1443-501 Spring 2002
Lecture #13
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Rotational Energy
Computation of Moments of Inertia
Parallel-axis Theorem
Torque & Angular Acceleration
Work, Power, & Energy of Rotational Motions
Remember the mid-term exam on Wednesday, Mar. 13. Will cover Chapters 1-10.
Today’s Homework Assignment is the Homework #5!!!
Rotational Energy
y
vi
mi
ri
q
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
Kinetic energy of a masslet, mi,
1
1
K i  mi vi2  mi ri 2 
2
2
moving at a tangential speed, vi, is
x
O
Since a rigid body is a collection of masslets,
the total kinetic energy of the rigid object is
By defining a new quantity called,
Moment of Inertia, I, as
K R   Ki 
I   mi ri2
i
i
The above expression
is simplified as
What are the dimension and unit of Moment of Inertia?
What do you think the
moment of inertia is?
1
1
2 
2 
m
r


m
r



i i
i i 
2 i
2 i

kg m 2
ML 
1
K R  I 
2
2
Measure of resistance of an object to
changes in its rotational motion.
What similarity do you see between
Mass and speed in linear kinetic energy are
rotational and linear kinetic energies? replaced by moment of inertia and angular speed.
Mar. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
2
Example 10.4
In a system consists of four small spheres as shown in the figure, assuming the radii are
negligible and the rods connecting the particles are massless, compute the moment of
inertia and the rotational kinetic energy when the system rotates about the y-axis at .
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
b
M
x
I   mi ri2  Ml 2  Ml 2  m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
Why are some 0s?
m
and the radii of the spheres are negligible.
1 2 1
K R  I  2 Ml 2  2  Ml 2 2
Thus, the rotational kinetic energy is
2
2


Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.
I   mi ri 2  Ml 2  Ml 2  mb2  mb2  2Ml 2  mb2 
i
Mar. 6, 2002




1
1
K R  I 2  2Ml 2  2mb2  2  Ml 2  mb2  2
2
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
3
Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume
the object consists of small volume elements with mass, Dmi.
2
2
I

lim
r
D
m

r

i
i
The moment of inertia for the large rigid object is
 dm
Dm 0
i
i
It is sometimes easier to compute moments of inertia in terms
How can we do this?
of volume of the elements rather than their mass
dm
Using the volume density, r, replace
I  rr 2 dV
r  ; dm  rdV The moments of
dm in the above equation with dV.
inertia becomes
dV

Example 10.5: Find the moment of inertia of a uniform hoop of mass M and radius R
about an axis perpendicular to the plane of the hoop and passing through its center.
y
O
Mar. 6, 2002
The moment
of inertia is
dm
R
x
I   r 2 dm  R 2  dm  MR 2
What do you notice
from this result?
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
4
Example 10.6
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis perpendicular to the rod and passing through its center of mass.

The line density of the rod is
y
dm  dx 
so the masslet is
dx
x
x
L
The moment
of inertia is
Will this be the same as the above.
Why or why not?
Mar. 6, 2002
M
dx
L
L/2
x2M
M 1 3 
2
I   r dm  
dx 
x 

L / 2
L
L  3  L / 2
L/2
M  L   L 

     
3L  2   2 
3
What is the moment of inertia
when the rotational axis is at
one end of the rod.
M
L
I   r dm  
2
L
0


3
 M  L3  ML2
  

3
L

 4  12
L
x2M
M 1 3 
dx 
x
L
L  3  0
 
M
M 3
ML2
3
L   0 

L 
3L
3L
3
Since the moment of inertia is resistance to motion, it makes perfect sense
for it to be harder to move when it is rotating about the axis at one end.
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
5
Parallel Axis Theorem
Moments of inertia for highly symmetric object is relatively easy if the
rotational axis is the same as the axis of symmetry. However if the axis of
rotation does not coincide with axis of symmetry, the calculation can still be
2
done in simple manner using parallel-axis theorem. I  I CM  MD
y
Moment of inertia is defined I   r 2 dm   x 2  y 2 dm (1)
(x,y)
Since x and y are
x  xCM  x' ;
y  yCM  y'
r
yCM
y
y’
One can substitute x and y in Eq. 1 to obtain
CM
(xCM,yCM)
D
xCM
x’
x
What does this
theorem
tell you?
Mar. 6, 2002


I   xCM  x'   yCM  y ' dm

2

2


2
2
2
2
 xCM
 yCM
dm

2
x
x
'
dm

2
y
y
'
dm

x
'

y
'
dm
CM
CM




 x' dm  0
x
 y ' dm  0
2
I

I

MD
Therefore, the parallel-axis theorem
CM
Since the x’ and y’ are the
distance from CM, by definition
Moment of inertia of any object about any arbitrary axis are the same as
the sum of moment
inertia
1443-501ofSpring
2002for a rotation about the CM and that of6 the
J. Yu, Lecture
CM about theDr.
rotation
axis. #13
Example 10.8
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis that goes through one end of the rod, using parallel-axis theorem.

