Chapter 10 Rotation 1 Rotation 1.1 Polar coordinates: general case The point P is located at (r, ), where r is the distance from the origin and is the measured counterclockwise from the reference line (the x axis).

We introduce the unit vectors given by

)cos,sin(ˆ

)sin,(cosˆ

r

The position vector (displacement vector) is given by

jir ˆsinˆcos)sin,cos( rrrr The velocity and acceleration are

ˆ)2(ˆ)(

ˆˆ2

rrrrr

rrr

a

v

or

rv

rrvr

ˆ

ˆ

v

v

)(1

2ˆ

ˆ

2

2

rdt

d

rrra

rrrar

a

a

((Note-1))

ˆˆ)cos,sin()sin,(cos)cossin,sincos(

)sin,cos(

rrrrrrrrr

rr

rv

r

sincoscoscossin,cossinsinsincos( 22 rrrrrrrrrr ra

or

)sincoscos2sin,cossinsin2cos( 22 rrrrrrrr r or

jrrrrirrrr ˆ)sincoscos2sin(ˆ)cossinsin2cos( 22 r

ˆ)2(ˆ)()ˆcosˆsin)(2()ˆsinˆ)(cos( 22 rrrrrjirrjirr r ((Note-2))

)sin,(cos re . )cos,sin( e

ee )cos,sin( r , ree )sin,(cos

er r

The velocity vector:

eeeerv rrrr rrr

The acceleration vector:

ee

eeeee

eeeee

ra

) 2()- (

)(

2

rrrr

rrrrr

rrrrr

r

rr

rr

((Mathematica)) Clear"Global`";

R rt Cost, rt Sint;

V DR, t FullSimplify

Cost rt rt Sint t,

Sint rt Cost rt tA DR, t, 2 FullSimplify

Cost rt t2 rt Sint 2 rt t rt t,

2 Cost rt t Sint rt t2 rt Cost rt t

er Cost, Sint;

e Sint, Cost;

A.er Simplify

rt t2 rtA.e Simplify

2 rt t rt tV.er Simplify

rtV.e Simplify

rt t 1.2 Circular motion (r = constant)

We consider a circular motion with r = constant. since 0r and 0r .

raa

ra

t

r

2

rvv

v

t

r

0

In summary, we have

vrvv t

dt

dvraa

r

vra

t

r

2

2

1.3 Angular velocity and angular acceleration

The case when the point P rotates about the origin in a circle of radius r (= constant). As the particle moves, the only coordinates that changes is . As the particle moves through , it moves through an arc length s.

dt

dvr

dt

dr

dt

sds

vrdt

dr

dt

dss

rdds

2

2

2

2

Here we define the angular velocity and angular acceleration as

dt

d angular velocity (rad/s)

2

2

dt

d

dt

d angular acceleration (rad/s2)

Then we have

rr

va

rdt

sda

rvdt

dsv

r2

2

2

2

1.4 Summary Rotational variables

Angular displacement displacement x Angular velocity velocity v Angular acceleration acceleration a

Average angular velocity tavg

Instantaneous angular velocity dt

d

Average angular acceleration tavg

Instantaneous angular acceleration 2

2

dt

d

dt

d

The direction of

The direction of is along the rotation axis. The sense of is defined by the right-hand rule. The thumb of the right hand gives the sense of .

2 The case of = constant

2

2

dt

d

dt

d = const

We solve this differential equation with the initial conditions:

0

0

)0(

)0(

The solution is as follows.

)(2

2

1

02

02

200

0

tt

t

Problem 10-13** (SP-10 Hint) (10-th edition)

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest, (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions? ((Solution)) 1 rev = 2 (rad)

0

/5.1

f

i srad

8024040 rev (rad)

is constant.

)(2

2

1

22

2

ifif

iif

if

tt

t

3 Kinetic energy of rotation

We assume that a particle (mass m) rotates around the z axis with an angular velocity .

The kinetic energy of the particle is given by

2222

2

1)(

2

1

2

1 mrrmmvKR

where r is distance perpendicular to the axis of rotation.

Now we consider an object as a collection of particles and assume that it rotates about a fixed axis (z-axis) with an angular velocity . Figure shows the rotating object and identifies particle element (mi) on the object located at a distance ri from the rotation axis.

