√

1+ a2et (2.180)

Next, the acceleration of the particle in reference frame F can be expressed in

terms of the intrinsic basis as

Fa = d

dt

(

Fv)

et + κ(

Fv)2

en (2.181)

Now we have thatd

dt

(

Fv)

= r0(θ − aθ2)e−aθ√

1+ a2 (2.182)

Furthermore,

κ(

Fv)2= 1

r0e−aθ√

1+ a2r2

0 θ2e−2aθ(1+ a2) = r0θ

2e−aθ√

1+ a2 (2.183)

The acceleration in reference frame F is then given as

Fa = r0e−aθ

√

1+ a2(θ − aθ2)et + r0θ2e−aθ

√

1+ a2en (2.184)

33

Question 2–15

A circular disk of radius R is attached to a rotating shaft of length L as shown in

Fig. P2-15. The shaft rotates about the vertical direction with a constant angular

velocity Ω relative to the ground. The disk, in turn, rotates about its center

about an axis orthogonal to the shaft. Knowing that the angle θ describes the

position of a point P located on the edge of the disk relative to the center of the

disk, determine the following quantities as viewed by an observer fixed to the

ground: (a) the angular velocity of the disk and (b) the velocity and acceleration

of point P .

A

L

O

P

R

θΩ

Figure P2-15


First, let F be a reference frame fixed to the ground. Then, we choose the

following coordinate system fixed in reference frame F :

Origin at Point AEx = Along Ω

Ey = Along AO at t = 0

Ez = Ex × Ey

Next, let A be a reference frame fixed to the horizontal shaft. Then, we choose

the following coordinate system fixed in reference frame F :

Origin at Point Aex = Along Ω

ey = Along AOez = ex × ey


Lastly, let B be a reference frame fixed to the disk. Then, choose the following

coordinate system fixed in reference frame B:

Origin at Point Oer = Along OPez = Same as Reference Frame Aeθ = ez × er

The geometry of the bases ex,ey ,ez and er ,eθ,ez is shown in Fig. (2.185).

In particular, using Fig. (2.185), we have that

ex = cosθer − sinθeθey = sinθer + cosθeθ

(2.185)

⊗

er

eθ

ex

eyez

θ

θ

Figure 2-3 Relationship Between Basis ex,ey ,ez and er ,eθ,ez for Ques-

tion 2–15

Now, since the shaft rotates with angular velocity Ω about the ey -direction

relative to the ground, the angular velocity of reference frame A in reference

frame F is given asFωA = Ω = Ωex (2.186)

Next, since the disk rotates with angular rate θ relative to the shaft in the ez-

direction, the angular velocity of reference frame B in reference frame A is

given asAωB = θez (2.187)

The angular velocity of reference frame B in reference frame F is then obtained

using the theorem of angular velocity addition as

FωB = FωA +AωB = Ωex + θez (2.188)

35

Then, using the relationship between ex,ey ,ez and er ,eθ,ez from Eq. (2.185),

we obtain FωB in terms of the basis er ,eθ,ez as

FωB = Ω(cosθer − sinθeθ)+ θez = Ω cosθer −Ω sinθeθ + θez (2.189)

Next, the position of point P is given as

rP = rO + rP/O (2.190)

Now, in terms of the basis ex,ey ,ez, the position of point O is given as

rO = Ley (2.191)

Also, in terms of the basis er ,eθ,ez we have that

rP/O = Rer (2.192)

Consequently,

rP = Ley + Rer (2.193)

The velocity of point P in reference frame F is then given as

FvP =Fddt(rP) =

Fddt(rO)+

Fddt

(

rO/P)

= FvO + FvP/O (2.194)

First, since rO is expressed in terms of the basis ex,ey ,ez and ex,ey ,ez is

fixed in reference frame A, we can apply the rate of change transport theorem

to rO between reference frames A and F as

FvO =Fddt(rO) =

Addt(rO)+ FωA × rO (2.195)

Now we have that

Addt(rO) = 0 (2.196)

FωA × rO = Ωex × Ley = LΩez (2.197)


FvO = LΩez (2.198)

Next, since rP/O is expressed in terms of the basis er ,eθ,ez and er ,eθ,ez is

fixed in reference frame B, we can apply the rate of change transport theorem

to rP/O between reference frames B and F as

FvP/O =Fddt

(

rP/O)

=Bddt

(

rP/O)

+ FωB × rP/O (2.199)


Now we have that

Bddt

(

rP/O)

= 0 (2.200)

FωB × rP/O = (Ω cosθer −Ω sinθeθ + θez)× Rer= Rθeθ + RΩ sinθez (2.201)


