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Rotation of a Rigid Body
Readings: Chapter 13
1
Newton’s second
law:
F  ma
How can we characterize the acceleration
during rotation?
- translational acceleration and
- angular acceleration
2
Angular acceleration
v2
v1
r1

Center of rotation
Both points have the same angular velocity
v1   r1
v2   r2


t
Linear acceleration:
v1 
a1 

r1
t
t
v2 
a2 

r2
t
t
Both points have the same angular acceleration


t
3
Rotation of Rigid Body:
Every point undergoes circular motion
with the same angular velocity and the
same angular acceleration


t
v  r
4
The relation between angular velocity and
angular acceleration is the same as the
relation between linear velocity and linear
acceleration


t
v
a
t
5
The Center of Mass
For Rigid Body sometimes it is convenient to describe the rotation about
the special point– the center of mass of the body.
Definition: The coordinate of the center of mass:
Rigid body consisting of two particles:
xcm
m1 x1  m2 x2

m1  m2
If
m1  m2
then
xcm
xcm
x1  x2

2
2  0  0.5  0.5

 0.1m
2.0  0.5
6
The Center of Mass
Definition: The coordinate of the center of mass:
 x m  xdm


i
xcm
i
i
M
 y m  ydm


i
ycm
M
i
i
M
M
7
The Center of Mass: Example
L
xcm
L
M
x dx M  xdx

2
xdm
L
L
1
1
L
L

2
0
0




x 0 

M
M
LM
L2
2L 2
The center of mass of a disk is the center O of the disk
O
8
Torque: Rotational Equivalent of Force
9
Torque
The rotation of the body is determined by the torque
  Fr sin 
10
Torque
  Fr sin 
  900
  00 or   1800
Torque is maximum if
Torque is 0 if
 0
1   3
1  4
11
  Fr sin 
Torque
Torque is positive if the force is trying to rotate the body counterclockwise
Torque is negative if the force is trying to rotate the body clockwise
F3
3  0
F4  4  0
F1   0
1
axis
F2
2  0
The net torque is the sum of the torques due to all applied forces:
 net   1   2   3   4
12
  Fr sin 
Torque: Example
Find the net torque
F5
F3
450
450
F4
1m
2m
F1
3m
30 0
4m
axis
600
F2
F1  F2  F3  F4  F5  5 N
 1  5  3sin(30)  7.5N  m
 2  5  4sin(60)  17.3N  m
 3  5  1sin(45)  3.5N  m
 4  5  2sin(45)  7.1N  m
 5  5  0  0N  m
 net   1   2   3   4   4  7.5  17.3  3.5  7.1  6.2 N  m
13
Torque: Relation between the torque and angular acceleration:
v
v 
a

r  r
t
t
F1
r
F  ma   mr

  Fr   mr 2
 net    mi ri   I
2
i
I   mi ri2
- moment of inertia
i
14
Moment of Inertia
Thin rod, about center
Thin rod, about end
Cylinder (or disk), about center
Cylindrical loop, about center
1
I  ML2
12
1
I  ML2
3
1
I  MR 2
2
I  MR2
15
Moment of Inertia: Parallel-axis Theorem
If you know the moment of Inertia about the center of mass (point O)
then the Moment of Inertia about point (axis) P will be
I P  I O  Md 2
P
d
O
I P   ( xi  x P )2 mi   ( xi  xO  xO  x P )2 mi 
i
i
  ( xi  xO  d )2 mi   ( ri  rO )2  2d ( ri  rO )  d 2  mi 
i
i
  ( ri  rO )2 mi  2d  ( ri  rO )mi  Md 2  I O  Md 2
i
i
16
I P  I O  Md 2
Parallel-axis Theorem: Example
Thin rod, about center of mass
1
I  ML2
12
2
1
1
1
1
 L
I
ML2  M   
ML2  ML2  ML2
12
4
3
 2  12
17
 net   I
 0
Equilibrium:
 net  0
m1  2kg
m2  10kg
n1
d  2m
x?
Massless rod
n2
Two forces (which can results in rotation) acting on the rod
n1  w1  m1 g
n2  w2  m2 g
Equilibrium:
 1  n1d  m1 gd
 2   n2 x   m2 gx
 net   1   2  0
m1 gd  m2 gx  0
m1
xd
 0.4m
m2
18
Rotational Energy:
1
1
1
2
2
2
K rot  m1v1  m2v 2  m3v 3  ... 
2
2
2
1
1
1
2 2
2 2
 m1 r1  m2 r2  m3 2 r32  ... 
2
2
2
1 2
1 2
2
2
2
  ( m1r1  m2 r2  m3 r3  ...)  I 
2
2
Conservation of energy (no friction):
1
1 2
2
Mgy  Mv  I  const
2
2
19
Kinetic energy of rolling motion
K rot
K rot
1
 I P 2
2
1 2
 I
2
I P  I cm  MR 2
vcm   R
K rot
Cylinder:
I cm 
1
MR 2
2
Cylindrical loop: I cm  MR
Solid sphere:
I cm 
2
2
MR 2
5
1
1
1
1
2
2 2
2
2
 I cm  MR  
I
v

Mv
cm
2
2
2 R2 cm cm 2
11
1
3
2
2
2
K rot 
Mvcm  Mvcm  Mvcm
22
2
4
1
1
2
2
2
K rot  Mvcm  Mvcm
 Mvcm
2
2
12
1
7
2
2
2
K rot 
Mvcm  Mvcm  Mvcm
20
25
2
10