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Work
Let us examine the work done by a torque applied to a system.
 
dW  F  ds  Ft ds  F sin  ds  F sin rd
dW    d
Power

This is a small amount of the total work done by a torque to move
an object a small angular distance d.


dW   d  d
P
 

dt
dt
dt
 
P   
Similar to
 
P  F v
Work – Kinetic Energy
d
d d
d
  I  I dt  I d dt  I d   d  Id

1 2 1
2
W   Id
W  I  I0  K R
2
2
0
Rolling motion (No Slipping)
The majority of motion we have been discussing is translational motion. We have recently
been exploring rotational motion. Now we will look at both together.
If a wheel is placed on a flat surface and a force is applied at the center of the wheel what will
it do? It will translate and rotate.
Why does it rotate?
It rotates due to the frictional force at the point of contact, that is in a direction opposite to
the direction the wheel would slip.
The rolling motion associated with this wheel can be modeled as if all parts of the wheel
rotate about the point of contact.
Using this model, what can we say about the velocities of point P, the center and point p’?
For this instant in time the point of contact would have zero velocity, the center of mass
would have a velocity vCM and the top of the wheel would have a velocity twice that of the
center of mass.
P’
v = 2vCM

vCM
v = vCM
v=0
P
f
P
How can we determine the translational velocity of the center of mass when we only know
the rotational speed of the wheel?
We can look at the translational distance covered by the wheel.
vcm
ds Rd 


 R
dt
dt

The angular velocity of the center about the
contact point is the same as the angular velocity
of the contact point around the center.
We can also look at the acceleration of the
center of mass.
acm
S

R
dvCM Rd 

 R

dt
dt
S
Let us also look at the kinetic energy associated with rolling.


1
1
1 2 1
2
2
2
K  I  I CM  MD   I CM   MR 2 2
2
2
2
2
1
1
2
2
K  I CM   MvCM Translational energy
2
2
Rotational Energy
Angular Momentum
Angular momentum describes the momentum of an object as it rotates about an axis.
This term is introduced to arrive at





dp  dp dr  the correct result. The end result is
 
 r F r
 r    p known and therefore this result is
dt
dt dt
more obvious when working the
 
 d r  p This resembles the impulsederivation in the opposite direction.
 

momentum equation.





dt
dr 
v
dt


v  mv  0
  
L  r  p L – Angular momentum [kg m2/s]
We are adding 0 to both sides.

 dL Impulse-angular
A net torque causes a change in the
  dt momentum equation angular momentum of a system.
  
L  r  p  rmv sin  This describes the angular momentum of a single particle .
L  rmv
This describes the angular momentum of a single particle , where r and v are
perpendicular. This would describe a particle moving in a circle.
If we want to look at a collection of particles, we can find the total angular momentum by
summing the angular momentums of each particle.
Li  ri mi vi  ri mi ri   ri 2 mi
Relates the angular momentum to the
angular velocity for the ith particle.


2
L   Li    ri mi 
L  I
i
 i




 dL d I 
d  


 dt  dt  I dt  I
Angular momentum of a rigid
object (continuous body).
Original definition of torque.
If there is no net torque on the system:

 dL
  dt  0
 
Li  L f

dL  0
L is constant

L  0
Angular momentum is conserved if there is no net torque on a system.