Monday, June 25, 2007
Dr. Jae hoon Yu
• Torque
• Vector Product
• Moment of Inertia
– Parallel Axis Theorem
• Torque and Angular Acceleration
• Rotational Kinetic Energy
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1

• Quiz Results
– Average: 66/100
– Top score: 90
• Last quiz this Thursday, June 28
– Early in the class
– Covers up to what we learn Wednesday, June 27
• Final exam
– Date and time: 8 – 10am, Monday, July 2
– Location: SH103
– Covers: Ch 8.4 – what we cover Thursday, June 28
• Don’t forget the end-of-semester Planetarium show
– 10:10am, Thursday, June 28
• Evaluation today
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
2

P d
Torque is the tendency of a force to rotate an object about an axis.
Torque, t , is a vector quantity.
f
F
Consider an object pivoting about the point P by the force F being exerted at a distance r from P . r d
2
The line of Action
The line that extends out of the tail of the force vector is called the line of action.
Moment arm
F
2
The perpendicular distance from the pivoting point
P to the line of action is called the 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 .
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu t  rF sin f 
Fd
 t  t
1


F
1 d
1
 t
2
F
2 d
2
3

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 and another force exerts F
2
R
1 exerts force F
1 on the core whose radius is R
2 to the right on the cylinder, downward on the cylinder. A)
What is the net torque acting on the cylinder about the rotation axis?
R
2
R
1
The torque due to F
1 t
1
 
R
1
F
1 and due to F
2
So the total torque acting on the system by the forces is
 t  t
1
 t
2
 t
2

R
2
F
2

R
1
F
1

R
2
F
2
Suppose F
1
=5.0 N, R
1
=1.0 m, F
2
= 15.0 N, and R
2
=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
 t  
R F
1 1

R F
2 2
     
The cylinder rotates in counter-clockwise.
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
4

z
Let’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens?
t rxF x
O r p
 F y
The disk will start rotating counter clockwise about the Z axis
The magnitude of torque given to the disk by the force F is t 
Fr
But torque is a vector quantity, what is the direction? sin

How is torque expressed mathematically? t  r

F
What is the direction?
The direction of the torque follows the right-hand rule!!
The above operation is called the
Vector product or Cross product
C

A

B
C

A B

A B sin

What is the result of a vector product?
Another vector
Monday, June 25, 2007
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-001, Summer 2007
C
  
Dr. Jaehoon Yu
Result?
A scalar
A B cos
5


Vector Product is Non-commutative What does this mean?
If the order of operation changes the result changes
Following the right-hand rule, the direction changes
Vector Product of two parallel vectors is 0.
A

B

B

A
A

B
 
B

A
C

A

B

A B sin
 
A B sin 0

0 Thus,
If two vectors are perpendicular to each other
A

B

A B sin
 
A B sin 90
 
A B

AB
A

A

0
Vector product follows distribution law
A

 

A

B

A

C
The derivative of a Vector product with respect to a scalar variable is d

A
 dt
B

 d A dt

B

A
 d B dt
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
6

The relationship between unit vectors, i , j and k i
 i
 j
 j
 k
 k

0 i
 j
  j
 i
 k j
 k k
 i
  k

  i
 k j
 i
 j
Vector product of two vectors can be expressed in the following determinant form i
A

B
 A x
B x j
A y
B y k
A z
B z
 i
A
B y y
A z
B z

A x j
B x
A z
B z
 k
A
B x x


A y
B z

A z
B y
 i


A x
B z

A z
B x
 j


A x
B y

A y
B x
 k
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
7
A y
B y

Rotational Inertia: Measure of resistance of an object to changes in its rotational motion.
A similar quantity?
For a group of particles
I

Equivalent to mass in linear motion.
i
 m i r i
2 For a rigid body
I
  r
2 dm
What are the dimension and unit of Moment of Inertia?
kg
 m
2
Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building.
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
8

M
In a system 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 angular speed w .
y m b
Since the rotation is about y axis, the moment of inertia about y axis, I y
, is l
O b l
M x
I
  m i r i
2 
Ml Ml
2 m 0
2
0
2 
2 Ml
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.
Thus, the rotational kinetic energy is K
R

1
2
I w
2 
1
2

2 Ml
2
 w
2 
Ml
2 w
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
  i m i r i
2 
Ml
2 
Ml
2  mb
2  mb
2
Monday, June 25, 2007