The total kinetic energy of the rotating object:

2222 )(2

1)(

2

1

2

1 i

iii

iii

iiR rmrmvmK

Rotational inertia (moment of inertia) I:

i

iirmI 2

The rotational kinetic energy: KR

2

2

1 IKR

Rotational inertia for the continuous body

dmrI 2

(1) From the textbook (Halliday, Resnick, and Walker)

(2) Table from the textbook (Serway and Jewett)

4. Calculation of moment of inertia 4.1 Example

Two balls with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axes passes through the center of mass.

Show that this moment of inertia is I = L2, where Mm

mM

.

((Solution))

The moment of inertia is given by

22 )( xLmMxI The derivative of I with respect to x has a local minimum

22min LL

Mm

mMI

at

mM

mL

m

ML

x

1

((Mathematica))

Clear"Global`"; I1 m L x2 M x2;

rule1 x L y, M m ;

I2 I1 m L2 . rule1 Expand;

eq1 SolveDI2, y 0, yy

1

1

I22 I2 . eq11 Simplify

1

PlotEvaluateTableI2, , 0, 1, 0.05, y, 0, 1,

PlotStyle TableHue0.1 i, Thick, i, 0, 10,

AxesLabel "xL", "1

m L2", Background LightGray

0.2 0.4 0.6 0.8 1.0xL

0.2

0.4

0.6

0.8

1.0

1

m L2

Fig. Plot of I/(mL2) as a function of x/L, where = M/m is changed between 0 and 1

( = 0.05). 4.2 Thin rod about axis through center perpendicular to length

We assume that the total mass of the rod (length L) is M. Then the mass between x and x+dx is

dxL

Mdm

The moment of inertia for thin rod is

12

]3

1[

22

2

2/0

32/

0

22/

2/

22/

2/

22/

2/

2

ML

xL

Mdxx

L

Mdxx

L

Mdx

L

MxdmxI L

LL

L

L

L

L

L

4.3 The square about perpendicular axis through center

2/

0

2/

0

2/

0

222/

0

2222

4

)(4))(

aa

aa

dydx

dyyxdx

dxdy

dxdyyx

dxdy

dxdyr

dm

dmr

or

24

22/

0

33

22

32

2/

0

2

2/0

322/

0

6

1)

24(

4]

2432[

4

)2

(

)83

1

2(

)2

(

]3

1[

aa

ax

axa

aa

aaxdx

a

yyxdxa

aa

a

Therefore the moment of inertia I is given by

2

6

1MaMI

((Mathematica))

Clear"Global`";

IntegrateIntegratex2 y2, y, 0, a 2,

x, 0, a 2a4

24 4.4 Solid cylinder

Moment of inertia for the solid cylinder (radius R, height h) (or disk) about central axis:

2

2

1MRIz

The mass: hRM 2 The moment of Inertia is given by

22

44

0

322

2

1

2422 MR

hR

MhRRhdrrhrdrdrhdmrI

R

4.5 Annular cylinder

The mass: hRRM )( 21

22

The moment of inertia is given by

)(2

)(

2

)(2

)(

][4

22

22

212

12

2

41

42

21

22

41

42

43

22

2

1

2

1

RRM

RR

RRM

RRh

MRRh

Rh

drrh

rdrdrhdmrI

RR

R

R

4.6 Solid sphere (a) Method-1

Moment of inertia for the solid sphere about any diameter: 2

5

2MRIz

dzzRR

MdzzR

R

MdzrrdMdIz

2223

222

3

42 )(8

3)(

3

42

1

2

1)(

2

1

where dzrdM 2 and 3

3

4RM

Then we have

253

0

2223

2223 5

2

15

8

4

3)(

4

3)(

8

3MRR

R

MdzzR

R

MdzzR

R

MI

RR

R

z

(b) Method-2: direct calculation

3

3

4Rdm

where is the density of the sphere.

