# Short Version : 10. Rotational Motion 短版: 10.轉動 ```Short Version :
10. Rotational Motion
Polar coord ( r,  )
10.1. Angular Velocity &amp; Acceleration
θ̂
r̂
Average angular velocity
ω

ˆ  ω
ˆ
ω
t
Right hand rule
 = angular displacement
( positive if CCW )
ω̂ // rotational axis
1 rad = 360 / 2  = 57.3
(Instantaneous) angular velocity

d
ˆ 
ˆ  ω
ˆ
ω  lim
ω
ω
t 0 t
dt

Angular speed:
Circular motion:
Linear speed:
s r
v
ω̂
d
dt

ds
r
dt

1ds v

r dt r
ω̂
Angular Acceleration
We shall restrict ourselves to rotations about a fixed axis.
(Instantaneous) angular acceleration
at
 d 
d2 
  lim


t 0 t
dt
d t2
ˆ
α  ω
Trajectory of point on rotating rigid body is a circle,
v / / θˆ
a
ar
i.e. r = const.
Its velocity v is always tangential:
v  r  θˆ
Its acceleration is in the plane of rotation (   ) :
ω̂
d ˆ
d θˆ
dv
r
θ  r
a
 at  a r
d
t
d
t
dt
d ˆ
Tangential component:
at  r
θ  r  θˆ
dt
d θˆ
v2
2
ar  r 
  rˆ   r  rˆ
dt
r
d θˆ
d

rˆ
dt
dt
  rˆ
Angular vs Linear
Table 10.1 Angular &amp; Linear Position, Velocity, &amp; Acceleration
Linear Quantity
Angular position 
Position x
Velocity
Acceleration
Angular Quantity
v
dx
dt
d v d2 x
a

dt
d t2
Eqs for Constant Linear Acceleration
1
 v0  v 
2
v  v0  a t
v
Angular velocity
Angular Acceleration
d
dt
d  d 


dt
d t2

Eqs for Constant Angular Acceleration
1
 0   
2
  0   t

1
x  x0  v0 t  a t 2
2
  0   0 t   t 2
1
2
v 2  v02  2a  x  x0 
 2   02 2   0 
10.2. Torque
Rotational analog of force
Torque :
τ   τˆ
  r F sin 
τ̂
 plane of r &amp; F

r to F right-hand rule.
[  ] = N-m ( not J )
  r
  sin 
10.3. Rotational Inertia &amp; the Analog of Newton’s Law
Linear acceleration:
F ma
Rotating baton
(massless rod of length R + ball of mass m at 1 end):
  R Ft
Tangential force on ball:
Ft  m at  m  R
  m  R2  I 
I mR
2
= moment of inertia
= rotational inertia
of the baton
Calculating the Rotational Inertia
Rotational inertia of discrete masses
I   m i ri 2
i
ri = perpendicular distance of mass i to the rotational axis.
Rotational inertia of continuous matter
I   r 2 d m   r 2   r  dV
r = perpendicular distance of point r to the rotational axis.
( r) = density at point r.
Example 10.4. Dumbbell
A dumbbell consists of 2 equal masses m = 0.64 kg
on the ends of a massless rod of length L = 85 cm.
Calculate its rotational inertia about an axis &frac14; of the
way from one end &amp; perpendicular to it.
GOT IT? 10.2
Would I
(a) increase
 L  2  3 L  2 
5
I  m        m L2
8
 4   4  

5
2
 0.64 kg   0.85 m   0.29 kg m2
8
(b) decrease
(c) stay the same
if the rotational axis were
(b)
(1) at the center of the rod
(a)
(2)
at one end?
Example 10.5. Rod
Find the rotational inertia of a uniform, narrow rod of mass M and length L
about an axis through its center &amp; perpendicular to it.
I   r 2 d m   r 2  dV
M 1 3

x
L 3
L /2

 L /2
M 2
L
12

L /2
x 2  dx  
 L /2
L /2
 L /2
x2
M
dx
L
Example 10.6. Ring
Find the rotational inertia of a thin ring of radius R and mass M about the ring’s axis.
2
  R2  R d
I   R2 d m
M R2

