Chapter 9
Rotation of rigid bodies
9.1
Angular Velocity and Acceleration
Angular quantities. Angular position is most conveniently describe in
terms of radians defined by
2π rad ≡ 360◦ = 1 rev
(9.1)
or
360◦
1
=
rev.
2π
2π
One can convert from one angular unit to another using,
!
"
2π rad
1 =
360◦
!
"
1rev
1 =
2π rad
!
"
360◦
1 =
1rev
1 rad ≡
(9.2)
For example,
180
◦
=
90◦ =
45◦ =
60◦ =
"
2π rad
1
= π rad = rev
180
◦
360
2
"
!
π
1
2π rad
= rad= rev
90◦
360◦
2
4
!
"
2π rad
π
1
45◦
= rad= rev
◦
360
4
8
"
!
π
1
2π
rad
= rad =
rev.
60◦
◦
360
6
12
◦
!
113
(9.3)
CHAPTER 9. ROTATION OF RIGID BODIES
114
Similarly to how linear average (velocity and acceleration)
∆x
∆t
∆v
≡
∆t
vavg ≡
aavg
(9.4)
and instantaneous (velocity and acceleration)
∆x
dx
=
∆t
dt
∆v
dv
d2 x
a ≡ lim
=
= 2
∆t→0 ∆t
dt
dt
v ≡
lim
∆t→0
(9.5)
was defined from linear position replaced, the average angular (velocity and
acceleration)
∆θ
∆t
∆ω
≡
∆t
ωavg ≡
αavg
(9.6)
and the instantaneous angular (velocity and acceleration)
dθ
∆θ
=
∆t
dt
∆ω
dω
d2 θ
α ≡ lim
=
= 2.
∆t→0 ∆t
dt
dt
ω ≡
lim
∆t→0
(9.7)
Example 9.1-9.2. The angular position θ of a 0.36 m-diameter flywheel
is given by
#
$
θ(t) = 2.0 rad/s3 t3 .
(9.8)
a) Find θ, in radians and in degrees, at t1 = 2.0 s and t2 = 5.0 s.
b) Find the total distance (not displacement) that a particle on the flywheel
rim moves over the time interval from t1 = 2.0 s to t2 = 5.0 s.
c) Find the average angular velocity, in rad/s and in rev/min over that interval.
d) Find the instantaneous angular velocity at t1 = 2.0 s and t2 = 5.0 s.
e) Find the average angular acceleration between t1 = 2.0 s and t2 = 5.0 s.
f ) Find the instantaneous angular acceleration at t1 = 2.0 s and t2 = 5.0 s.
115
CHAPTER 9. ROTATION OF RIGID BODIES
a) The angular position in degrees is
θ(2.0 s) =
≈
θ(5.0 s) =
≈
"
360◦
2.0 rad/s (2.0 s) = 16 rad
2πrad
◦
◦
◦
920 = 2 × 360 + 200
!
"
#
360◦
3$
3
2.0 rad/s (5.0 s) = 250 rad
2πrad
◦
◦
◦
14330 = 39 × 360 + 290 .
3$
#
!
3
(9.9)
b) The distance that a particle moves is
s =
d
(θ(t2 ) − θ(t1 ))
2
◦
◦
= (0.18 m) (14000 − 920 )
!
2πrad
360◦
"
= 42 m
(9.10)
c) The average angular velocity is
θ(t2 ) − θ(t1 )
t −t
! 2 ◦1
"!
"
14000 − 920◦
2πrad
=
= 78 rad/s
5.0 s − 2.0 s
360◦
ωavg =
or
ωavg
!
1 rev
= 78 rad/s
2π rad
"!
60 s
1 min
"
= 740 rev/min.
(9.11)
(9.12)
d) The instantaneous velocity is
ω=
or
$
dθ #
= 6.0 rad/s3 t2
dt
#
$
6.0 rad/s3 (2.0 s)2 = 24 rad/s
#
$
ω(5.0 s) = 6.0 rad/s3 (5.0 s)2 = 150 rad/s.