FvP/O = Rθeθ + RΩ sinθez (2.202)

The velocity of point P in reference frameF is then obtained by adding Eq. (2.198)

and Eq. (2.202) as

FvP = FvO + FvP/O = LΩez + Rθeθ + RΩ sinθez (2.203)

Simplifying Eq. (2.203), we obtain

FvP = Rθeθ + (L+ R sinθ)Ωez (2.204)

The acceleration of point P in reference frame F is obtained in the same

manner as was used to obtain the velocity in reference frame F . First, we have

from Eq. (2.203) thatFvP = FvO + FvP/O (2.205)

Now, since FvO is expressed in the basis ex,ey ,ez and ex,ey ,ez is fixed

in reference frame A, the acceleration of point O in reference frame F can

be obtained by applying the rate of change transport theorem to FvO between

reference frames A and F as

FaO =Fddt

(

FvO

)

=Addt

(

FvO

)

+ FωA × FvO (2.206)

Now we have that

Addt

(

FvO

)

= 0 (2.207)

FωA × FvO = Ωex × LΩez = −LΩ2ey (2.208)

Then, adding Eq. (2.207) and Eq. (2.208), we obtain FaO as

FaO = −LΩ2ey (2.209)

Next, since FvP/O is expressed in terms of the basis er ,eθ,ez, the acceleration

of point P relative to point O in reference frame F is obtained by applying the

rate of change transport theorem to FvP/O between reference frames B and Fas

FaP/O =Fddt

(

FvP/O

)

=Bddt

(

FvP/O

)

+ FωB × FvP/O (2.210)

37

Bddt

(

FvP/O

)

= Rθeθ + RΩθ cosθez (2.211)

FωB × FvO = (Ω cosθer −Ω sinθeθ + θez)× (Rθeθ + RΩ sinθez)= RΩθ cosθez − RΩ2 cosθ sinθeθ

−RΩ2 sin2θer − Rθ2er (2.212)


FωB×FvO = −(Rθ2+RΩ2 sin2θ)er −RΩ2 cosθ sinθeθ+RΩθ cosθez (2.213)

Then, adding Eq. (2.211) and Eq. (2.213), we obtain FaP/O as

FaP/O = −(Rθ2 + RΩ2 sin2θ)er + (Rθ − RΩ2 cosθ sinθ)eθ + 2RΩθ cosθez(2.214)

Finally, adding Eq. (2.209) and Eq. (2.214), we obtain FaP as

FaP = −LΩ2ey−(Rθ2+RΩ2 sin2θ)er +(Rθ−RΩ2 cosθ sinθ)eθ+2RΩθ cosθez(2.215)

Now it is seen from Eq. (2.215) that some of the terms in FaP are expressed

in the basis ex,ey ,ez while other terms are expressed in the basis er ,eθ,ez.However, using Eq. (2.185), we can obtain an expression for FaP in terms of

a single basis. Now, while it is possible to write FaP in terms of the basis

ex,ey ,ez, it is preferable (and simpler) to write both quantities in terms of

the basis er ,eθ,ez. First, substituting the expression for ey from Eq. (2.185)

into Eq. (2.215), we obtain FaP in terms of er ,eθ,ez as

FaP = −LΩ2(sinθer + cosθeθ)− (Rθ2 + RΩ2 sin2θ)er

+ (Rθ − RΩ2 cosθ sinθ)eθ + 2RΩθ cosθez(2.216)

Simplifying Eq. (2.216) gives

FaP = −(LΩ2 sinθ + Rθ2 + RΩ2 sin2θ)er

+ (Rθ − LΩ2 cosθ − RΩ2 cosθ sinθ)eθ

+ 2RΩθ cosθez

(2.217)


Question 2–16

A disk of radius R rotates freely about its center at a point located on the end of

an arm of length L as shown in Fig. P2-16. The arm itself pivots freely at its other

end at point O to a vertical shaft. Finally, the shaft rotates with constant angular

velocity Ω relative to the ground. Knowing that φ describes the location of a

point P on the edge of the disk relative to the direction OQ and that θ is formed

by the arm with the downward direction, determine the following quantities as

viewed by an observer fixed to the ground: (a) the angular velocity of the disk

and (b) the velocity and acceleration of point P .