The line density of the rod is
y
dm  dx 
so the masslet is
CM
dx
M
L
M
dx
L
L/2
x2M
M 1 3 
2
  r dm  
dx 
x 

L / 2
L
L  3  L / 2
x The moment of I CM
inertia about
3
3
the CM
M  L   L  
L/2
x
L
M  L3  ML2
 

       
3L  2   2   3L  4  12
2
Using the parallel axis theorem
I  I CM
ML2  L 
ML2 ML2 ML2
2
D M 
  M 


12  2 
12
4
3
The result is the same as using the definition of moment of inertia.
Parallel-axis theorem is useful to compute moment of inertia of a rotation of a
rigid object with complicated shape about an arbitrary axis
Mar. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
7
Torque
Torque is the tendency of a force to rotate an object about some axis.
Torque, t, is a vector quantity.
F

r
P
Line of
Action
d2
d
Moment
arm
F2
Consider an object pivoting about the point P
by the force F being exerted at a distance r.
The line that extends out of the tail of the force
vector is called the line of action.
The perpendicular distance from the pivoting point
P to the line of action is called Moment arm.
Magnitude of torque is defined as the product of the force
exerted on the object to rotate it and the moment arm.
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force vectorially. The convention for sign of the torque is positive if
rotation is in counter-clockwise and negative if clockwise.
Mar. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
t  rF sin   Fd
t  t
1
t 2
 Fd  F2 d 2
8
Example 10.9
A one piece cylinder is shaped as in the figure with core section protruding from the larger
drum. The cylinder is free to rotate around the central axis shown in the picture. A rope
wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder,
and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A)
What is the net torque acting on the cylinder about the rotation axis?
R1
F1
The torque due to F1
t1  R1F1 and due to F2
So the total torque acting on
the system by the forces is
R2
t  t
1
t 2  R2 F2
 t 2  R1F1  R2 F2
F2
Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque
about the rotation axis and which way does the cylinder rotate from the rest?
Using the
above result
Mar. 6, 2002
t   R F  R F
1 1
2 2
 5.0 1.0  15.0  0.50  2.5 N  m
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
The cylinder rotates in
counter-clockwise.
9
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating in a circle.
What forces do you see in this motion?
m
The tangential force Ft and radial force Fr
Ft  mat  mr
The tangential force Ft is
The torque due to tangential force Ft is
What do you see from the above relationship?
What does this mean?
t  Ft r  mat r  mr 2
t  I
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2nd law of motion in rotation.
How about a rigid object?
The external tangential force dFt is dFt  dmat  dmr
dFt
The torque due to tangential force Ft is dt  dFt r  r 2 dm 
dm
2
t


r
dm  I
The total torque is


r
Contribution from radial force is 0, because its
What is the contribution due
line of action passes through the pivoting
O
to
radial
force
and
why?
1443-501 Spring 2002 point, making the moment arm 0.
Mar. 6, 2002
10
Dr. J. Yu, Lecture #13
Example 10.10
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is
free to rotate about the pivot in the vertical plane. The rod is released from rest in the
horizontal position what is the initial angular acceleration of the rod and the initial linear
acceleration of its right end?
The only force generating torque is the gravitational force Mg
L/2
t  Fd  F
Mg
L
L
 Mg  I
2
2
L
Since the moment of inertia of the rod I  r 2 dm 
0
0
when it rotates about one end
L
We obtain

MgL MgL 3g


2
2ML 2 L
2I
3
Mar. 6, 2002
L
3
2


M
x
ML


x 2dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
11
Work, Power, and Energy in Rotation
F

dq r
ds
Let’s consider a motion of a rigid body with a single external
force F exerted on the point P, moving the object by ds.
The work done by the force F as the object rotates
through infinitesimal distance ds=rdq in a time dt is
dW  F  d s  F sin  rdq
O
What is Fsin?
What is the work done by
radial component Fcos?
The tangential component of force F.
Zero, because it is perpendicular to the
displacement.
Since the magnitude of torque is rFsin,
dW  tdq
P
The rate of work, or power becomes
dW tdq

 t
dt
dt
How was the power
defined in linear motion?
 d 
 d  dq 

I




dt
d
q
dt





The rotational work done by an external force
equals the change in rotational energy.
t  I  I 
The work put in by the external force then
dW  t dq  Id
Mar. 6, 2002
q
W  
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
q
f
t dq   Id 

1 2 1 2
I i  I f
2
2 12
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Similar Quantity
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Mar. 6, 2002
Linear
Mass
Rotational
Moment of Inertia
M
Distance
I   r 2 dm
L
Angle q (Radian)
v 
dr
dt
 
dq
dt
a 
dv
dt
 
d
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Torque
Work
t  I
W 
P  F v
P  t
p  mv
L  I
K 
1
mv 2
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #13
Rotational
qf
q
td q
i
KR 
1
I 2
2
13