### Parallel Axis Theorem and Torque

```PHY205
1.
Ch15: // Axis Theorem and Torque
Recall main points:
•
Expression of the Kinetic energy of a rigid body in terms of Kcm
and Icm
•
Parallel Axis Theorem
•
Torque and Cross Product
•
Angular acceleration from the torque
Slide # 1
PHY205
Ch15: // Axis Theorem and Torque
Kinetic energy of a rigid body in terms of Kcm and Icm
In the 3rd slide of Ch 13 we showed that the total kinetic energy of a system is equal to
the energy of the Center of Mass (CM) plus the energy RELATIVE to the CM:
K 
1
2
2
1. Main Points
M vC M 

i
1
2
m i vi /C M
CM

2
vCM
In our case since we study RIGID bodies, any motion relative to the CM MUST be a
rotation around an axis through the CM. And therefore, since :
v i / C M  ri / cm i 
w e ca n re w rite th e se co n d te rm o f th e Kin e tic e n e rgy a b o ve . W e ge t:

i
1
2
m i vi / C M
2
1
1

2
2 
   m i ri / cm   
2

 2
 m r
2
i i / cm

2

1
2
I CM 
2
So th a t w e ca n re w rite th e to ta l kin e tic e n e rgy a s:
K 
1
2
M v cm 
2
1
2
I CM   K  K CM 
2
1
2
I CM 
2
This is a general result that should be used when the work-energy theorem, or
the conservation of mech. energy, is used for solving problems involving rigid
bodies rotating . Note that any motion can be viewed as the CM motion PLUS a
rotation AROUND an axis thru the CM
Slide # 2
PHY205
Ch15: // Axis Theorem and Torque
Optional: Direct proof of the preceding result:
K /O 

i
K /O 

1. Main Points
i
1
2
1
2
m i vi
2
m i vi  vi
N o w w e e xp re ss v i in te rm s o f th e ve lo city v i / cm
re la tive to th e ce n te r o f m a ss a n d th e v e lo city v cm O F th e C M
K /O 

i
K /O 

i
K /O 

i
K /O 
1
K /O 
1
2
2
1
2
1
2
1
2

 
m i v cm  v i / cm  v cm  v i / cm


m i v cm  v cm  v i / cm  v i / cm  2 v i / cm  v cm

m i v cm  v i / cm  2 v i / cm  v cm
2
 m v
i
M v cm 
2
2
2
cm


1
2
m i v i / cm  2
1
 m r
2
i i / cm
2
2

2
2


 m v   v
i
d
dt
i / cm
cm
 m r   v
i i / cm
cm
By definition of the center of mass (see CM chapter) the last term is zero. Note also
that we replaced vi/cm by ri/cm since we are dealing with a rigid body. We obtain the
fundamental result:
1
2
K / o  K CM  I /CM 
Slide # 3
2
PHY205
Ch15: // Axis Theorem and Torque
Parallel Axis Theorem via K=Kcm+1/2Iw2
1. Main Points
Using the result on the previous slides, we can arrive at the very useful parallel
axis theorem:
Slide # 4
PHY205
Ch15: // Axis Theorem and Torque
Torque and Cross product:
1. Main Points
The torque is defined as:
 /O  r  F
Right Hand
Rule:
In this course we only look at rotations around a axis
of fixed direction, therefore, from studying the right
hand rule for a second, we see that the only
components of r and F that matter are the x and y
y
components (see picture at right). In the problems
O
you are given, the forces are always in the X-Y plane
and the r should always be the vector r=(x,y) and thus
 is the angle in the x-y plane. The magnitude of the
torque around the z-axis is:
 / O  rF sin
F
r

m
x
Slide # 5
PHY205
Ch15: // Axis Theorem and Torque
1. Main Points
Angular acceleration from the torque, using Newton’s 2nd law:
Slide # 6
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