L

O

P

Q

θ φ

Ω

Figure P2-16




Origin at OEx = Along OQ when θ = 0

Ez = Orthogonal to plane of shaft and arm and out of page at t = 0

Ey = Ez × Ex

39

Next, let A be a reference frame fixed to the vertical shaft. Then choose the


Origin at Oex = Exez = Orthogonal to plane of shaft and arm

ey = ez × ex

We note that ez and Ez are equal when t = 0. Next, let B be a reference frame

fixed to the rod OQ. Then choose the following coordinate system fixed in

reference frame B:

Origin at Oux = Along OQuz = Orthogonal to plane of shaft, arm, and disk

uy = uz × ux

Finally, let D be a reference frame fixed to the disk. Then choose the following

coordinate system fixed in reference frame D:

Origin at Oir = Along OQiz = −eziφ = uz × ur

The geometry of the bases ex,ey ,ez, ux,uy ,uz, and ir , iφ, iz is shown in

Fig. 2-4.

⊗

ex

ey

ir

iφ

ux

−uxuy

θ

θ

φ

φ

iz (into page)

ez (out of page)

Figure 2-4 Geometry of bases ex,ey ,ez, ux,uy ,uz, and ir , iφ, iz for

Question 2–16.

The angular velocity of reference frame A in reference frame F is then given as

FωA = −Ω = −Ωex (2.218)


where the negative sign arises from the fact that the positive sense of Ω is

vertically upward while the direction Ex is downward. Next, the angular velocity

of reference frame B in reference frame A is given as

AωB = θuz (2.219)

Next, the angular velocity of reference frame D in reference frame B is given as

BωD = φiz = −φuz (2.220)

The angular velocity of the disk as viewed by an observer fixed to the ground is

then obtained from the angular velocity addition theorem as

FωD = FωA +AωB + BωD = −Ωex + θuz − φuz (2.221)

Now from the geometry of the bases, it is seen that

ex = cosθux − sinθuy (2.222)

which implies that

FωD = −Ω(cosθux − sinθuy)+ (θ − φ)uz= −Ω cosθux +Ω sinθuy + (θ − φ)uz

(2.223)

Now because point P (i.e., the point for which we want the velocity) is fixed to

the disk, it is helpful to obtain an expression for FωD in terms of the basis

ir , iφ, iz. In order to obtain such an expression, it is first important to see

from Fig. 2-4 that

−ux = cosφir − sinφiφ =⇒ ux = − cosφir + sinφiφ (2.224)

uy = sinφir + cosφiφ (2.225)

where it is observed that diagramatically it is first easier to determine −ux in

terms of ir and iφ and then take the negative sign of the result. Consequently,

FωD = −Ω cosθ(− cosφir + sinφiφ)+Ω sinθ(sinφir + cosφiφ)+ (θ − φ)uz= Ω(cosθ cosφ+ sinθ sinφ)ir +Ω(cosθ sinφ− sinθ cosφ)iφ + (θ − φ)iz= Ω cos(θ −φ)ir +Ω sin(θ −φ)iφ + (θ − φ)iz

(2.226)

where we have used the two trigonometric identities

cosθ cosφ+ sinθ sinφ = cos(θ −φ) (2.227)

cosθ sinφ− sinθ cosφ = sin(θ −φ) (2.228)

41

Now we know that the position of point P is given as

rP = rQ + rP/Q (2.229)

where

rQ = Lux (2.230)

rP/Q = Rir (2.231)

Because the basis ux,uy ,uz is fixed in reference frame B, we can apply the

transport theorem to rQ between reference frames B and F as

FvQ =FdrQ

dt=BdrQ

dt+ FωB × rQ (2.232)

Now we have

FωB = FωA +AωB = −Ωex + θuz = −Ω cosθux +Ω sinθuy + θuz (2.233)

Furthermore,

BdrQ

dt= 0 (2.234)

FωB × rQ = (−Ω cosθux +Ω sinθuy + θuz)× Lux (2.235)

which gives

FωB×rQ = (−Ω cosθux+Ω sinθuy + θuz)×Lux = Lθuy −LΩ sinθuz (2.236)

Therefore,FvQ = Lθuy − LΩ sinθuz (2.237)

Next, because rP/Q is fixed in reference frame D, we can apply the transport

theorem to rP/Q between reference frames D and F as

FvP/Q =Fddt

(

rP/Q)

=Dddt

(

rP/Q)

+ FωD × rP/Q (2.238)

Now we have

Dddt

(

rP/Q)

= 0 (2.239)

FωB × rP/Q = [Ω cos(θ −φ)ir +Ω sin(θ −φ)iφ + (θ − φ)iz]×Rir (2.240)

which implies that

FωB × rP/Q = [Ω cos(θ −φ)ir +Ω sin(θ −φ)iφ + (θ − φ)iz]× Rir

= R(θ − φ)iφ − RΩ sin(θ −φ)iz(2.241)


Therefore,FvP/Q = R(θ − φ)iφ − RΩ sin(θ −φ)iz (2.242)

The velocity of point P in reference frameF is then obtained by adding Eqs. (2.237)

and (2.242) as

FvP = Lθuy − LΩ sinθuz ++R(θ − φ)iφ − RΩ sin(θ −φ)iz (2.243)

It is noted that this last expression can be converted to an expression in terms

of a single basis using the relationships between the bases as given in Fig. 2-4.

