Rotational Form of Newton’s second law
  I
Very similar to other second law.
Motion of Rolling Objects
A rolling object has rotational kinetic energy and translational kinetic energy.
K  Ktrans  K rot
 12 mvCM  12 ICM  2
2
Why does the object roll (and not slide)? Frictional forces exert a torque on the object.
Example 8.13 The acceleration of a rolling ball.
The rotational form of Newton’s second law is
  Ia
The torque on the ball is due to friction
  rf
So
  I
rf  I
I
r
f 
We can use Newton’s second law to find the linear acceleration of the ball. As we usually do,
take the +x-axis is along the incline.
F
x
 max
mg sin  f  ma
Use the expression for the frictional force to find,
mg sin  f  ma
I
mg sin 
 ma
r
But the acceleration of the ball is related to its angular acceleration, a = r.
I
 ma
r
Ia
mg sin   2  ma
r
Ia
mg sin   2  ma
r
mg sin 
a
m  I r2
mg sin  
For a uniform, solid sphere, I = (2/5)MR2 and for a thin ring, I = MR2. Which has the larger
acceleration?
A solid sphere rolls down a hill that has a height h. What is its speed at the bottom?
Use conservation of energy. Since the ball rolls without slipping, the frictional force doesn’t do
any work. Its displacement is zero in the definition W = F r cos .
U1  K1  U 2  K 2
mgy1  0  0  12 mv 2  12 I 2
The translational speed of the ball is related to its rotational speed, v = r.
mgy1  12 mv 2  12 I 2
mgh  12 mv 2  12 I (v / r ) 2
 12 (m  I r 2 )v 2
v
2mgh
m  I r2
For the solid sphere, I = (2/5)MR2
v
2mgh
2mgh
2 gh



2
m I r
m  (2 5)m
(7 5)
This is less than the answer we found when we ignored rolling,
10
7
gh
.
Angular momentum
We introduced the idea of linear momentum in chapter 7. We had
 dp
 F  dt
A similar expression exits for rotational motion
dL
  dt
The net external torque acting on a system is equal to the rate of change of the angular
momentum of the system. The angular momentum
L  I
is the tendency of a rotating object to continue rotating with the same angular speed and in the
same direction. Angular momentum is measured in kg-m2/s.
If the net torque is zero, we have the conservation of angular momentum
L  0
Li  L f
If the rotational inertia of the system changes, its angular speed will change to compensate.
Angular momentum is a vector. (So is the torque!) The direction is given by the righthand rule.
Our seasons are a consequence of the conservation of angular momentum.
We have completed our study of rotational motion. Try to see how it is analogous to (the more
familiar) linear motion. Here is a summary:
Description
Linear
x
x
vx
ax
x
v x ,av 
t
x
v x  lim
t 0 t
v
a x ,av  x
t
v
ax  lim x
t  0 t
position
displacement
Rate of change of position
Rate of change of velocity
Average rate of change of position
Instantaneous rate of change of position
Average rate of change of speed
Instantaneous rate of change of speed
Rotational





 av 
t

  lim
t 0 t

 av 
t

  lim
t 0 t
 f  i  t
v fx  vix  a x t
Equations of uniform acceleration
x  v x t  12 a x (t )
2
  t  12  (t ) 2
x  12 (v fx  vix )t
  12 ( f  i )t
v fx  vix  2ax
 f 2  i 2  2
2
2
Inertia
m
I   mi ri2
Influence that causes acceleration
F
  Fr  F r 
Newton’s second law
Work
Kinetic energy
Momentum
Newton’s second law
Condition for conservation of momentum
Conservation of momentum
F
F
x
 ma x
y
 ma y
i
  I
W  F r cos
W   
mv 2


p  mv
 dp
 F  dt

F
 0
 
p1  p 2
I 2


L  Iω
dL
  dt
1
2
1
2
  0
L1  L2