2
0

2
0
d

M
2 R
 R2  d m
 M R2
Pipe of radius R &amp; length L :
I 
L
0
 R3

2
0
R  R d d z
2

M
2 R L
M
2 L  M R 2  R 2 d m

2 R L
I = MR2 for any thin ring / pipe
Example 10.7. Disk
Find the rotational inertia of a uniform disk of radius R &amp; mass M about an axis
through its center &amp; perpendicular to it.
I   r2 d m

M
 R2
d m   r d d r
I 

2
0
d

R
0
3
r dr
1 
   2   R 4 
4 

1
M R2
2


R
0
r 2 dM
dM    2 r d r 
Table 10.2. Rotational Inertia
Parallel - Axis Theorem
Parallel - Axis Theorem:
I  I cm  M d 2
Ex. Prove the theorem
for a set of particles.
Example 10.9. Into the Well
A solid cylinder of mass M &amp; radius R is mounted on a frictionless horizontal axle over a well.
A rope of negligible mass is wrapped around the cylinder &amp; supports a bucket of mass m.
Find the bucket’s acceleration as it falls into the well.
Let downward direction be positive.
Bucket:
Cylinder:

Fnet  mg  T  m a
a
 T R  I   I
R
mg  I
a
ma
2
R
g
a
1
I
mR 2

g
M
1
2m

T I
a
R2
GOT IT? 10.4.
Two masses m is connected by a string that passes over a frictionless pulley of mass M.
One mass rests on a frictionless table; the other vertically.
Is the magnitude of the tension force in the vertical section of the string
(a) greater than,
(b) equal to,
or (c) less than
in the horizontal? Explain.
(a):
There must be a net torque to increase the
pulley’s clockwise angular velocity.
10.4. Rotational Energy
Rotational kinetic energy = sum of kinetic energies of all mass elements,
taken w.r.t the rotational axis.
Set of particles:
K
1
1
2
m
v

mi ri 2  2


i
i
2 i
2 i
dK 

1
I 2
2
1
1
2
2
dm
v

dm

r
  
  
2
2
K rot   dK  
1
1
2
 r  d m   2
2
2
K rot 
1
I 2
2

r2 d m
Energy &amp; Work in Rotational Motion
Work-energy theorem for rotations:
2
W    d  K rot 

1
1
1
I  2f  I i2
2
2
10.5. Rolling Motion
V = velocity of CM.
ui = velocity relative to CM.
Composite object:
Ktotal 

1
mi v i2

2 i

1
2
m
V

u


 i
i
2 i
1
1
M V 2   mi ui2
2
2 i

1
mi  V 2  ui2  2V  ui 

2 i
 mi ui 
i
d
mi  ri  R cm   0

dt i
Ktotal  Kcm  Kinternal
Moving wheel:
K total 
1
1
M V 2   u 2 dm
2
2
Ktotal 

1
1
M V 2   2  r 2 dm
2
2
1
1
M V 2  I cm  2
2
2
 is w.r.t. axis thru cm
1
1
 M V 2  I cm  2
2
2
Moving wheel:
Ktotal
Rolling wheel:
X  R
V  R
V = velocity of CM
 is w.r.t. axis thru CM
Example 10.12. Rolling Downhill
A solid ball of mass M and radius R starts from rest &amp; rolls down a hill.
Its center of mass drops a total distance h.
Find the ball’s speed at the bottom of the hill.
Initially:E0
 Ktrans 0  K rot 0  U 0  M g h
Finally:
E  Ktrans  K rot  U

1
1
M v2  I  2
2
2
1
12
 v 
 M v2   M R2   
2
25
 R
E  E0
 v
10
gh
7
2


7
M v2
10
2g h
sliding ball
Note: v is independent of M &amp; R
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