43

Question 2–17

A particle slides along a track in the form of a spiral as shown in Fig. P2-17. The

equation for the spiral is

r = aθ

where a is a constant and θ is the angle measured from the horizontal. Deter-

mine (a) expressions for the intrinsic basis vectors et , en, and eb in terms any

other basis of your choosing, (b) determine the curvature of the trajectory as a

function of the angle θ, and (c) determine the velocity and acceleration of the

collar as viewed by an observer fixed to the track.

r

O

P

θ

Figure P2-17


First, let F be a reference frame fixed to the spiral. Then, choose the following


Origin at OEx = To the Right

Ez = Out of Page

Ey = Ez × Ex

Next, letA be a reference frame fixed to direction of OP . Then, choose the


Origin at Point Oer = Along OPEz = Out of The Page

eθ = Ez × er


Determination of Intrinsic Basis

The position of the particle in terms of the basis er ,eθ,Ez is given as

r = rer = aθer (2.244)

Furthermore, the angular velocity of reference frame A in reference frame F is

given asFωA = θEz (2.245)

The velocity of the particle in reference frame F is then obtained using the rate

of change transport theorem as

Fv =Fdr

dt=Adr

dt+ FωA × r (2.246)

Now we have that

Adr

dt= aθer

FωA × r = θEz × aθer = aθθeθ

(2.247)

Adding the two expressions in Eq. (2.247), the velocity of the particle in refer-

ence frame F is obtained as

Fv = aθer + aθθeθ (2.248)

Eq. (2.248) can be re-written as

Fv = aθ (er + θeθ) (2.249)

The speed of the particle in reference frame F is then obtained from Eq. (2.249)

asFv = ‖Fv‖ = aθ

√

1+ θ2 = aθ(

1+ θ2)1/2

(2.250)

Then, the tangent vector in reference frame F is obtained as

et =FvFv

(2.251)

Then, using Fv from Eq. (2.249) and Fv from Eq. (2.250), we obtain the tangent

vector in reference frame F as

et =aθ (er + θeθ)

aθ√

1+ θ2= er + θeθ√

1+ θ2=(

1+ θ2)−1/2

(er + θeθ) (2.252)

Next, the principle unit normal vector in reference frame F is obtained as

en =Fdet/dt

‖Fdet/dt‖(2.253)

45

Now, using the rate of change transport theorem, we can compute Fdet/dt in

reference frame F asFdet

dt=Adet

dt+ FωA × et (2.254)

Using the expression for et from Eq. (2.252), we have that

Adet

dt= −1

2

(

1+ θ2)−3/2

(2θθ) (er + θeθ)+(

1+ θ2)−1/2

θeθ (2.255)

Eq. (2.255) simplifies to

Adet

dt= θ

(

1+ θ2)−3/2

(−θer + eθ) (2.256)

Next, the second term in Eq. (2.254) is obtained as

FωA × et = θEz ×(

1+ θ2)−1/2

(er + θeθ) (2.257)

Eq. (2.257) simplifies to

FωA × et = θ(1+ θ2)−1/2 (−θer + eθ) (2.258)

Eq. (2.258) can be re-written as

FωA × et = θ(1+ θ2)(1+ θ2)−3/2 (−θer + eθ) (2.259)

Then, adding Eq. (2.256) and Eq. (2.259), we obtain

Fdet

dt= θ

(

1+ θ2)−3/2 (

2+ θ2)

(−θer + eθ) (2.260)

Then the magnitude of Fdet/dt is obtained as

∥

∥

∥

∥

∥

Fdet

dt

∥

∥

∥

∥

∥

= θ(

1+ θ2)−3/2 (

2+ θ2)√

1+ θ2 (2.261)

Then, dividing Eq. (2.260) by Eq. (2.261), we obtain the principle unit normal in

reference frame F as

en =−θer + eθ√

1+ θ2(2.262)

Finally, the principle unit bi-normal vector in reference frame F is obtained as

eb = et × en =er + θeθ√

1+ θ2× −θer + eθ√

1+ θ2= ez (2.263)