(9.13)
ω(2.0 s) =
(9.14)
e) The average angular acceleration is
ω(5.0 s) − ω(2.0 s)
5.0 s − 2.0 s
126 rad/s
=
= 42 rad/s2 .
3.0s
αavg =
(9.15)
CHAPTER 9. ROTATION OF RIGID BODIES
116
f) The instantaneous acceleration is
α=
or
$
dω #
= 12.0 rad/s3 t
dt
#
$
12.0 rad/s3 (2.0 s) = 24 rad/s2
#
$
α(5.0 s) = 12.0 rad/s3 (5.0 s) = 60 rad/s2 .
(9.16)
α(2.0 s) =
9.2
(9.17)
Rotation with constant angular acceleration
The angular velocity and acceleration can be also thought of as vector quantities pointing along the axis of rotation with direction determined by the
right-hand rule. Then, at the level of magnitudes there is the following equivalence of linear and angular quantities
x → θ
vx → ωz
ax → αz .
(9.18)
and so for motion with constant acceleration we have
a(t) = a → α(t) = αz
v(t) = v0 + ax t → ω(t) = ω0z + αz t
1
1
x = x0 + v0x t + ax t2 → θ(t) = θ0 + ω0z t + αz t2
2
2
(9.19)
as well useful relations
2
2
vx (t)2 = v0x
+ 2ax (x − x0 ) → ωz (t)2 = ω0z
+ 2αz (θ(t) − θ0 )
1
1
x(t) − x0 = (v0x + vx (t)) t → θ(t) − θ0 =
(ω0z + ωz (t)) t. (9.20)
2
2
Example 9.3. You have finished watching a movie Blu-ray and the disk
is slowing to a stop. The disc’s angular velocity at t = 0 is 27.5 rad/s, and
its angular acceleration is a constant −10.0 rad/s2 . A line PQ on the disc’s
surface lies along the +x-axis at t = 0 s.
117
CHAPTER 9. ROTATION OF RIGID BODIES
a) What is the disc’s angular velocity at t = 0.300 s?
b) What angle does the line PQ make with the +x-axis at this time?
a) The disc’s angular velocity is
ωz (t) = ω0z + αz t
#
$
ωz (0.300 s) = 27.5 rad/s + −10.0 rad/s2 (0.300 s) = 24.5 rad/s(9.21)
b) The angular displacement is given by
1
θ(t) = θ0 + ω0z t + αz t2
2
θ(t) = 0 + (27.5 rad/s) (0.300 s) +
and so the line PQ makes angle
$
1#
−10.0 rad/s2 (0.300 s)2 = 7.80(9.22)
rad
2
(7.80 rad − 2π rad )
360◦
= 87◦
2πrad
(9.23)
with +x-axis.
Example. Supposed the disc in Example 9.3 was initially spinning at
twice the rate (i.e. 55.0 rad/s) and slowed down at twice the rate (i.e. −20.0 rad/s2 ).
a) Compared to the situation in Example 9.3 how long would it take the
disc to come to a stop?
(i) the same amount of time;
(ii) twice as much time;
(iii) four times as much time;
(iv) one half as much time;
(v) one fourth as much time
CHAPTER 9. ROTATION OF RIGID BODIES
118
b) Compared to the situation in Example 9.3 through how many revolutions would the disc rotate before coming to a stop?
(i) the same number of revolutions;
(ii) twice as many revolutions;
(iii) four times as many revolutions;
(iv) one half as many revolutions;
(v) one fourth as many revolutions.
9.3
Relating linear and angular kinematics
The linear displacement of a given point P on rotating body at distance
r from the axis of rotation is related to angular displacement through the
following relation
s = rθ,
(9.24)
where θ must be measured in radians.
This suggests that the (magnitude of) linear velocity of the point P is
given by
% %
% %
% ds %
% %
% % = r % dθ %
(9.25)
% dt %
% dt %
CHAPTER 9. ROTATION OF RIGID BODIES
119
or
v = rω.
