PHYS 1443 – Section 003
Lecture #17
Wednesday, Oct. 29, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Rolling Motion of a Rigid Body
Torque
Moment of Inertia
Rotational Kinetic Energy
Torque and Vector Products
Remember the 2nd term exam (ch 6 – 11), Monday, Nov. 3!
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how

 i



f
would you define the angular displacement?
How about the average angular speed?
And the instantaneous angular speed?

 f  i
t f  ti
  lim
t 0

And the instantaneous angular
acceleration?
  lim
t f  ti
t 0

t
 d

t
dt
 f  i
By the same token, the average angular
acceleration


f
i

t
 d

dt
t
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.
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
Rotational Kinematics
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
Wednesday, Oct. 29, 2003
 f i  t
1 2
 f   i  i t   t
2
    2  f  i 
2
f
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
i
3
Rolling Motion of a Rigid Body
What is a rolling motion?
A more generalized case of a motion where the
rotational axis moves together with the object
A rotational motion about the moving axis
To simplify the discussion, let’s
make a few assumptions
1.
2.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
The object rolls on a flat surface
Let’s consider a cylinder rolling without slipping on a flat surface
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
s  R
R  s
s=R
Wednesday, Oct. 29, 2003
Thus the linear
speed of the CM is
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
vCM 
ds
d
R
 R
dt
dt
Condition for “Pure Rolling”
4
More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is
P’
CM
aCM
dvCM
d

R
 R
dt
dt
2vCM 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.
vCM
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
CM is moving at the same speed at all times.
Why??
A rolling motion can be interpreted as the sum of Translation and Rotation
P’
CM
P
vCM
P’
vCM
CM
v=0
vCM
Wednesday, Oct. 29, 2003
+
v=R
v=R
2vCM
P’
=
P
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
CM
vCM
P
5
Torque
Torque is the tendency of a force to rotate an object about an axis.
Torque, t, is a vector quantity.
F
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.
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
t  rF sin f  Fd
t  t
1
t 2
 F1d1  F2 d2
6
Example for Torque
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
Wednesday, Oct. 29, 2003
t  R F  R F
1 1
2 2
 5.0 1.0 15.0  0.50  2.5N  m
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
The cylinder rotates in
counter-clockwise.
7
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating on 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  I
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
dt dFt r  r 2 dm
The torque due to tangential force Ft is
dm
The total torque is t    r 2 dm I
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?
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002 point, making the moment arm 0.
8
Dr. Jaehoon Yu
Example for Torque and Angular Acceleration
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?
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 
2I
MgL 3 g

2ML2 2 L
3
Wednesday, Oct. 29, 2003
L
3


M
x
ML2


2
x dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
9
Moment of Inertia
Measure of resistance of an object to
changes in its rotational motion.
Equivalent to mass in linear motion.
Rotational Inertia:
For a group
of particles
I   mi ri
i
What are the dimension and
unit of Moment of Inertia?
2
For a rigid
body
ML 
2
I   r 2 dm
kg m
2
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
Example for Moment of Inertia
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
m
l
M
x
b
2

m
r
I  i i  Ml 2  Ml 2  m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
1 2 1
K R  I  2 Ml 2  2  Ml 2 2
2
2
Why are some 0s?

Thus, the rotational kinetic energy is

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.
2
I   mi ri  Ml 2  Ml 2  mb2  mb2  2 Ml 2  mb2
i
Wednesday, Oct. 29, 2003


1
1
K R  I 2  2 Ml 2  2mb2  2  Ml 2  mb2  2
2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
11
Rotational Kinetic Energy
y
vi
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
K i  mi vi2  mi ri 2 
Kinetic energy of a masslet, mi,
2
2
moving at a tangential speed, vi, is
mi
ri

