Lab 8: Rotational Dynamics and Moment of Inertia
Only 4 more to go!!
Rotational motion and linear motion are very similar. Many of the quantities we
discuss in rotation have linear motion counterparts:
Linear Motion
Rotational Counterpart
Position: s, d
Angular position: 
Displacement:  x = xF - xi
Angular displacement:
  = F - i
Velocity: v = x / t
Angular velocity:
= / t
Acceleration: a = v/ t
Angular acceleration:
 =  / t
Mass: m
Moment of inertia: I
Force
Torque
Newton's 2nd law: F = ma
Newtons’s 2nd law:  = I 
We can relate these rotational quantities to those analogues linear ones:
Linear or tangential velocity:
v=r
Tangential or linear acceleration: a = r 
Torque: This is the rotational analogue to a linear force. Torque
causes rotations. Torque is related to force by the following
equation:
 = F r sin 
= F r sin 
sin  = d/r so then torque
becomes:  = F d
We refer to d as the
lever arm.
F
d
r
What if  = 90o? The torque becomes:  = F r

Moment of Inertia is the rotational analogue of mass, m
I (point mass) = Mr2
I (ring) = Mr2
I (disk) = ½ Mr2
What is the moment of inertia for the “dumbells” shown below?
M
r
r
M
I (point mass) = Mr2
Moments of inertia add, so our total I for the two masses is:
I = I1 + I2 = Mr2 + Mr2 = 2Mr2
What is the effect on I if a triple r? r  3r
I = I1 + I2 = M(3r)2 + M(3r)2 = 9Mr2 + 9Mr2 = 18 Mr2
We increased I by a factor of 9!!
How do we find the moment of inertia for the disk in this situation?
After drawing the FBD, we can write Newton’s 2nd Law
for the block:
r


 F  ma  mg  T  ma
T
m
Next we write Newton’s 2nd law for the rotating disk,
and since the disk is rotating we need to use the rotational
form of Newton’s 2nd law:


  I 
  I
Which force is causing the disk to raotate? It’s the force
Associated with the tension, T in the string so we can write:
 I  rT I
mg
remember t hat linear accelerati on : a  r   
a
rT I
r
a
r
If we use the equation for the block we can solve for tension, T:
T  mg  ma
a
rT I
r
a
And we get: r ( mg  ma)  I
r
mr 2 ( g  a)
Now solve for I: I 
a
Substitute this equation back into :
If instead we solve for 1/a from this equation, we get:
1  I  1 1
  2  
a  gr  m g