22)2( 322 dxdzxdxdzxxdmr

where

RR

R

R

R

zRzRR

R

zRdzzRdzxdzdxxdzdxdzx0

222

_

222

_

04

0

3

_

3 )(2

1)(

4

1]

4

1[

22

22

or

50

5324

0

4224

15

4]

5

1

3

12[

2

1)2(

2

1RzzRzRzzRRdz R

R

Then we have

2

3

52

5

2

3

415

42

RR

R

dm

dmr

M

I

or

2

5

2MRI

((Mathematica))

Clear"Global`";

IntegrateIntegratex3, x, 0, R2 z2 ,

z, R, R4 R5

15 4.8 Sphere (using the spherical co-ordinate)

x

y

z

q

dq

f df

r

drr cosq

r sinq

r sinq df

rdq

er

ef

eq

The moment of inertia I for the sphere can be calculated as

2

5

3

2

00

3

0

43

3

222

5

23

8

54

3

sin4

3

34

)sin(sin

R

R

R

dddrrR

R

ddrdrr

M

I

R

with the use of the spherical co-ordinate.

3

4sin

0

3

d

4.8 Solid cone

Show that the moment of inertia of a uniform solid cone about an axis through its center is given by I = (3/10)MR2. The cone has mass M and altitude h. The radius of its circular base is R.

The density of the cone is

hR

M

hR

M

V

M2

2

3

3

1

From the geometry, we have

R

r

h

z or z

h

Rr

The moment of inertia;

dzzh

MR

dzzh

R

hR

Mdzr

hR

Mdzr

hR

MrdzrrdmdI

45

2

44

4

24

24

2222

2

3

2

3

2

33

2

1)(

2

1)(

2

1

or

25

5

2

0

45

2

10

3

52

3

2

3MR

h

h

MRdzz

h

MRI

h

4.9 Moment of inertia for spherical shell

The mass per unit area is given by

24 R

M

The moment of inertia for the spherical shell is

22

0

342

0

2

3

2

3

4

2sin2

4)sin2()sin( MR

MRdR

R

MRdRRI

5 Parallel-axis theorem

The moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the center of mass, plus the mass of the body times the square of the distance between the two axes.

2MdII CM

We consider the rotation axis passing through the origin, and the axis passing through the center of mass. This axis is parallel to the rotation axis. We consider a plane which is perpendicular the rotation axis. The projection of the center of mass on this plane is denoted by CM’ (red dot). In the above figure, the vectors R, r, and d are on this plane.

drR The moment of inertia around the origin is

dmdmdmdmdmI )(2)()2[( 2222222 drrddrdrdrR

Note that

0dmr

From the definition of d,

0)(

dmdm

dmdm

dm

dm

rdR

dR

Rd

Then we have

222 )( MdIdmI CM dr

where ICM is the moment of inertia around the center of the mass. ((Note)) Perpendicular axis theorem

The perpendicular axis theorem for planar objects can be demonstrated by looking at the contribution to the three axis moments of inertia from an arbitrary mass element. From the point mass moment, the contributions to each of the axis moments of inertia are

dmyxI

dmxI

dmyI

z

y

x

)( 22

2

2

Then we have

yxz III

for planar object.

((Example)) (a) Circular disk (mass M and radius R)

Ix = Iy = 2

4

1MR

Iz = 2

2

1MR

(b) Circular hoop (mass M and radius R)

Ix = Iy = 2

2

1MR

Iz = 2MR 6 Torque The torque is defined by

Frτ

In this figure

dF The moment arm of F: d: the component of r perpendicular to the force F. It is called The line of action of F: the component of r parallel to the force F. The torque is a vector. The direction of the torque vector is perpendicular to the plane formed of r and F.

With the choice that counterclockwise torques are positive and clockwise torques are negative.

7 Newton’s second law of rotation

When r = constant,

vrv

raa

ra

t

r

2

Torque

zrmazrFFrFrr r ˆˆ)ˆˆ()ˆ( Frτ

Since

ra

we have

IImrz 2 8. Example 8.1 Example-1: Angular acceleration of a solid cylinder subject

We consider a solid cylinder, free to rotate about a fixed axis coinciding with its geometric axis, subject to an applied torque. The torque is provided by a mass hanging from a string wrapped around the cylinder.

