Download Wednesday, Nov. 6, 2002

PHYS 1443 – Section 003
Lecture #15
Wednesday, Nov. 6, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Rolling Motion of a Rigid Body
Total Kinetic Energy of a Rolling Rigid Body
Kinetic Energy of a Rolling Sphere
Torque and Vector Product
Properties of Vector Product
Angular Momentum
Today’s homework is homework #15 due 12:00pm, Wednesday, Nov. 13!!
Wednesday, Nov. 6, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
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
Wednesday, Nov. 6, 2002
Linear
Mass
Rotational
Moment of Inertia
M
I   r 2 dm
Displacement
r
Angle  (Radian)
v
dr
dt

d
dt
a
dv
dt
 
d
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Torque
Work
  I
f
W   d 
i
P  F v
P  
p  mv
L  I
K
1
mv 2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Rotational
KR 
1
I 2
2
2
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, Nov. 6, 2002
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”
3
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, Nov. 6, 2002
+
v=R
v=R
2vCM
P’
=
P
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
CM
vCM
P
4
Total Kinetic Energy of a Rolling Body
What do you think the total kinetic
energy of the rolling cylinder is?
P’
CM
Since it is a rotational motion about the point
P, we can writ the total kinetic energy
1
K  I P 2
2
2vCM
vCM
Where, IP, is the moment of
inertia about the point P.
Using the parallel axis theorem, we can rewrite

P

1
1
1
1
2
2
2
2
2 2
K  I P  I CM  MR   I CM   MR 
2
2
2
2
1
1
Since vCM=R, the above
2
2
K  I CM   MvCM
relationship can be rewritten as
2
2
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
Wednesday, Nov. 6, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
5
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down a hill without slipping.
R

h
1
1 2 2
2
K  I CM   MR 
2
2

vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
2
1
 vCM  1
2
 I CM 
  MvCM
2
 R  2
1  I CM
 2
  2  M vCM
2 R

Since the kinetic energy at the bottom of the hill must
be equal to the potential energy at the top of the hill
1  I CM
 2


M
 2
vCM  Mgh
K
2 R

2 gh
2
1

I
/
MR
CM
PHYS 1443-003, Fall 2002
vCM 
Wednesday, Nov. 6, 2002
Dr. Jaehoon Yu
6
Example 11.1
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.
What must we know first?
R

h
I CM   r 2 dm 

vCM 
2
vCM

The linear acceleration
of the CM is
Wednesday, Nov. 6, 2002
2 gh
2 
1  I CM / MR
10
gx sin 
7
aCM
2
MR 2
5
Thus using the formula in the previous slide
vCM
Since h=xsin,
one obtains
The moment of inertia the
sphere with respect to the CM!!
2 gh

1 2 / 5
Using kinematic
relationship
2
vCM
5

 g sin 
2x 7
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
gh
7
2
vCM
 2aCM x
What do you see?
Linear acceleration of a sphere does
not depend on anything but g and .
7
Example 11.2
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.
What are the forces involved in this motion?
Gravitational Force, Frictional Force, Normal Force
Newton’s second law applied to the CM gives
n
f
M
h
Mg
F
F

x
 Mg sin   f  MaCM
y
 n  Mg cos  0
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 CM
We know that
I CM 
2
MR 2
5
aCM  R
We
obtain
2
MR 2
I CM 
2
f 
 aCM   MaCM
5
R  R  R  5


Substituting f in
dynamic equations
Wednesday, Nov. 6, 2002
 fR  I CM 
7
Mg sin   MaCM
5
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
aCM 
5
g sin 
7
8
Torque and Vector Product
z
O
r
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

  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 quantity is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Wednesday, Nov. 6, 2002
  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
9
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, Nov. 6, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
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
 j
Ax
Az
Bx
Bz
k
Ax
Ay
Bx
By
 Ay Bz  Az B y  i   Ax Bz  Az Bx  j  Ax B y  Ay Bx k
Wednesday, Nov. 6, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
11
Example 11.3
Two vectors lying in the xy plane are given by the equations A=2i+3j and
B=-i+2j, verify that AxB= -BxA
(2,3)
(-1,2)
B
A



A B  2i  3 j   i  2 j  4i  j  3 j  i
Since i  j  k
 
B  A   i  2 j  2i  3 j   3i  j  4 j  i
A B  4k  3 j   i  4k  3k  7k
Using the same unit vector relationship as above
B  A  3k  4k  7k
Therefore, AxB= -BxA
Now prove this using determinant method
Wednesday, Nov. 6, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
12