Curvature of Trajectory in Reference Frame F

First, we know thatFdet

dt= κFven (2.264)

Taking the magnitude of both sides, we have that

∥

∥

∥

∥

∥

Fdet

dt

∥

∥

∥

∥

∥

= κFv (2.265)

Solving for κ, we have that

κ = ‖Fdet/dt‖

Fv(2.266)

Substituting the expression for ‖Fdet/dt‖ from Eq. (2.261) and the expression

for Fv from Eq. (2.250) into Eq. (2.266), we obtain κ as

κ = θ(

1+ θ2)−3/2 (

2+ θ2)√

1+ θ2

aθ√

1+ θ2= 2+ θ2

a(1+ θ2)3/2(2.267)

Velocity and Acceleration of Particle

The velocity of the particle in reference frame F is given in intrinsic coordinates

asFv = Fvet (2.268)

Using the expression for Fv from Eq. (2.250), we obtain Fv as

Fv = aθ√

1+ θ2et (2.269)

Furthermore, the acceleration in reference frame F is obtained in intrinsic co-

ordinates asFa = d

dt

(

Fv)

et + κ(

Fv)2

en (2.270)

Differentiating Fv from Eq. (2.250), we have that

d

dt

(

Fv)

= aθ√

1+ θ2 + aθ(1+ θ2)−1/22θθ (2.271)


d

dt

(

Fv)

= a(

1+ θ2)−1/2 [

θ(

1+ θ2)

+ θ2θ]

(2.272)

Next, using the expression for κ from Eq. (2.267), we have that

κ(

Fv)2= 2+ θ2

a(1+ θ2)3/2

(

aθ√

1+ θ2)

= a(2+ θ2)θ2

√1+ θ2

(2.273)

47

Substituting the results of Eq. (2.272) and Eq. (2.273) into Eq. (2.270), we obtain

the acceleration of the particle in reference frame F as

Fa = a(

1+ θ2)−1/2 [

θ(

1+ θ2)

+ θ2θ]

et +a(2+ θ2)θ2

√1+ θ2

en (2.274)

Simplifying Eq. (2.274) gives

Fa = a√1+ θ2

[(

θ(1+ θ2)+ θ2θ)

et + (2+ θ2)θ2en

]

(2.275)


Question 2–19

A particle P slides without friction along the inside of a fixed hemispherical bowl

of radius R as shown in Fig. P2-19. The basis Ex,Ey ,Ez is fixed to the bowl.

Furthermore, the angle θ is measured from the Ex-direction to the directionOQ,

where point Q lies on the rim of the bowl while the angle φ is measured from

the OQ-direction to the position of the particle. Determine the velocity and

acceleration of the particle as viewed by an observer fixed to the bowl. Hint:

Express the position in terms of a spherical basis that is fixed to the direction

OP ; then determine the velocity and acceleration as viewed by an observer fixed

to the bowl in terms of this spherical basis.

Ex

Ey

Ez

O

P

Q

R

θφ

Figure P2-18


Let F be a reference frame fixed to the bowl. Then choose the following coordi-

nate system fixed in reference frame F :

Origin at OEx = Given

Ey = Given

Ez = Ex × Ey = Given

Next, let A be a reference frame fixed to the plane defined by the points O and

Q and the direction Ez. Then choose the following coordinate system fixed in

reference frame A:

Origin at Oer = Along OQez = Ezeθ = ez × er

49

Finally, let B be a reference frame fixed to the direction OP . Then choose the

following coordinate system fixed in reference frame B:

Origin at Our = Along OPuθ = eθuφ = ur × uθ

The relationship between the bases Ex,Ey ,Ez and er ,eθ,ez is shown in

Fig. 2-5 while the relationship between the bases er ,eθ,ez and ur ,uθ,uφis shown in Fig. 2-6.

⊗

ereθ

Ex

Ey

ez,Ezθ

θ

Figure 2-5 Relationship between bases Ex,Ey ,Ez and er ,eθ,ez for Ques-

tion 2–19.

er

uruφ

ez

uθ,eθφ

φ

Figure 2-6 Relationship between bases er ,eθ,ez and ur ,uθ,uφ for Ques-

tion 2–19.