The moment of inertia for the cylinder is

2

2

1MRI

The free-body diagram

maTmg

Ra

RTI

From these equations, we have

2

Mm

mga

,

R

gM

m

m

2

8.2 ((Example 2)) Serway 10-61

A long uniform rod of length L and mass M is pivoted about a horizontal frictionless pin through one end. The rod is released from rest in a vertical position. At the instant the rod is horizontal, find (a) its angular velocity, (b) the magnitude of its angular acceleration of its center of mass, (c) the x and y components of the acceleration of its center of mass, and (d) the components of the reaction force at the point.

((Solution)) The moment of inertia of the rod around the pin,

222 )2

(12

1

3

1 LMMLMLI

from the parallel axis theorem. The energy conservation law:

UKE R where KR is the rotational energy and U is the potential energy

2

2

12

IE

LMgE

f

i

or

if EE L

g32

(a) The angular velocity : L

g3

(b) The angular acceleration L

g

2

3

L

g

ML

LMg

LMgI

2

3

3

12

2

2

(c) Acceleration of the center of mass along the (-y) direction.

4

3

2

gLat

(d) Centripetal acceleration of the center of mass:

2

3

2

3

22 gL

L

gLar

(e) The component of the reaction force along the (-y axis)

4

4

3

1

1

MgF

MgMaFMg t

(f) The component of the reaction force toward the center

2

32

MgMaF r

8.3 Example Example from Tipler and Mosca Chapter 9 problems 9-96 and 9-97

A uniform cylinder of mass M and radius R is at rest on a block of mass m, which in turn rests on a horizontal, frictionless table. If a horizontal force F is applied to the block,

it accelerates and the cylinder rolls without slipping. (a) Find the acceleration of the block. (b) Find the angular acceleration of the cylinder. Is the cylinder rotating clockwise or counterclockwise? (c) What is the cylinder's linear acceleration relative to the table? Let the direction of F be the positive direction. (d) What is the linear acceleration relative to the block

((Solution)) We draw the two free body diagrams.

((Solution)) The moment of inertia for the cylinder is

2

2

1MRI for cylinder

We note that

BGBGcBcG xRxxx

where G denotes the ground. This means that

BC aRa

For the cylinder, we have

MgNC

CMaf

fRI or MRf2

1

For the block, we have

mgNN CB

BmafF Then we have

MgNC , gmMNB )( ,

mM

Faa CB 3

33

,

mM

FaC 3

,

mM

F

R 3

21

mM

FRaa BC 3

2

(a) mM

FaB 3

3

(b) mM

F

R 3

21

. is in the counterclockwise direction.

(c) mM

FaC 3

.

(d) mM

FRaa BC 3

2

; the acceleration of the cylinder relative to the block

9 Work - energy theorem for rotation

ddrFFrdFdsddW )cos(coscossF where

)cos( rF The work done by the rotation

dId

d

dId

dt

d

d

dId

dt

dIdIddW

with

d

d

dt

d

d

d

dt

d

Work-rotational energy theorem

Rif KIIdIdW 22

2

1

2

1

This equation has exactly the same mathematical form as the work-energy theorem for translation. If an object is both rotating and translating, we have

RT KKW 10 Power in rotational motion The power delivered to a rotating object is

dt

d

dt

dWP

This expression is analogous to P = Fv in the case of translation motion. 11 Analogies between translational and rotational motion

x v a

2

2

1 IKR 2

2

1mvKT

I m F=ma

F Power= Power = Fv Work = d Work = F dx

Momentum p = mv Angular momentum L = I12. Homework (Hint) and SP problems 12.1 Problem 10-51 (SP-10) (10-th edition)

In Fig., block 1 has mass m1 = 460 g, block has mass m2 = 500 g, and the pulley, which is mounted on a horizontal axle with negligible friction, has radius R = 5.00 cm. When released from rest, block 2 falls 75.0 cm in 5.00 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T1 and (c) tension T2? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia?

