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Chap. 9: Rotational
Motion
1
I. Rotational Kinematics II. Rotational Dynamics - Newton’s Law for
Rotation III. Angular Momentum Conservation (Chap.
10)
Rotational Motion
Rotational Motion I. Rotational Kinematics
II. Rotational Dynamics - Netwton’s
Law for Rotation
III.Angular Momentum Conservation
2
Rotational Motion
1. Remember how Newton’s Laws for
translational motion were studied:
1. Kinematics (x = x0 + v0 t + ½ a t 2 )
2. Dynamics (F = m a)
3. Momentum Conservation
2. Now, we repeat them again, but for
rotational motion:
1. Kinematics (q, w, a)
2. Dynamics (t = I a)
3. Angular Momentum
OUTLOOK
3
Rotational Motion
Newton’s Laws for Rotation
at
In e t
1st part
[s–2]
3rd part
[N m]
2nd part
[kg m2]
Krot = (1/2) I w2 (K = (1/2) m v2)
5
Rotational Motion
Angular Quantities
Kinematical variables to describe the rotational motion:
Angular position, velocity and acceleration
6
Rotational Motion
Angular Quantities
Kinematical variables to describe the rotational motion:
Angular position, velocity and acceleration
)rad/s(lim
)rad/s(lim
)rad(
2
0
0
dt
d
ave
t
dt
d
ave
t
R
l
t
t
w
a
wa
q
w
qw
q
q = q2 – q1
l = arc of length
7
Rotational Motion
Vector Nature of
Angular Quantities Kinematical variables to describe the rotational motion:
Angular position, velocity and acceleration
c.w. or c.c.w. rotation (like +x or –x direction in 1D)
Vector natures!
)rad/s(
)rad/s(
)rad(
2 k̂
dt
dk̂
k̂ dt
dk̂
R
l
wa
qw
q
x
y
z
R.-H. Rule
>0 or <0
8
Rotational Motion
Vector Nature of
Angular Quantities Kinematical variables to describe the rotational motion:
Angular position, velocity and acceleration
c.w. or c.c.w. rotation (like +x or –x direction in 1D)
Vector natures!
9
Rotational Motion
Example 1 (a = constant) A bicycle wheel has an initial angular velocity of 1.50 rad/s. Its
angular acceleration is constant and equal to 0.300 rad/s2.
a) What is its angular velocity at t = 2.50 s?
b) Through what angle has the wheel turned between t = 0 and
t = 2.50 s? y
x
12
Rotational Motion
Example 1 (a = constant) A bicycle wheel has an initial angular velocity of 1.50 rad/s. Its
angular acceleration is constant and equal to 0.300 rad/s2.
a) What is its angular velocity at t = 2.50 s?
b) Through what angle has the wheel turned between t = 0 and
t = 2.50 s?
s.rad 252
s) 502)(srad 3000(srad 501
a)
2
0
.
...
tαωω zzz
rad. 694
s) 502)(srad 3000(s) 502)(srad 501(
21 b)
22
21
2
0
.
....
tαtωθ zz
y
x
13
Rotational Motion
Example 1 (a = constant) A bicycle wheel has an initial angular velocity of 1.50 rad/s. Its
angular acceleration is constant and equal to 0.300 rad/s2.
a) What is its angular velocity at t = 2.50 s?
b) Through what angle has the wheel turned between t = 0 and
t = 2.50 s?
s.rad 252
s) 502)(srad 3000(srad 501
a)
2
0
.
...
tαωω zzz
rad. 694
s) 502)(srad 3000(s) 502)(srad 501(
21 b)
22
21
2
0
.
....
tαtωθ zz
y
x
14
Rotational Motion
What is the resultant angular velocity of
the wheel, as seen by an outside observer?
r a d / s 0.5 0w h e e l w
r a d / s 0.3 5t u r n t a b l e w
turntable
wheel
x
y
Let’s see how move. 15
Rotational Motion
y
z
x : out of page
r a d / s 0.3 5t u r n t a b l e w
r a d / s 0.5 0w h e e l w
t u r n t a b l ew h e e lr e s u l t a n t www
Easy Way: Vector Analysis
18
Rotational Motion
Linear and Angular Quantities
RR
R
Ra
Ra
d
dR
d
da
R
Rl
dωt R
l θ
222
rad
tan
tan
)( (3)
(2)
)(
)(like (1)
ww
a
w
w
w
v
v
v
v
tt
t
t
atan
arad
(Increase/decrease the speed..)
(Maintain the circular motion..)
l = arc of length
19
Rotational Motion
Physics in Throwing a Discus
R = 80 cm
w = 10.0 rad/s
a = 50.0 rad/s2
atan = ?
arad = ? 23
Rotational Motion
Example 1x (a = constant) (1) What is the angular speed of rotation of the Earth?
1 rotation (2p rad) per 24 hours (1440 min)
rotational speed in rotation per minute (rpm) or rad/s
(2) An old 33 1/3 rpm record player starts from rest and
reaches operating speed in 2.00 seconds. Through what
angle did it turn in those 2.00 seconds?
t = 2.00 s, w0 = 0, w = 33 1/3 rpm
a from Eq. (2)
Use Eq. (1) with q0 = 0
(3) A computer hard drive rotates at 5400 rpm. What
angular acceleration will get it up to speed in just 150
revolutions starting from rest?
Use Eq. (3) with q – q0 = 150 rev. = ? rad )(2 ( 3 ) 0
2
0
2 qqaww
t
t t
aww
awqq
0
2
00
(2)
2
1 (1)
26
Rotational Motion
Example 1x (cont’d)
(4) A hard drive reaches 5400 rpm in 3.20 seconds. What
was the average angular speed assuming constant
angular acceleration?
