PHYS 1443 – Section 002
Lecture #18
Wednesday, Nov. 12, 2008
Dr. Jae Yu
•
Rotational Dynamics
•
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Torque
Vector Product
Moment of Inertia
Rotational Kinetic Energy
Work, Power and Energy in Rotation
Rolling Motion of a Rigid Body
Relationship between angular and linear quantities
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
1
• Quiz results
Announcements
– Class average: 3.9/6
• Equivalent to 65/100
• Previous quizzes: 53/100, 30/100, 56/100
– Top score: 6/6
• Reading assignment: CH 10.10
• Third term exam
– 1 – 2:20pm, Wednesday, Nov. 19, in SH103
– Covers CH 9 – CH10
– Jason will do a summary session on Monday, Nov. 17
• No colloquium today…
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
2
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
The line
of Action
d2
d1
Moment arm
F2
Consider an object pivoting about the point P by
the force F being exerted at a distance r from P.
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 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.
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
t  rF sin f  Fd1
t  t
1
t 2
 F1d1  F2 d2
3
Example for Torque
A one piece cylinder is shaped as shown 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?
The torque due to F1
R1
R2
t1  R1F1 and due to F2
So the total torque acting on
the system by the forces is
t  t
1
t 2  R2 F2
 t 2   R1F1  R2 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
t 
 R1F1  R2 F2
 5.0 1.0 15.0  0.50  2.5 N  m
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
The cylinder rotates in
counter-clockwise.
4
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
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 operation is called the
Vector product or Cross product
What is the result of a vector product?
Another vector
Wednesday, Nov. 12, 2008
t  rF
C  A B
ur ur
ur ur
ur
C  A  B  A B sin 
What is another vector operation we’ve learned?
ur ur
ur ur
Scalar product C  A  B  A B cos 
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
Result? A scalar
5
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
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,
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, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
6
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, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
7
Moment of Inertia
Rotational Inertia:
For a group
of particles
Measure of resistance of an object against
changes in its rotational motion. Equivalent
to MASS in linear motion.
I   mi ri
i
2
For a rigid
body

I   r 2 dm

What are the dimension and
2
2
kg m
ML
unit of Moment of Inertia?
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Dependent on the axis of rotation!!!
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
8
Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume that
the object consists of small volume elements with mass, Dmi.
I  lim
The moment of inertia for the large rigid object is
Dmi 0
It is sometimes easier to compute moments of inertia in terms
of volume of the elements rather than their mass
dm
Using the volume density, r, replace
r
dm in the above equation with dV.
dV
 r Dm   r 2dm
2
i
i
i
How do we do this?
dm  rdV The moments of
inertia becomes
I   rr 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
O
The moment
of inertia is
dm
R
x
Wednesday, Nov. 12, 2008
What do you notice
from this result?
I   r 2 dm  R 2  dm  MR 2
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
9
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.
M


The line density of the rod is
y
L
so the masslet is
dm  dx  M dx
L
dx
x
x
L
The moment
of inertia is
M  L   L 

     
3L  2   2 
3
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?
Wednesday, Nov. 12, 2008
L/2
2
M 1 3 
x
M
x 
I   r dm  
dx 

L / 2
L  3  L / 2
L
L/2
2
3
 M  L3  ML2
  

3
L
12

 4
L
M 1
I  r 2 dm   x M dx   x3 
0
L 3 0
L
M
M 3
ML2
3
L   0  L  

3L
3L
3

2
L


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.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
10
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 the radial force Fr
Ft  mat  mr
The torque due to tangential force Ft is t  Ft r  mat r  mr 2  I
The tangential force Ft is
What do you see from the above relationship?
What does this mean?
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?
point, making the moment arm 0.
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
11
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
Mg
t  Fd  F
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
L
MgL 3 g
MgL


We obtain  
2
2L
2ML
2I
3
Using the relationship
between tangential and
angular acceleration
Wednesday, Nov. 12, 2008
3g
at  L 
2
3


M
x
ML2


2
x dx      
3
 L  3  0
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
12
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
2
2 

m
v
Kinetic energy of a masslet, mi,
Ki

m
r
i i
i i 
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, Nov. 12, 2008
1 
K R  I
2
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
13
Example for Moment of Inertia
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
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
I   mi ri2  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?

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
2
2
I   mi ri 2  Ml 
Ml 2 mb2 mb 2  2 Ml  mb
i
Wednesday, Nov. 12, 2008

1
1
K R  I 2  2 Ml 2  2mb2  2  Ml 2  mb2  2
2
2
PHYS 1443-002, Fall 2008 Dr. Jaehoon
Yu
14
Work, Power, and Energy in Rotation
f
d r
O
ds
Let’s consider the motion of a rigid body with a single external
force F exerting on the point P, moving the object by ds.
The work done by the force F as the object rotates through
the infinitesimal distance ds=rd is
ur r
dW  F  d s   F cos( 2  f )  rd  F sin f rd
What is Fsinf?
What is the work done by
radial component Fcosf?
Since the magnitude of torque is rFsinf,
The rate of work, or power, becomes
The tangential component of the force F.
Zero, because it is perpendicular to the
displacement.
dW  rF sin f d  td
P
The rotational work done by an external force
equals the change in rotational Kinetic energy.
How was the power
dW t d
 t defined in linear motion?


dt
dt
 d 
t

I

I




 d  d   I  d 
 I




d



 d  dt 
 dt 
dW  t d  Id
t d  Id
The work put in by the external force then
1
1


W   t d   Id  I 2f  I i2
2
215
Wednesday, Nov. 12, 2008
PHYS 1443-002, Fall 2008 Dr. Jaehoon
f

Yu
f