((Solution)) m1 = 0.46 kg, m2 = 0.5 kg, R = 0.05 m, h = 0.75 m, t = 5 sec Free-body diagram

,

Ra

amTgm

TTRR

aII

amgmT

222

12

111

)(

Determination of a:

22 /06.0

2sm

t

ha

2

22

2121

22

11

/2.1

0139.0)]()([

87.4)(

54.4)(

sradR

a

kgma

RmmammgI

NgamT

NgamT

12.2 Problem 10-71 (HW-10, Hint) (10-th edition)

In Fig., two 6.20 kg blocks are connected by a massless string over a pulley of radius 2.40 cm and rotational inertia 7.40 x 10-4 kg m2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the puley’s axis is frictionless. When this system is released from rest, the pulley turns through 1.30 rad in 91.0 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley’s angular acceleration, (b) the magnitude of either block’s acceleration, (c) string tension T1, and (d) string tension T2?

((Solution)) M = 6.2 kg, R = 0.024 m, I = 7.40 x 10-4 kg m2, = 1.30 rad in t = 91 ms.

Determination of the angular acceleration and the acceleration a

2

2

1t

2

2

t

RRa

Equations of motion:

R

aIIRTT )( 21 or

221 R

aITT

MaTMg 1

12.3 Problem 10-93 (10-th edition)

A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.50 kg m2. A massless cord wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 3.0 N is applied to the block as shown in Fig., what is the magnitude of the angular acceleration of the wheel? Assume the cord does not slip on the wheel.

((Solution)) m = 0.05 kg, I = 0.05 kg m2, r = 0.2 m, M = 2.0 kg, P = 3N.

Newton’s second law

MaTPFx

ra

rTI

From these equations,

2

2

2

22

/6.4

1

sradr

aI

MrP

T

IMr

Pr

I

Tra

12.4 Problem 10-67*** (SP-10) (10-th edition)

Figure shows a rigid assembly of a thin hoop (of mass m and radius R = 0.150 m) and a thin radial rod (of mass m and length L = 2.00 R). The assembly is upright, but if we give it a slight nudge, it will rotate around a horizontal axis in the plane of the rod and hoop, through a lower end of the rod. Assuming that the energy given to the assembly in such a nudge is negligible, what would be the assembly’s angular speed about the rotation axis when it passes through the upside-down (inverted) orientation?

((My solution)) R = 0.150 m, L = 2R

22222

2

83.10])2(12

1[]

2

1)3([

)3(

)3(2

1

mRRmmRmRRmI

RRmgE

RRmgIE

i

f

The energy conservation law

)3()3(2

1 2 RRmgRRmgI

or

sradI

Rmg

RmgI

/8.9)16(

)8(2

1 2

((Note)) The moment of inertia for the circular hoop

2

0

2

2

1)(

2)cos(2 MRRd

R

MRI

Appendix A

The expression of the velocity of the particle rotating around the axis

From the above geometry, we have

sinrR

Rr

Then

sinr

tR

t

r

where

dt

d

Taking into account of the direction of the velocity, we get

rωv B. ((Richard Feynman, Rotation in space, Feynman Lectures on Physics)) Is a torque in three dimension a vector?

B1. Definition of vector

We now consider the one coordinate system is rotated by a fixed angle , such that the axis z- and z’- are the same.

We assume that the vector a is given by

YYXXyyxx eaeaeaea ˆˆˆˆ a

yxY

yxX

eee

eee

ˆ)(cosˆ)sin(ˆ

ˆ)(sinˆ)(cosˆ

or

YXy

YXx

eee

eee

ˆ)(cosˆ)(sinˆ

ˆ)sin(ˆ)(cosˆ

Then we have

)ˆcosˆ(sin)ˆsinˆ(cos

ˆˆˆˆ

YXyYXx

yyxxYYXX

eeaeea

eaeaeaea

or

YyxXyx

YYXX

eaaeaa

eaea

ˆ)cossin(ˆ)sincos(

ˆˆ

or

cossin

sincos

yxY

yxX

aaa

aaa

or

y

x

Y

X

a

a

a

a

cossin

sincos

or

Y

X

x

y

a

a

a

a

cossin

sincos

In the three dimensional notation, we have

z

y

x

Z

Y

X

a

a

a

a

a

a

100

0cossin

0sincos

In general, any vector transforms into the new system in the same way as do ax, ay, and az. For convenience, we use the notation

'

'

'

zZ

yY

xX

aa

aa

aa

B2. Torque Torque is defined by

kji

kji

Frτ zyx

zyx FFF

zyx

where

xyz

zxy

yzx

yFxF

xFzF

zFyF

Now we consider how the torque transforms under the rotation of the coordinate by the angle around the z axis. The torque in the new coordinate is