(9.26)
and the tangential (or parallel) component of linear acceleration is given by
a|| =
dω
dv
=r
= rα.
dt
dt
(9.27)
and the centripetal (or perpendicular) component of linear acceleration is
given by
v2
a⊥ =
= ω 2r.
(9.28)
r
Example 9.4. An athlete whirls a discus in a circle of radius 80.0 cm.
At a certain instant, the athlete is rotating at 10.0 rad/s and the angular
speed is increasing at 50.0 rad/s2 . At this instant, find the tangential and
centripetal components of the acceleration of the discus and the magnitude of
the acceleration.
Components of linear acceleration are given by
#
$
a|| = rα = (0.800 m) 50.0 rad/s2 = 40.0 m/s
a⊥ = ω 2r = (10.0 rad/s)2 (0.800 m) = 80.0 m/s
and so the magnitude is
&
&
a = a2|| + a2⊥ = (40.0 m/s)2 + (80.0 m/s)2 = 89.4 m/s.
(9.29)
(9.30)
Example 9.5. You are designing an airplane propeller that is to turn
at 2400 rpm (or revolutions per minute). The forward airspeed of the plane
is to be 75.0 m/s, and the speed of the tips of the propeller blades through the
air must not exceed 270 m/s. (This is about 80% of the speed of the sound
in air. If the speed of the propeller tips were greater than this, they would
produce a lot of noise.) (a) What is the maximum possible propeller radius?
(b) With this radius, what is the acceleration of the propeller tip?
CHAPTER 9. ROTATION OF RIGID BODIES
120
(a) The tangential component of the velocity of the tip must at most
&
2
2
vmax = vtip
− vplane
(9.31)
which is also given by
v = rω
and thus
rmax ω =
or
rmax =
=
&
2
2
vtip
− vplane
(9.32)
(9.33)
&
2
2
vtip
− vplane
ω&
(270 m/s)2 − (75.0 m/s)2
(2400 rev/min) (1 min/60 s) (2π rad/1 rev )
259 m/s
= 1.03 m.
=
251 rad
(9.34)
(b) The centripetal acceleration of the tip is
a⊥ = ω 2 rmax
= (251 rad)2 (1.03 m) = 6.5 × 10 m/s2 = 6600 g.
(9.35)
Example. Information is stored on a disc in a coded pattern of tiny pits.
The pits are arranged in a track that spirals outwards towards the rim of the
disc. As the disc spins inside a player, the track is scanned at a constant
linear speed. How must the rotation speed of the disc change as the player’s
scanning head moves over the track?
(i) The rotation speed must increase.
(ii) The rotation speed must decrease.
(iii) The rotation speed must stay the same.
121
CHAPTER 9. ROTATION OF RIGID BODIES
9.4
Energy in Rotational Motion
When a rigid body rotates around a fixed axis different parts
of it move with different linear velocities and so have different
kinetic energies energies.
1
1
Ki = mi vi2 = mi ri2 ω 2 .
2
2
(9.36)
The total (rotational) kinetic energy is then
'
K =
Ki
i
'1
mi ri2 ω 2
2
)
1 ('
mi ri2 ω 2
=
2
1 2
=
Iω
2
=
where
I≡
'
mi ri2
(9.37)
(9.38)
is known as the moment of inertia. Note that the rotational kinetic energy looks very similar to the translational kinetic energy
1 2
Iω
2
↔
1 2
mv .
2
(9.39)
CHAPTER 9. ROTATION OF RIGID BODIES
122
Example 9.6. A machine part consists of three discs linked by light-weight
struts.
(a) What is this body’s moment of inertia about an axis through the center
of the disc A, perpendicular to the plane of the diagram?
(b) What is its moment of inertia about an axis through the centers of disks
B and C?
