1. Rotational Kinematics Examples
1. A bowling ball of mass 7.0 kg and radius 10.5 cm is rolled down an
alley 20.0 m long. After release, the ball hits the pins in a time of 3.6 s.
Assume the roll of the ball is constant.
(a) Find the rotational inertia of the ball.
m = 7.0 kg r = 10.5 cm d = m t = 3.6 s
rotational inertia of a sphere = 2/5 mr2
(from Figure 8-20 on p. 223)
I = 2/5 mr2 = 2/5 ( 7.0 kg )( m )2
I = kg m2

2. (b) Find the translational and angular velocities of the ball.
m = 7.0 kg r = 10.5 cm d = m t = 3.6 s
v = ? and ω = ? d
20.0 m
v = =
= v = m/s t
3.6 s
v = r ω r r v
5.56 m/s
ω = =
= ω = rad/s r
0.105 m

3. (c) Find the total kinetic energy of the ball.
m = 7.0 kg r = 10.5 cm d = m t = 3.6 s
v = m/s
ω = rad/s I = kg m2
Total KE = Translational KE + Rotational KE
Translational KE = ½ m v2
= ½ ( 7.0 kg )( 5.56 m/s )2
= J
Rotational KE = ½ I ω2
= ½ ( kg m2 )( 52.9 rad/s )2
= J
Total KE = J

4. 2. A baseball can be considered a solid sphere of radius 3.7 cm. Let’s
say a baseball is thrown so that the translational speed is 31.0 m/s and
its angular velocity (about its center of mass) is 180 rad/s. What
percent of the ball’s kinetic energy is rotational KE?
r = 3.7 cm v = m/s ω = 180 rad/s
Translational KE = KEt = ½ m v2
Rotational KE = KEr = ½ I ω2
part
Rotational KE
x 100%
percent =
x 100% =
whole
Total KE
KEr
x 100% =
KEt + KEr

5. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
KEr
x 100%
percent =
KEt + KEr
Seeming obstacle: we do not know the mass of
the ball
Strategy:
Assign a mass:
Let m = 2.0 kg
and then complete the problem

6. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
m = 2.0 kg
KEr
x 100%
percent =
KEt + KEr
KEt = ½ ( 2.0 kg )( 31.0 m/s )2
= J
KEr = ½ I ω2
need to calculate I
Isphere = 2/5 m r2 = 2/5 ( 2.0 kg )( m )2
= kg m2

7. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
m = 2.0 kg
I = kg m2
KEr
x 100%
percent =
KEt + KEr
KEt = ½ ( 2.0 kg )( 31.0 m/s )2
= J
KEr = ½ I ω2
= ½ ( kg m2 )( 180 rad/s )2
= J

8. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = J
KEr = J
m = 2.0 kg
I = kg m2
KEr
x 100%
percent =
KEt + KEr
17.7 J
x 100% =
( ) J
= x 100 %
= %

9. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
KEr
x 100%
percent =
KEt + KEr
Would the answer be different if the mass were
different (but all other values the same)?
Actually, no; the answer will be the same
To show this, the calculations will be done again,
but mass will be a general value m

10. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
m = m
KEr
x 100%
percent =
KEt + KEr
KEt = ½ ( m )( 31.0 m/s )2
= ( m ) J
KEr = ½ I ω2
need to calculate I
Isphere = 2/5 m r2 = 2/5 ( m )( m )2
= ( m ) kg m2

11. 2. r = 3.7 cm v = m/s ω = 180 rad/s
KEt = ½ m v2
KEr = ½ I ω2
m = m
I = m
KEr
x 100%
percent =
KEt + KEr
KEt = ½ ( m )( 31.0 m/s )2
= ( m ) J
KEr = ½ I ω2
= ½ ( m kg m2 )( 180 rad/s )2
= ( 8.87 m ) J

12. KEt = ( m ) J
KEr = ( 8.87 m ) J
KEr
x 100%
percent =
KEt + KEr
for clarity, will ignore units
8.87 m
x 100%
percent =
480.5 m m
8.87
8.87 m
x 100%
x 100% =
percent =
489.37 m
same result
as before
= x 100%
= 1.81%

13. 3. A ball of radius 8.0 cm and mass 3.5 kg is rolling across the floor
towards a ramp. It rolls onto the ramp up to a point, where it stops
and begins to roll back. In terms of the ball’s velocity v and g, how
high above the floor will the center of mass be when the ball
momentarily stops at its highest point on the ramp?
r = 8.0 cm m = 3.5 kg v h
Use energy arguments
KEt = ½ mv2
PEtop = KEbottom
= KEt + KEr
KEr = ½ Iω2
mgh = ½ mv2 + ½ Iω2
I = ? ω = ?

14. 3. r = 8.0 cm m = 3.5 kg v h
mgh = ½ mv2 + ½ Iω2
I = ? ω = ? v
For sphere, I = 2/5 mr2
v = r ω
ω = r
KEr = ½ Iω2 = ½ ( 2/5 mr2)(v/r)2
= ½ ( 2/5 ) mr2 (v2/r2 )
KEr = 1/5 m v2
radius of ball doesn’t matter;
(independent of radius)

15. 3. r = 8.0 cm m = 3.5 kg v h mgh = ½ mv2 + ½ Iω2 KEt = ½ mv2
= ½ mv mv2
KEr = ½ Iω2
1/5
= mv mv2
KEr = mv2
5/10
2/10
1/5
mgh = mv2
7/10
gh = v2
(independent of mass)
7/10
same height for small marble or large bowling ball, if they are traveling at the same initial speed
h = 7v2 / 10g

16. with its lower end attached to a frictionless axis that is mounted on
4. A thin uniform rod is initially positioned in the vertical direction,
with its lower end attached to a frictionless axis that is mounted on
the floor. The rod has a length of 2.00 m and is allowed to fall,
starting from rest. Find the tangential speed of the free end of the rod,
just before the rod hits the floor after rotating through 90o.
L = m
v = speed of center
of mass of rod
vt = tangential speed
of end of rod
Use energy arguments x
PEtop = KEbottom h x
vt = ?
= KEt + KEr v 1
see p. 223:
I = mL2
mgh = ½ mv2 + ½ Iω2 3

17. L = 2r ( v = vcenter of mass )
v = speed of center
of mass of rod
vt = tangential speed
of end of rod
mgh = ½ mv2 + ½ Iω2 1
I = mL2
ω = v/r 3 x h 1
mgh = ½ mv2 +
½ ( mL2)(v/r)2 x
vt = ? 3 v
L = 2r ( v = vcenter of mass ) 1
mgh = ½ mv2 +
½ ( m(2r)2)(v/r)2 3
mgh = ½ mv2 +
½ ( 4/3 m r2)(v/r)2 2
mgh = ½ mv mv2 3 7
gh = v2 6

18. ; speed of the 4. L = 2.00 m v = speed of center of mass of rod
vt = tangential speed
of end of rod 7
gh = v2 6 x h
v2 = gh 6 x 7
vt = ? v
v = gh 6 7
v = (9.8 m/s2)(1.00 m) 6
= v = m/s 7
; speed of the
end of the rod is greater
This is the speed of the center of mass of the rod
What is the same for the center of the rod and the end
is the rotational speed ω

19. 4. L = m
v = speed of center
of mass of rod
vt = tangential speed
of end of rod v
ω = x r h x
vt = ?
2.90 m/s v
ω =
1.00 m
v = m/s
ω = rad/s
For end of rod, L = 2.00 m
v = r ω
vt = L ω
= ( 2.00 m )( 2.90 rad/s )
vt = m/s