''''''

'''

''' '''

'''

kji

kji

Frτ zyx

zyx FFF

zyx

'''

'''

'''

''

''

''

xyz

zxy

yzx

FyFx

FxFz

FzFy

Using

z

y

x

z

y

x

100

0cossin

0sincos

'

'

'

z

y

x

z

y

x

F

F

F

F

F

F

100

0cossin

0sincos

'

'

'

We can show that

zxyz

yxzxy

yxyzx

FyFx

FxFz

FzFy

'''

'''

'''

''

cossin''

sincos''

or

''

'

'

100

0cossin

0sincos

z

y

x

z

y

x

This means that the torque is a vector. ((Mathematica))

A Cos, Sin, 0,

Sin, Cos, 0, 0, 0, 1Cos, Sin, 0,Sin, Cos, 0, 0, 0, 1

A MatrixForm

Cos Sin 0Sin Cos 0

0 0 1

F Fx, Fy, FzFx, Fy, Fz

R x, y, zx, y, z

Fn A.F

Fx Cos Fy Sin,Fy Cos Fx Sin, Fz

Rn A.R

x Cos y Sin, y Cos x Sin, z

CrossR, FFz y Fy z, Fz x Fx z, Fy x Fx y

n CrossRn, Fn Simplify

Fz y Fy z Cos Fz x Fx z Sin,

Fz x Fx z Cos Fz y Fy z Sin,Fy x Fx y

APPENDIX C

The kinetic rotation energy of the earth The kinetic rotation energy for the earth is given by

2

2

1 IK

where I is the moment of inertia of the earth and w is the angular velocity,

2

5

2EE RMI ,

v

RT E

22 , ERv

ME is the mass of the earth, RE is the radius and T is the period (24 hours). The rotation energy K can be rewritten as

2222 2

5

1

5

1

5

2

2

1

T

RMvMRMK E

EEEE

The value of K can be estimated as

K = 2.56537 x 1029 J. ((Mathematica))

Clear"Global`";

rule1 ME 5.9736 1024, RE 6.372 106 , hour 3600;

K 2

5

1

2ME

2 RE

T1

2 . T1 24 hour . rule1

2.56537 1029

APPENDIX D: How to determine the moment of inertia (experimentally) Walter Lewin; https://www.youtube.com/watch?v=cB8GNQuyMPc The concept of moment of inertia is demonstrated by rolling a series of cylinders down an inclined plane.

Suppose that either hollow cylinder or solid cylinder roll down on the incline with the angle . There is no slipping on the surface of the incline. First we calculate the acceleration using the Newton’s second law for the translation and rotation.

The Torque around the central axis is

CMIfR

where the force f is considered. The Newton’s second law for the movement for the center of mass,

0cos MgN and

fMgMa sin where M is the mass of the cylinder and is the angle of incline. The condition for no slipping is given by

Raa cm

Then we get

aR

I

R

If CMCM

2

The acceleration a is obtained by

21

sin

MR

Ig

aCM

We consider an annular cylinder (hollow cylinder) about the central axis. The inner radius is R1 and the outer radius is R2. The moment of inertia around the central axis is given by

)(2

22

21 RR

MICM

Noting that 2RR , we have the expression for the acceleration as

sin

3

2sin

3

2

1

sin2

22

21

22

gx

g

R

RMR

Ig

aCM

Thus the acceleration a depends only on the ratio 21 / RRx (≤1). It does not depend on the mass (M), the length of cylinder (L), and the kind of materials for the hollow cylinder (such as copper, aluminum, wood, and so on). In other words, a is a universal function of the ratio x.

x R1 R2

a gsin

0.2 0.4 0.6 0.8 1.0

0.55

0.60

0.65

Fig. )sin/( ga vs the rario 21 / RRx for the hollow cylinder. (a) x =0, (R1=0): solid cylinder (or disk) about the central axis

sin3

2ga

(b) x = 1 (R1 = R2): hoop about the central axis

sin2

1ga

Then the acceleration a for the solid cylinder is larger than that for the hoop.