2

Ml
2  mb
2

K
R

1
2
I w
2 
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1
2

2 Ml
2 
2 mb
2
 w
2 

Ml
2  mb
2
 w
2
9

Moments of inertia for large objects can be computed, if we assume that the object consists of small volume elements with mass, D m i
.
The moment of inertia for the large rigid object is
It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass
I
 lim
D m i

0 i
 r i
2
D m i

How do we do this?
 r
2 dm
Using the volume density, r , replace d m in the above equation with dV.
r  dm dV dm
 r dV
The moments of inertia becomes
I
  r r
2 dV
Example: 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 dm
The moment of inertia is
I
  r
2 dm

R
2
 dm 
MR 2
O R
Monday, June 25, 2007 x What do you notice from this result?
The moment of inertia for this object is the same as that of a point of mass M at the distance R.
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
10

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.
y
L x dx x
The line density of the rod is so the masslet is dm


 dx

M

L
M
L
The moment of inertia is
I
  r
2 dm  

L
L
/
/
2
2 x
2
M dx
L

M
3 L







L
2
3

L
2
3


 dx

M
L

M
3 L


L
3
4




1
3 x
3


L / 2

L / 2

ML
2
12
What is the moment of inertia when the rotational axis is at one end of the rod.
Will this be the same as the above.
Why or why not?
I
  r
2 dm

M
3 L
 
0
L

 
3 
0
 x
2
M
L

M dx
3 L

M
L



1
3 x
3
ML 2


L
0
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.
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
11

Moments of inertia for highly symmetric object is easy to compute 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 done in simple manner using y parallel-axis theorem
Moment of inertia is defined
.
I

I

 r
2
I
CM dm


 
MD x
2 
2
2

(1)
(x
, y) Since x and y are x
 x
CM
 x ' y
 y
CM
 y ' x
CM r
D x x’
CM
(x
CM, y
CM
) x
One can substitute x and y in Eq. 1 to obtain

I

 x 2
CM



 x
CM y 2
CM


 x ' dm
2

2 y
CM x
CM

 x ' y ' dm

2

 dm
2 y
CM
 y ' dm
Since the x’ and y’ are the distances from CM, by definition
Therefore, the parallel-axis theorem
I


2 x
CM
 2 y
CM

 dm
 
 x '
2  y '
2
 dm
 x ' dm

0
 
 x ' 2  y ' 2
 dm

MD
2 
 y ' dm

0
I
CM
What does this theorem tell you?
Moment of inertia of any object about any arbitrary axis are the same as
PHYS 1443-001, Summer 2007
the CM about the rotation axis.
12

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.
y
CM
L x dx x
The line density of the rod is   so the masslet is
M
L dm
  dx

M
L dx
The moment of inertia about the CM
I
CM
  r
2 dm

M
3 L







L
2
3

 

L
L
/
/
2
2
L
2 x
2
M dx

L
M
L


1
3 x
3


L / 2

L / 2
3




M
3 L


L
3
4



ML
2
12
2
Using the parallel axis theorem
I

I
CM

D
2
M

ML
2

12
L
2
M

ML
2
12

ML
2
4

ML
2
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
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
13

r
F t
F r m
Let’s consider a point object with mass m rotating on a circle.
What forces do you see in this motion?
The tangential force F t
The tangential force F t and the radial force F r is
The torque due to tangential force F t is t
F
 t

F t ma t r

What do you see from the above relationship?
t 
I

 mr
 ma t r
 mr
2
 
I

What does this mean?
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2 nd law of motion in rotation.
How about a rigid object?
dF t r dm
The external tangential force dF t
The torque due to tangential force F t
The total torque is  t    r
2 dm is

I
 is
O
Monday, June 25, 2007
What is the contribution due to radial force and why?
dF t d t
 dma t
 dF t r

 dmr

 r 2 dm


Contribution from radial force is 0, because its line of action passes through the pivoting
14
Dr. Jaehoon Yu

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 are the initial angular acceleration of the rod and the initial linear acceleration of its right end?
L/2
The only force generating torque is the gravitational force Mg
Mg t 
Fd

F
L
2

Mg
L
2

I

Since the moment of inertia of the rod when it rotates about one end
We obtain
 
MgL
2 I

MgL
2 ML
2
I


0
 L r
2 dm

0
 L x
2
 dx




M
L 




 x
3
3 


0
L
3 g
2 L
Using the relationship between tangential and angular acceleration a t
3

L
 
3 g
2
What does this mean?

ML
2
3
The tip of the rod falls faster than an object undergoing a free fall.
Monday, June 25, 2007 PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
15