chapter 10 rotation - bingwebbingweb.binghamton.edu/~suzuki/generalphysnote_pdf/ln10v... · 2017....
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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.