The position of the particle is then given as

r = Rur (2.276)

Next, the angular velocity of reference frame A in reference frame F is given as

FωA = θez (2.277)

Furthermore, the angular velocity of reference frame B in reference frame A is

given asAωB = −φuθ (2.278)

where the negative sign on AωB is due to the fact that the angle φ is mea-

sured positively about the negative uθ-direction (see Fig. 2-6). Then, applying

the angular velocity addition theorem, we have

FωB = FωA +AωB = θez − φuθ (2.279)


Now we can obtain an expression for FωB in terms of the basis ur ,uθ,uφ by

expressing ez in terms of ur and uφ as

ez = sinφur + cosφuφ (2.280)

Consequently,

FωB = θ(sinφur + cosφuφ)− φuθ = θ sinφur − φuθ + θ cosφuφ (2.281)

Then, the velocity of point P in reference frame F is obtained by applying the

transport theorem between reference frames B and F as

Fv =Fdr

dt=Bdr

dt+ FωB × r (2.282)

Now we have

Bdr

dt= 0 (2.283)

FωB × r = (θ sinφur − φuθ + θ cosφuφ)× Rur

= Rθ cosφuθ + Rφuφ (2.284)

Therefore,Fv = Rθ cosφuθ + Rφuφ (2.285)

The acceleration of point P in reference frame F is obtained by applying the

transport theorem to Fv between reference frames B and F as

Fa =F ddt

(

Fv)

=B ddt

(

Fv)

+ FωB × Fv (2.286)

Now we have

B ddt

(

Fv)

= R(θ cosφ− θφ sinφ)uθ + Rφuφ (2.287)

FωB × Fv = (θ sinφur − φuθ + θ cosφuφ)× (Rθ cosφuθ + Rφuφ)

= Rθ2 cosφ sinφuφ − Rθφ sinφuθ

−Rφ2ur − Rθ2 cos2φur (2.288)

Adding these last two equations and simplifying gives

Fa = −(Rφ2 + Rθ2 cos2φ)ur

+ (Rθ cosφ− 2Rθφ sinφ)uθ

+ (Rφ+ Rθ2 cosφ sinφ)uφ

(2.289)

51

Question 2–20

A particle P slides along a circular table as shown in Fig. P2-20. The table is

rigidly attached to two shafts such that the shafts and table rotate with angular

velocity Ω about an axis along the direction of the shafts. Knowing that the

position of the particle is given in terms of a polar coordinate system relative to

the table, determine (a) the angular velocity of the table as viewed by an observer

fixed to the ground, (b) the velocity and acceleration of the particle as viewed

by an observer fixed to the table, and (c) the velocity and acceleration of the

particle as viewed by an observer fixed to the ground.

r

A

B

OP

θ

Ω

Figure P2-19




Origin at 0

Ex = Along OBEz = Vertically Upward

Ey = Ez × Ex

Next, let A be a reference frame fixed to the table. Then choose the following

coordinate system fixed in reference frame A:

Origin at 0

ex = Along OBez = Orthogonal to Table and = Ez When t = 0

ey = ez × ex

Finally, let B be a reference frame fixed to the direction of OP . Then choose the

following coordinate system fixed in reference frame B:

Origin at 0

er = Along OBez = Same as in Reference Frame Beθ = ez × er



r = rer (2.290)

Now because the position is expressed in terms of the basis er ,eθ,ez and

er ,eθ,ez is fixed in reference frame B, the velocity of the particle as viewed by

an observer fixed to the ground is obtained by applying the transport theorem

between reference frames B and F as

Fv =Fdr

dt=Bdr

dt+ FωB × r (2.291)

First, the angular velocity of B in F is obtained from the angular velocity addi-

tion theorem asFωB = FωA +AωB (2.292)

Now we have

FωA = Ω = Ωex (2.293)AωB = θez (2.294)

which implies thatFωB = Ωex + θez (2.295)

Next, because the position is expressed in terms of the basis er ,eθ,ez, the

unit vector ex must also be expressed in terms of the basis er ,eθ,ez. The

relationship between the bases ex,ey ,ez and er ,eθ,ez is shown in Fig. 2-7.

Using Fig. 2-7, it is seen that

ereθ

ex

ey

ez

θ

θ

Figure 2-7 Geometry of bases ex,ey ,ez and er ,eθ,ez for Question P2–20.

ex = cosθer − sinθeθ (2.296)

ey = sinθer + cosθeθ (2.297)

Therefore,

FωB = Ω(cosθer − sinθeθ)+ θez = Ω cosθer −Ω sinθeθ + θez (2.298)

53

Now the two terms required to obtain Fv are given as

Fdr

dt= rer (2.299)

FωB × r = (Ω cosθer −Ω sinθeθ + θez)× rer = r θeθ + rΩ sinθez(2.300)

