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Thursday, June 22, 2006
Dr. Jae hoon Yu
CM and the Center of Gravity
Fundamentals on Rotational Motion
Rotational Kinematics
Relationship between angular and linear quantities
Rolling Motion of a Rigid Body
Torque
Torque and Vector Product
Moment of Inertia
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
1

Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2

CM
The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry, if object’s mass is evenly distributed throughout the body.
How do you think you can determine the CM of objects that are not following a plum-bob.
Axis of
One can use gravity to locate CM.
symmetry
1. Hang the object by one point and draw a vertical line symmetric?
2. Hang the object by another point and do the same.
3. The point where the two lines meet is the CM.
Center of Gravity
Since a rigid object can be considered as a collection of small masses , one can see the total gravitational force exerted on the object as
D m i
F g
  i
F i
  i
D m g

Mg
D m i g
What does this equation tell you?
The net effect of these small gravitational forces is equivalent to a single force acting on a point ( Center of Gravity ) with mass M.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
3

We’ve learned that the CM of a system can represent the motion of a system.
Therefore, for an isolated system of many particles in which the total mass
M is preserved, the velocity, total momentum, acceleration of the system are
Velocity of the system v
CM
 dr
CM dt

1  m r i i




1
M
 m i dr i dt

 m v i i
M
Total Momentum of the system p
CM

Mv
CM

M
 m v i i
M
  m v i i
  p
 p tot
Acceleration of the system a
CM
 dv
CM dt
 d



1 dt M
 m v i i




1
M
 m i dv i dt

 m a i i
M
External force exerting on the system
If net external force is 0
Thursday, June 22, 2006

F ext

Ma
CM
  m a i i
 dp tot dt

F
 ext
0
 dp tot p
 tot const dt
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
What about the internal forces?
System’s momentum is conserved.
4

Linear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it.
Is this still true for rotational motions?
No, because different parts of the object have different linear velocities and accelerations.
Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide.
The arc length is l

R q
Therefore the angle, q q  l
, is . And the unit of
R the angle is in radian. It is dimensionless!!
One radian is the angle swept by an arc length equal to the radius of the arc.
Since the circumference of a circle is 2 p r,
360
 
2 p r / r

2 p
The relationship between radian and degrees is
Thursday, June 22, 2006
1 rad
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

360

/ 2 p
 o
180 3.14

180


57.3
o
/ p
5

A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10 -4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m?
Thursday, June 22, 2006
(a) One radian is 360 o /2 p . Thus

4
3 10 rad


3 10

4 rad o
360 2 p rad


 
0.017
(b) Since l =r q and for small angle
 o arc length is approximately the same as the chord length.
l
 r q 
100 m
  
4 rad


PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

2 m

3 cm
6

The first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant angular acceleration:
Angular displacement under constant angular acceleration:
One can also obtain
 f
   i
 t q f

2 f

 q  i
 i
2
 i t

1
2
 t
2

2

 q f
 q i

Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
7

Using what we have learned in the previous slide, how would you define the angular displacement?
D q  q  f q i
How about the average angular speed?
Unit? rad/s
And the instantaneous angular speed?
Unit? rad/s
By the same token, the average angular acceleration
Unit? rad/s 2
And the instantaneous angular acceleration?
Unit? rad/s 2
  q f t f

 q i
 t i
 lim
D t

0
D 
D t

D q
D t
  lim
D t

0
D q
D t
 
 f t f
  i
 t i

 d q dt
D 
D t
 d
 dt q f q i
When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
8

A wheel rotates with a constant angular acceleration of 3.50 rad/s 2 . If the angular speed of the wheel is 2.00 rad/s at t i
=0, a) through what angle does the wheel rotate in 2.00s?
Using the angular displacement formula in the previous slide, one gets

 q  f q i
2 .
00

2 .
00


 t
1
2

1
 t
2
2
3 .
50

 
2
11 .
0
2 p rev .

1 .
75 rev .

11 .
0 rad
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
9

What is the angular speed at t=2.00s?
Using the angular speed and acceleration relationship
 f
  i
  t 
2 .
00

3 .
50

2 .
00

9 .
00 rad / s
Find the angle through which the wheel rotates between t=2.00 s and t=3.00 s.
Using the angular kinematic formula q  f
At t=2.00s
At t=3.00s
q t

2 q t
Angular displacement
Thursday, June 22, 2006


3

D q
2.00 3.00

 q

 q
2
1
2

1
2
3.50
 
3.00

2

10 .
8 rad
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu q i


  t


11.0
rad
21.8
rad
10 .
8
2 p
1
2 rev .

 t
2
1 .
72 rev .
10

What do we know about a rigid object that rotates about a fixed axis of rotation?
Every particle (or masslet) in the object moves in a circle centered at the axis of rotation.
When a point rotates, it has both the linear and angular components in its motion.
The arc-length is
What is the linear component of the motion you see?
Linear velocity along the tangential direction.
How do we related this linear component of the motion with angular component?
l