(c) What is the body’s kinetic energy if it rotates about the axis through A
with angular speed ω = 4.0 rad/s?
a) The moment of inertia about an axis through the center of disk A is
2
2
2
IA = mA rA|A
+ mB rB|A
+ mC rC|A
= (0.30 kg) (0)2 + (0.10 kg) (0.50 m)2 + (0.20 kg) (0.40 m)2 = 0.057 kg
(9.40)
· m2
b) The moment of inertia about an axis through the center of disk B is
2
2
2
IBC = mA rA|BC
+ mB rB|BC
+ mC rC|BC
= (0.30 kg) (0.4 m)2 + (0.10 kg) (0)2 + (0.20 kg) (0)2 = 0.048 kg(9.41)
· m2
c) The kinetic energy is
$
1
1#
KA = I A ω 2 =
0.057 kg · m2 (4.0 rad/s)2 .
2
2
(9.42)
Form more complicated objects the moment of inertia must be obtained
by volume integration. Here we summarize some useful results
CHAPTER 9. ROTATION OF RIGID BODIES
123
Example 9.7. We wrap a light, nonstretching cable around a solid cylinder of mass 50 kg and diameter 0.120 m, which rotates in frictionless bearings
about a stationary horizontal axis. We pull the free end of the cable with a
constant 9.0 − N force for a distance of 2.0 m; it turns the cylinder as it unwinds without slipping. The cylinder is initially at rest. Find its final angular
speed and the final speed of the cable.
The total work done by the pulling force is
W = F s = (9.0 N) (2.0 m) = 18 J.
CHAPTER 9. ROTATION OF RIGID BODIES
124
According to the work-energy theorem
where
W = Kf − Ki
1 2
Iω − 0
W =
2 f
*
2W
ωf =
I
1
1
I = mR2 = (50kg) (0.060 m)2 = 0.090 kg·m2
2
2
and thus the final angular speed of the cylinder is
+
2 · 18 J
ωf =
= 20 rad/s.
0.090 kg·m2
and the final linear speed of the cable is
vf = ωf R = (20 rad/s) (0.060 m) = 1.2 m/s.
Example 9.8. We wrap a light, non-stretching cable around a solid
cylinder with mass M and radius R. The cylinder rotates with negligible
friction about a stationary horizontal axis. We tie the free and of the cable
to a block of mass m and release the block from rest at distance h above the
floor. As the block falls, the cable unwinds without stretching or slipping.
Find expression for the speed of the cylinder as the block strikes the floor.
CHAPTER 9. ROTATION OF RIGID BODIES
125
Since there are no other (non-gravitational) forces acting on the system
the total (mechanical) energy must be conserved
1
1
mgh = mv 2 + Iω 2
2
2
(9.43)
where
v
R
and the moment of inertia for the disc is
ω=
1
I = MR2 .
2
By plugging Eqs. (9.44) and (9.45) into (9.43) we get
!
"( )
1 2 1 1
v 2
2
mv +
MR
mgh =
2
2 2
R
!
"
m M
mgh =
+
v2
2
4
or
v=
9.5
+
mgh
.
+ M4
m
2
(9.44)
(9.45)
(9.46)
(9.47)
Parallel-Axis Theorem
There are (infinitely) many axis passing through a given object, and thus
there are infinitely many moments of inertia. However, some of them are
related to each other. Consider two axis of rotation:
• axis O: passing through the center of mass and
• axis D: parallel to the first one, but displaced by distance d.
Let’s choose z-axis to point along the direction of the first axis and with
origin at the center of mass. Then the O axis of rotation has coordinates
(x, y, z) = (0, 0, z)
(9.48)
and the D axis of rotation has coordinates
(x, y, z) = (a, b, z)
where
d=
√
a2 + b2 .
(9.49)
(9.50)
CHAPTER 9. ROTATION OF RIGID BODIES
126
Then the two moments of inertia are given by
' #
$
IO =
mi x2i + yi2
(9.51)
i
and
ID =
'
i
#
$
mi (xi − a)2 + (yi − b)2 .