Therefore, the velocity of the particle in reference frame F is

Fv = rer + r θeθ + rΩ sinθez (2.301)

Next, the acceleration of the particle as viewed by an observer fixed to the

ground is given from the transport theorem as

Fa =Fddt

(

Fv)

=Bddt

(

Fv)

+ FωB × Fv (2.302)

Now we have

Bddt

(

Fv)

= rer + (r θ + r θ)eθ +[

rΩ sinθ + r(Ω sinθ +Ωθ cosθ)]

ez(2.303)

FωB × Fv = (Ω cosθer −Ω sinθeθ + θez)× (rer + r θeθ + rΩ sinθez)

= r θΩ cosθez − rΩ2 cosθ sinθeθ + rΩ sinθez

−rΩ2 sin2θer + r θeθ − r θ2er

= −(r θ2 + rΩ2 sin2θ)er + (r θ − rΩ2 cosθ sinθ)eθ

+ (rΩ sinθ + r θΩ cosθ)ez (2.304)

Adding these last two equations, we obtain the acceleration as viewed by an

observer fixed to the ground as

Fa = (r − r θ2 − rΩ2 sin2θ)er

+ (2r θ + r θ − rΩ2 cosθ sinθ)eθ

+[

r Ω sinθ + 2(rΩ sinθ + rΩθ cosθ)]

ez

(2.305)


Question 2–21

A slender rod of length l is hinged to a collar as shown in Fig. P2-21. The collar

slides freely along a fixed horizontal track. Knowing that x is the horizontal

displacement of the collar and that θ describes the orientation of the rod relative

to the vertical direction, determine the velocity and acceleration of the free end

of the rod as viewed by an observer fixed to the track.

l

xO

Pθ

Figure P2-20


Let F be a reference frame fixed to the horizontal track. Then, choose the

following coordinate system fixed in reference frame F :

Origin at O at t = 0

Ex = To the Right

Ez = Into Page

Ey = Ez × Ex

Next, let A be a reference frame fixed to the rod. Then, choose the following

coordinate system fixed in reference frame A:

Origin at Oer = Along OPez = Into Page

eθ = Ez × er

We note that the relationship between the basis Ex,Ey ,Ez and er ,eθ,ez is

given as

Ex = sinθer + cosθeθEy = − cosθer + sinθeθ

(2.306)

Using the bases Ex,Ey ,Ez and er ,eθ,ez, the position of point P is given

as

rP = rO + rP/O = xEx + ler (2.307)

55

Next, the angular velocity of reference frame A in reference frame F is given as

FωA = θez (2.308)

The velocity of point P in reference frame F is then given as

FvP =Fddt(rO)+

Fddt

(

rP/O)

= FvO + FvP/O (2.309)

Now since rO is expressed in the basis Ex,Ey ,Ez and Ex,Ey ,Ez is fixed, we

have that

FvO =Fddt(rO) = xEx (2.310)

Next, since rP/O is expressed in the basis er ,eθ,ez and er ,eθ,ez rotates

with angular velocity FωA, we can apply the rate of change transport theorem

to rP/O between reference frame A and reference frame F as

FvP/O =Fddt

(

rP/O)

=Addt

(

rP/O)

+ FωA × rP/O (2.311)

Now we have that

Addt

(

rP/O)

= 0 (2.312)

FωA × rP/O = θez × ler = lθeθ (2.313)

Adding Eq. (2.312) and Eq. (2.313) gives

FvP/O = lθeθ (2.314)

Therefore, the velocity of point P in reference frame F is given as

FvP = xEx + lθeθ (2.315)

Next, the acceleration of point P in reference frame F is obtained as

FaP =Fddt

(

FvP

)

(2.316)

Now we have thatFvP = FvO + FvP/O (2.317)

where

FvO = xEx (2.318)FvP/O = lθeθ (2.319)

Consequently,

FaP =Fddt

(

FvO

)

+Fddt

(

FvP/O

)

(2.320)


Now, since FvO is expressed in the basis Ex,Ey ,Ez, we have that

FaO =Fddt

(

FvO

)

= xEx (2.321)

Furthermore, since FvP/O is expressed in the basis er ,eθ,ez and er ,eθ,ezrotates with angular velocity FωA, we can obtain FaP/O by applying the rate of

change transport theorem between reference frame A and reference frame Fas

FaP/O =Fddt

(

FvO

)

=Addt

(

FvO

)

+ FωA × FvO (2.322)

Now we have that

Addt

(

FvO

)

= lθeθ (2.323)

FωA × FvO = θez × lθeθ = −lθ2er (2.324)