(5) A dentist’s drill accelerates to 1800 rpm in 2.50 seconds.
what is its angular acceleration?
(6) The angular velocity changes from 47.0 rad/s to –47.0
rad/s in 2.00 seconds. What is the angular acceleration?
28
Rotational Motion
Example 2
A grinding wheel turns at a constant angular acceleration
of 60.0 rad/s2 from 24.0 rad/s for 2.00 sec. Then, a
circuit breaker trips. It turns through 432 rad as it coasts
to a stop at a constant angular acceleration.
Find:
(a) the total angle between t = 0
and the time it stopped;
(b) the time it stopped;
(c) the angular acceleration as it
slowed down.
Also sketch q-t graph.
30
Rotational Motion
Visualization via Graph
432 rad
q1 rad
2 sec T sec
t
q
a = ? rad/s2
a = 60.0 rad/s2
w0 = 24.0 rad/s
?
? 31
Rotational Motion
Problem 3: (25 points)
A grinding wheel turns at a varying angular acceleration of a(t) = [30.0 rad/s3] t for 2.00
sec. Assume the initial angular speed of 20.0 rad/s. Then, a circuit breaker trips. It turns
through 400 rad as it coasts to a stop at a constant angular acceleration.
a. (5 pts) Find the total angle (qtotal) between t = 0 and the time it stopped.
b. (10 pts) Find the time (ttotal) it stopped. Find the angular acceleration as it slowed down.
c. (10 pts) Sketch qt, w-t, a-t graphs.
Practice Problem 1
36
Rotational Motion
Practice Problem 2 A solid cylinder (radius R = 2 m and height H = 5 m) turns at a
constant angular acceleration of 60.0 rad/s2 from 24.0 rad/s for
2.00 sec. Then, a circuit breaker trips. It turns through 432 rad
as it coasts to a stop at a constant angular acceleration.
(a) Find the total angle between t = 0
and the time it stopped.
(b) Find the time it stopped.
(c) Find the angular acceleration
as it slowed down.
(d) Find the speed (v) of point P
at t = 2.00 sec.
(e) Sketch the motion of point P
in v-t graph. Also sketch q-t graph.
H 0.4H
w
37
Rotational Motion
Newton’s Laws for Rotation
at
In e t
1st part
[s–2]
3rd part
[N m]
2nd part
[kg m2]
Krot = (1/2) I w2 (K = (1/2) m v2)
38
Rotational Motion
Rotational Kinetic Energy and Inertia
particles) of group afor particle) singlefor
inertia of(moment inertia of(moment
(b) (a)
2
1)(
2
1
2
1
2
1
2
1
)(2
1 )(
2
1
2
1
222
321
2
3
2
2
2
1
2222
2
ii
2
i
ii
Rm I R mI
RmvmmmvmvmvmK
mRRmmvK
w
ww
m1
m2
m3 39
Rotational Motion
Two blocks with masses m1 = 35.0 kg and m2 = 38.0 kg are
connected by a rope that hangs over a pulley. The pulley is a
uniform disk of radius R0 = 0.300 m and mass M = 4.80 kg.
Initially m1 is on the ground and m2 rests h = 2.50 m above the
ground. Then the system is released. Find the speed of m2 just
before it strikes the ground. Use conservation of mechanical energy.
Assume that the pulley bearing is frictionless and
that the rope does not slip on the pulley rim.
45
M
Example 1
Rotational Motion
Parallel-axis Theorem
d1
2
0
2
0
2
0
2
0
2
0
21
11
32
1
2
1
22
21
MR
RMMR
RMMR
III
d
d
d2
50
Rotational Motion
2(a) Express the moment of inertia of the array of point
objects about the y-axis in terms of m, M, X1, X2, and/or
Y.
X1 X2
Y
52
Rotational Motion
2(b) Consider a helicopter rotor blade as a long thin rod. If
each of the three blades is 3.75 m long and has a mass of
160 kg, calculate the moment of inertia of the three blades
about the axis of the rotation.
53
Rotational Motion
2(c) A meter stick (mass M = 0.500 kg and length L = 1.00 m) is
hung from the wall by means of a metal pin through the hole,
and used as a pendulum. Express the moment of inertia of the
stick about the pin (= the axis of the rotation) in terms of M, L,
and x.
54
Rotational Motion
2(d) A door (solid rectangular thin plate) of mass M = 15.0 kg
is free to rotate on about hinge line. Calculate the moment of
inertia of the door about the hinge line.
H
W
55
Rotational Motion
2(e) A solid disk (mass M = 3.00 kg and radius R = 20.0 cm) is
hung from the wall by means of a metal pin through the hole,
and used as a pendulum. Calculate the moment of inertia of the
disk about the pin (= the axis of the rotation).
56
Rotational Motion
2(f) A ball (solid sphere) of mass M and radius R on the end of
a thin rod (mass m and length l). Express the moment of
inertia of the system of the rod and the ball about the A-B axis
(thin rod; mass m and length l) in terms of M, R, m, and l.
l
m
M R
m
l
Hin
t
57
58
Moment-of-Inertia Calculation
[Key Concept] We divide the rod into many small segments (particles). I Itotal = S[I] Integral
Figure 9.24
Uniform Solid Cone
Moment-of-Inertia Calculation
59
[Key Concept] We divide the cone into many thin, solid disks. I Itotal = S[I] Integral