R q So the tangential speed v is v  dl dt
 d dt
 r d q dt
The direction of  follows a right-hand rule.
 r

What does this relationship tell you about the tangential speed of the points in the object and their angular speed?:
Although every particle in the object has the same angular speed, its tangential speed differs proportional to its distance from the axis of rotation.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 rotation, the higher the tangential speed.
11

A rotating carousel has one child sitting on a horse near the outer edge and another child on a lion halfway out from the center. (a) Which child has the greater linear speed? (b) Which child has the greater angular speed?
(a) Linear speed is the distance traveled divided by the time interval. So the child sitting at the outer edge travels more distance within the given time than the child sitting closer to the center. Thus, the horse is faster than the lion.
(b) Angular speed is the angle traveled divided by the time interval. The angle both the children travel in the given time interval is the same.
Thus, both the horse and the lion have the same angular speed.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
12

How many different linear accelerations do you see in a circular motion and what are they?
Two
Tangential, a t
, and the radial acceleration, a r
.
Since the tangential speed v is v
 r

The magnitude of tangential acceleration a t is a t
 dv dt
 d dt
 r d
 dt
 r

What does this relationship tell you?
Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation.
The radial or centripetal acceleration a r is a r
 v r
2

 
2 r
 r

2
What does this tell you?
The father away the particle is from the rotation axis, the more radial acceleration it receives. In other words, it receives more centripetal force.
Total linear acceleration is a  a
2  t a r
2 
 
2  r
 
2  r

2  
4
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
13

(a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration?
First, figure out what the angular speed of the merry-go-around is.
  1 rev
4.0
s
 2 p
4.0
s
Using the formula for linear speed v
 r
 
1.2
m

1.6
/

1.9 /

1.6
/
Since the angular speed is constant, there is no angular acceleration.
Tangential acceleration is a t
 r
 
1.2
m

0 rad s
2  m s
Radial acceleration is a r
 r

2

1.2
m
 
1.6
/

2  m s
2
Thus the total acceleration is a
 a t
2  a r
2

0
  
2 
3.1 /
2
2
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
14

Audio information on compact discs are transmitted digitally through the readout system consisting of laser and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since the speed of readout system is constant, it reads out the same number of pits and flats in the same time interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the linear speed is 1.3 m/s, find the angular speed of the disc in revolutions per minute when the inner most
(r=23mm) and outer most tracks (r=58mm) are read.
Using the relationship between angular and tangential speed v
 r
 r

23 mm r

58 mm


 v r

1 .
3 m / s
23 mm

1 .
3 m / s
58 mm

1 .
3
23

10

3

56 .
5 rad / s

9 .
00 rev / s

5 .
4

10
2 rev / min

1 .
3
58

10

3


22 .
4 rad / s


  i
 f

2

2 .
1

10
2 rev / min


540

210
 rev / min
2

375 rev / min b) The maximum playing time of a standard music
CD is 74 minutes and 33 seconds. How many revolutions does the disk make during that time?
q f
 q  i
 t c) What is the total length of the track past through the readout mechanism?

0

375
60 rev / s

4473 s

2 .
8

10
4 rev l
 v t
D t 
1 .
3 m / s

4473 s

5 .
8

10
3 m d) What is the angular acceleration of the CD over the 4473s time interval, assuming constant  ?
Thursday, June 22, 2006



 f
  i

D t
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu



22 .
4

56 .
5
 rad
4473 s
7 .
6

10

3 rad / s
2
/ s
15

What is a rolling motion?
To simplify the discussion, let’s make a few assumptions
A more generalized case of a motion where the rotational axis moves together with the object
A rotational motion about the moving axis
1. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc
2. The object rolls on a flat surface
Let’s consider a cylinder rolling without slipping on a flat surface
R q s s=R q
Thursday, June 22, 2006
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is s

R q
Thus the linear speed of the CM is v
CM
 ds dt

R d q dt

R

16
Dr. Jaehoon Yu

The magnitude of the linear acceleration of the CM is a
CM
 dv
CM dt

R d
 dt

R

P’
2v
CM
As we learned in the rotational motion, all points in a rigid body moves at the same angular speed but at a different linear speed.
CM v
CM
CM is moving at the same speed at all times.
P
At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM
Why??
A rolling motion can be interpreted as the sum of Translation and Rotation
P’ v
CM P’ v=R 
P’
2v
CM
CM v
CM
P v
CM
Thursday, June 22, 2006
+ v=R 
CM v=0
P
= CM
P v
CM
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
17

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 . r d
2
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 Moment arm.
Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.
t  rF sin f 
Fd
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.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
 t  t
1


F
1 d
1
 t
2
F
2 d
2
18

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
1
F
1 The torque due to F
1 t
1
 
R
1
F
1 and due to F
2 t
2

R
2
F
2
R
2
So the total torque acting on the system by the forces is
 t  t
1
 t
2
 
R
1
F
1

R
2
F
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 

5 .
0


R F
1 1
1 .
0


R F
2 2
15 .
0

0 .
50

2 .
5 N
 m
The cylinder rotates in counter-clockwise.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
19

z
Let’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens?
t rxF r p
The disk will start rotating counter clockwise about the Z axis x
O q F y
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 f
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
Vector product or Cross product
C

A

B
C

A

B

A B sin q
What is the result of a vector product?
Another vector
Thursday, June 22, 2006
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
C

A
Result?