Adding Eq. (2.323) and Eq. (2.324) gives

FaP/O = −lθ2er + lθeθ (2.325)

Then, adding Eq. (2.321) and Eq. (2.325), we obtain the velocity of point P in

reference frame F asFaP = xEx − lθ2er + lθeθ (2.326)

Finally, substituting the expression for Ex from Eq. (2.306), we obtain FaP in

terms of the basis er ,eθ,ez as

FaP = x(sinθer + cosθeθ)− lθ2er + lθeθ = (x sinθ − lθ2)er + (x cosθ + lθ)eθ(2.327)

57

Question 2–23

A particle slides along a fixed track y = − ln cosx as shown in Fig. P2-23 (where

−π/2 < x < π/2). Using the horizontal component of position, x, as the vari-

able to describe the motion and the initial condition x(t = 0) = x0, determine

the following quantities as viewed by an observer fixed to the track: (a) the arc-

length parameter s as a function of x, (b) the intrinsic basis et,en,eb and the

curvature κ, and (c) the velocity and acceleration of the particle.

y = − ln cosx

Ex

Ey

O

P

Figure P2-21


For this problem, it is convenient to use a reference frame F that is fixed to the

track. Then, we choose the following coordinate system fixed in reference frame

F :

Origin at OEx = Along OxEy = Along OyEz = Ex × Ey


r = xEx − ln cosxEy (2.328)

Now, since the basis Ex,Ey ,Ez does not rotate, the velocity in reference frame

F is given asFv = xEx + x tanxEy (2.329)

Using the velocity from Eq. (2.329), the speed of the particle in reference frame

F is given as

Fv = ‖Fv‖ = x√

1+ tan2 x = x secx (2.330)


Arc-length Parameter as a Function of x

Now we recall the arc-length equation as

d

dt

(

Fs)

= Fv = x√

1+ tan2 x = x secx (2.331)

Separating variables in Eq. (2.331), we obtain

Fds = secxdx (2.332)

Integrating both sides of Eq. (2.332) gives

Fs − Fs0 =∫ x

x0

secxdx (2.333)

Using the integral given for secx, we obtain

Fs − Fs0 = ln [secx + tanx]xx0= ln

[

secx + tanx

secx0 + tanx0

]

(2.334)

Noting that Fs(0) = Fs0 = 0, the arc-length is given as

Fs = ln [secx + tanx]xx0(2.335)


Fs = ln

[

secx + tanx

secx0 + tanx0

]

(2.336)

Intrinsic Basis

Next, we need to compute the intrinsic basis. First, we have the tangent vector

as

et =FvFv

= x(Ex + tanxEy)

x secx= 1

secxEx +

tanx

secxEy (2.337)

Now we note that secx = 1/ cosx. Therefore,

tanx

secx= sinx (2.338)

Eq. (2.337) then simplifies to

et = cosxEx + sinxEy (2.339)

Next, the principle unit normal is given as

Fdet

dt= κFven (2.340)

59

Differentiating et in Eq. (2.339), we obtain

Fdet

dt= −x sinxEx + x cosxEy (2.341)

Consequently,∥

∥

∥

∥

∥

Fdet

dt

∥

∥

∥

∥

∥

= x = κFv (2.342)

which implies that

en =Fdet/dt∥

∥

∥

Fdet/dt∥

∥

∥

= −x sinxEx + x cosxEy

x= − sinxEx + cosxEy (2.343)

Then, using Fv from Eq. (2.330), we obtain the curvature as

κ = x

x secx= 1

secx= cosx (2.344)

Finally, the principle unit bi-normal is given as

eb = et × en = (cosxEx + sinxEy)× (− sinxEx + cosxEy) = Ez (2.345)

Velocity and Acceleration in Terms of Intrinsic Basis

Using the speed from Eq. (2.330), the velocity of the particle in reference frame

F is given in terms of the instrinsic basis as

Fv = x secxet (2.346)

Next, the acceleration is given in terms of the intrinsic basis as

Fa = d

dt

(

Fv)

et + κ(

Fv)

en (2.347)

Now, using Fv from Eq. (2.330), we obtain d(Fv)/dt as

d

dt

(

Fv)

= x secx + x2 secx tanx = secx[

x + x2 tanx]

(2.348)

Also, using κ from Eq. (2.344) we obtain

κ(

Fv)

= cosx(x secx)2 = x2 secx (2.349)

The acceleration of the particle in reference frame F is then given as

Fa = secx[

x + x2 tanx]

et + x2 secxen (2.350)


Documents

Question 2–11 Solution to Question 2–11