1443-501 Spring 2002
Lecture #15
Dr. Jaehoon Yu
1.
2.
3.
4.
Mid-term Results
Mid Term Problem Recap
Rotational Motion Recap
Rolling Motion of a Rigid Body
Today’s Homework Assignment is the Homework #6!!!
Mid Term Distributions
46.4
19.7
Certainly better than before. You
are getting there. Homework and
examples must do good for you.
Mar. 25, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
2
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Similar Quantity
Mass
Mass
Length of motion
Distance
Speed
Acceleration
Force
Linear
Rotational
Moment of Inertia
M
I   r 2 dm
L
Angle  (Radian)
v 
dr
dt
 
d
dt
a 
dv
dt
 
d
dt
Force F  ma
Torque
Work
Power
Momentum
Work
Work
Kinetic Energy
Kinetic
Mar. 25, 2002
W 

xf
xi
Fdx
  I
W 
P  F v
P  
p  mv
L  I
K 
1
mv 2
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
Rotational
f

d
i
KR 
1
I 2
2
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
Mar. 25, 2002
Thus the linear
speed of the CM is
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
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
Why??
speed while the point at P’ has twice the speed of CM
CM is moving at the same speed at all times.
A rolling motion can be interpreted as the sum of Translation and Rotation
P’
CM
P
Mar. 25, 2002
vCM
P’
vCM
CM
v=0
vCM
+
v=R
v=R
2vCM
P’
=
P
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
CM
vCM
P
5
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
Using the parallel axis theorem, we can rewrite
P
K
Since vCM=R, the above
relationship can be rewritten as
What does this equation mean?
1
1
1
I P 2  I CM  2  MR 2 2
2
2
2
1
1
2
2
K  I CM   MvCM
2
2
Rotational kinetic
energy about the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Mar. 25, 2002
Where, IP, is the moment of
inertia about the point P.
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
6
Kinetic Energy of a Rolling Sphere
R

h

vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
Mar. 25, 2002
Let’s consider a sphere with radius R
rolling down a hill without slipping.
1
1
2
K  I CM   MR 2 2
2
2
2
1
1
v

2
K  I CM  CM   MvCM
2
2
 R 
1I
 2
  CM2  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
K 
1  I CM
 2
 2  M vCM  Mgh
2 R

2 gh
2
1443-501 Spring 20021  I CM / MR
vCM 
Dr. J. Yu, Lecture #15
7
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
vCM 
Since h=xsin,
one obtains
2
vCM

The linear acceleration
of the CM is
Mar. 25, 2002
2
MR 2
5
Thus using the formula in the previous slide
2 gh

2
1  I CM / MR
10
gx sin 
7
aCM
The moment of inertia the
sphere with respect to the CM!!
Using kinematic
relationship
2
vCM
5

 g sin 
2x
7
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
2 gh

1 2 / 5
10
gh
7
2
vCM
 2aCM x
What do you see?
Linear acceleration of a sphere does
not depend on anything but g and .
8
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
 n  Mg cos  0
 CM  fR
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
 I CM 
y
We know that
2
MR 2
5
 R
I CM 
aCM
Mar. 25, 2002
We
obtain
2
MR 2
I 
 aCM  2
f  CM  5

  MaCM
R
R  R  5
Substituting f in
dynamic equations
Mg sin  
1443-501 Spring 2002
Dr. J. Yu, Lecture #15
7
5
MaCM ; aCM  g sin 
5
7
9