O
x
Since a rigid body is a collection of masslets, the total kinetic energy of the
rigid object is
1
1
2 
2 
K R   Ki   mi ri     mi ri 
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Wednesday, Oct. 29, 2003
1 
K R  I
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
12
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, mi.
2
2
I

lim
r

m

r

i
i
The moment of inertia for the large rigid object is
 dm
m 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
The moments of
2
r
I

r
r
dV
dm

r
dV

dm in the above equation with dV.
inertia
becomes
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
O
The moment
of inertia is
dm
R
x
Wednesday, Oct. 29, 2003
I   r 2 dm  R 2  dm  MR 2
What do you notice
from this result?
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
13
Example for Rigid Body Moment of Inertia
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
so the masslet is
dx
x
L
x
The moment
of inertia is
dm  dx 
M
L
M
dx
L
L/2
2
M 1 3 
x
M
x 
I   r dm  
dx 

L / 2
L  3  L / 2
L
L/2
2
M  L   L 

     
3L  2   2 
3
3
 M  L3  ML2
  

3
L
12

 4
L
M 1 3 
x2M
What is the moment of inertia
I  r dm  
dx 
3 x 
0
L
L

0
when the rotational axis is at
M
M 3
ML2
3
one end of the rod.
L   0  L  

3L
3L
3
Will this be the same as the above.
Since the moment of inertia is resistance to motion, it makes perfect sense
Why or why not?
for it to be harder to move when it is rotating about the axis at one end.

L
2

Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu

14
Parallel Axis Theorem
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
2
done in simple manner using parallel-axis theorem. I  I CM  MD
y
2
Moment of inertia is defined I   r dm   x 2  y 2 dm (1)
(x,y)
x  xCM  x' y  yCM  y'
Since x and y are
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?
Wednesday, Oct. 29, 2003


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




Since the x’ and y’ are the
 x' dm  0  y' dm  0
x distance from CM, by definition
Therefore, the parallel-axis theorem
2
2
 dm   x'2  y'2 dm  MD 2  I CM
I  xCM
 yCM
Moment of inertia of any object about any arbitrary axis are the same as
the sum of moment
of inertia
for a rotation about the CM and that 15
of the
PHYS 1443-003,
Fall 2002
Dr. Jaehoon
CM about the rotation
axis.Yu
Example for Parallel Axis Theorem
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
x
L
x The moment of
inertia about
the CM
M
L
M
dx
L
L/2
2
M 1 3 
x
M
I CM   r dm  
x 
dx 

L / 2
L
3
L

 L / 2
3
3
M  L   L   M  L3  ML2
  

       
3L  2   2   3L  4  12
2
L/2
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
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
16
Torque and Vector Product
z
O
r
trxF
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
p
The disk will start rotating counter clockwise about the Z axis
y
The magnitude of torque given to the disk by the force F is

t  Fr sin f
F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
What is the direction?
The direction of the torque follows the right-hand rule!!
The above quantity is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Wednesday, Oct. 29, 2003
t  rF
C  A B
C  A  B  A B sin 
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
C  A  B  A B cos
Result? A scalar
17
Properties of Vector Product
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
A B  B  A
A  B  B  A
Vector Product of two parallel vectors is 0.
 A B sin 0  0
C  A  B  A B sin 
Thus,
A A  0
If two vectors are perpendicular to each other
A B  A B sin   A B sin 90  A B  AB
Vector product follows distribution law


A B  C  A B  A C
The derivative of a Vector product with respect to a scalar variable is


d A B
dA
dB

 B  A
dt
dt
dt
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
18
More Properties of Vector Product
The relationship between
unit vectors, i, j and k
i i  j  j  k  k  0
i  j   j i  k
j  k  k  j  i
k  i  i  k  j
Vector product of two vectors can be expressed in the following determinant form
i
A  B  Ax
Bx
j
k
Ay
Az  i
By
Bz
Ay
Az
By
Bz
Ax
 j
Bx
Ax
Az
k
Bx
Bz
Ay
By
 Ay Bz  Az B y  i   Ax Bz  Az Bx  j  Ax B y  Ay Bx k
Wednesday, Oct. 29, 2003
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
19
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Wednesday, Oct. 29, 2003
Linear
Mass
M
Distance
L
Rotational
Moment of Inertia
I   r 2 dm
Angle  (Radian)
v
dr
dt

d
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
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Rotational
f

td 
i
KR 
1
I 2
2
20