B

A scalar
A B cos
20 q

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 q
A B sin 0

0 Thus,
If two vectors are perpendicular to each other
A

B

A B sin q 
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
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
21

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
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
22
A y
B y

I
 i
 m i r i
2
I
  r
2 dm
kg
 m
2
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
23

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  .
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

2 
1
2

2 Ml
2


2 
Ml
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
  i m i r i
2 
Ml
2 
Ml
2  mb
2  mb
2
Thursday, June 22, 2006

2

Ml
2  mb
2

K
R

1
2
I

2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

1
2

2 Ml
2 
2 mb
2


2 

Ml
2  mb
2


2
24

Moments of inertia for large objects can be computed, if we assume 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
  r
How can we do this?
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 x
Thursday, June 22, 2006
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 2006
Dr. Jaehoon Yu
25

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.
Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
26

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 radial force F r is
The torque due to tangential force F t is t 
F
 t
F t r ma t

 ma t
What do you see from the above relationship?
t 
I
 r mr

 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
Thursday, June 22, 2006
What is the contribution due to radial force and why?
Dr. Jaehoon Yu 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
27

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
I

0
 L r
2 dm

0
 L x
2
 dx




M
L 




 x
3
3 


0
L

ML
2
3
We obtain
 
MgL
2 I

MgL
2 ML
2
3

3 g
2 L
Thursday, June 22, 2006
Using the relationship between tangential and angular acceleration a t

L
 
3 g
2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
What does this mean?
The tip of the rod falls faster than an object undergoing a free fall.
28

v i m i
y r i q
What do you think the kinetic energy of a rigid object that is undergoing a circular motion is?
Kinetic energy of a masslet, m i moving at a tangential speed, v
, i
, is
K i

1
2 m i v i
2 
1
2 m i r i
2

O x
Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is
K
R
  i
K i

1
2
The above expression is simplified as i
 m i r i
2
 
Since moment of Inertia, I, is defined as
K
I
R


1
 i

1 m i
2
2
I
 r i



2
 i
 m i r i
2



 

Thursday, June 22, 2006 PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
29

What do you think the total kinetic energy of the rolling cylinder is?
Since it is a rotational motion about the point
P, we can write the total kinetic energy
P’
2v
CM
K

1
2
I
P

2
Where, I
P
, is the moment of inertia about the point P.
CM v
CM
Using the parallel axis theorem, we can rewrite
P
K

1
2
I
P

2
Since v
CM
=R  , the above relationship can be rewritten as

1
2
I

CM

MR
2


2 
1
2
I
CM

2 
1
2
MR
2

2
K

1
2
I
CM

2 
1
2
2
Mv
CM
What does this equation mean?
Rotational kinetic energy about the CM
Total kinetic energy of a rolling motion is the sum of the rotational kinetic energy about the CM
Thursday, June 22, 2006
Dr. Jaehoon Yu
Translational Kinetic energy of the CM
And the translational
30

h
R
Let’s consider a sphere with radius R rolling down a hill without slipping.
 q
Since v
CM
=R  v
CM
What is the speed of the
CM in terms of known quantities and how do you find this out?
K


1
2
1
2
I
I
CM

2
CM

1
2


 v
CM
R



2
MR
2

2

1
2
2
Mv
CM

1
2
I
CM
R
2

M

2 v
CM
Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill
K

1
2
I
CM
R
2

M v
2
CM

Mgh
Thursday, June 22, 2006 v
CM
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

2 gh
1

I
CM
/ MR
2
31

For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method.
h f
M
Mg n q
What are the forces involved in this motion?
Gravitational Force, Frictional Force, Normal Force
Newton’s second law applied to the CM gives


F x
F y


Mg n
 sin
Mg q  cos q f


Ma
0
CM
Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torque t
CM
 fR

I
CM

I
We know that
CM a
CM


2
5
R
MR

2
Thursday, June 22, 2006
We obtain f
Substituting f in
 dynamic equations
I
CM

R

Mg
2
5
MR
R
2 sin q  a
CM
7
5
R
PHYS 1443-001, Summer 2006

Ma
CM
2
5
Dr. Jaehoon Yu
Ma
CM a
CM

5
7 g sin q
32