(9.52)
The second equation can be expanded as
' #
$
ID =
mi (xi − a)2 + (yi − b)2
(9.53)
i
=
'
i
=
'
i
#
$
mi x2i − 2xi a + a2 + yi2 − 2yi b + b2
'
'
#
$
#
$'
mi x2i + yi2 − 2a
mi xi − 2a
mi xi + a2 + b2
mi
i
i
i
The first term is the moment of inertia around center of mass (9.51), the
second and third terms a proportional to center of mass and thus vanish
in the chosen coordinate system, and the forth term can be rewritten using
total mass
'
M=
mi
(9.54)
i
and (9.50). The resulting expression is known as the parallel-axis theorem:
ID = IO + Md2 .
(9.55)
Example 9.9. A part of mechanical linkage has a mass of m = 3.6 kg.
Its moment of inertia IP about an axis l = 0.15 m from its center of mass is
IP = 0.132 kg · m2 . What is the moment of inertia Icm about a parallel axis
through the center of mass?
From the parallel-axis theorem we get
Icm = Ip − ml2
= 0.132 kg · m2 − (3.6 kg) (0.15 m)2 = 0.051 kg · m2
(9.56)
CHAPTER 9. ROTATION OF RIGID BODIES
9.6
127
Moment-of-Inertia Calculations
In general moment-of-inertia is defined by replacing summation with integration, i.e.
,
'
I=
r 2 mi
→
I = r 2 dm.
(9.57)
i
In the case when the density (or mass per unit volume)
dm
dV
is independent of position the moment of inertia
,
I = r 2 ρdV
ρ=
or
I=
,
r 2 ρ dx dy dz
(9.58)
(9.59)
(9.60)
where the limits of integration are set by the geometry of the body.
Example 9.10. Consider a hollow cylinder of uniform mass density ρ
with length L, inner radius R1 , and outer radius R2 . Find its moment of
inertia about its axis of symmetry.
Integration on the x-y plane is most convenient done in polar coordinates
x(r, θ) = r cos θ
y(r, θ) = r sin θ
(9.61)
CHAPTER 9. ROTATION OF RIGID BODIES
and so one needs to multiply by determinant of Jacobian
! ∂x ∂y "
∂r
∂r
drdθ
dxdy = det ∂x
∂y
∂θ
∂θ
!
"
cos θ
sin θ
= det
drdθ
−r sin θ r cos θ
#
$
= r cos2 θ + r sin2 θ drdθ
= rdrdθ.
128
(9.62)
Then the moment of inertia is given by
, L , R2 , 2π
I =
r 2 ρ (rdrdθdz)
0
R1
= 2πρL
,
0
R2
r 3 dr
R1
R14 R24
= 2πρL
−
4
4
#
$
πρL 4
R1 − R24 .
=
2
!
"
(9.63)
Similarly on can calculate the volume of the cylinder
, L , R2 , 2π
V =
r 2 ρ (rdrdθdz)
0
= 2πL
R1
0
R2
,
rdr
R1
and so the total mass
"
! 2
R1 R22
−
= 2πL
2
2
# 2
$
2
= πL R1 − R2
#
$
M = V ρ = πLρ R12 − R22 .
(9.64)
(9.65)
From (9.63) and (9.65) we get
$#
$
πρL # 2
R1 − R22 R12 + R22
2
$
1 # 2
=
M R1 + R22 .
2
I =
(9.66)
CHAPTER 9. ROTATION OF RIGID BODIES
129
Example 9.11. Find the moment of inertia of a solid sphere of uniform
mass density ρ (like a billiard ball) about an axis through its center.
Then the moment of inertia is given by
I =
,
R
−R
=
=
=
2π
0
,
R
,
,
√
0
√
R2 −z 2
r 2 ρ (rdrdθdz)
R2 −z 2
r 3 drdz
−R 0
.
, R- 2
2
(R − z 2 )
2πρ
dz
4
−R
,
$
πρ R # 4
R − 2R2 z 2 + z 4 dz
2 −R
!
! 3"
"
πρ
2R
2R5
4
R (2R) − 2R
+
2
3
5
5
8πρR
.
15
= 2πρ
=
,
(9.67)