Transcript 投影片 1 - National Cheng Kung University

10. Rotational Motion
1.
2.
3.
4.
5.
Angular Velocity & Acceleration
Torque
Rotational Inertia & the Analog of Newton’s Law
Rotational Energy
Rolling Motion
Examples of rotating objects:
• Planet Earth.
• Wheels of your bike.
• DVD disc in the player.
• Circular saw.
• Pirouetting dancer.
• Spinning satellite.
How should you engineer the blades
so it’s easiest for the wind to get the
turbine rotating?
Ans. blade mass toward axis
Polar coord ( r,  )
10.1. Angular Velocity & Acceleration
θˆ
rˆ
Average angular velocity
ω

ˆ  ω
ˆ
ω
t
 = angular displacement
( positive if CCW )
ωˆ // rotational axis
 in radians
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
ωˆ
d
dt

ds
r
dt
 in radians

1ds v

r dt r
ωˆ
Example 10.1. Wind Turbine
A wind turbine’s blades are 28 m long & rotate at 21 rpm.
Find the angular speed of the blades in rad / s,
& determine the linear speed at the tip of a blade.
  21 rpm 
 21 rev / min   2 rad / rev 
60 s / min
v  r    28 m 2.2 rad / s   62 m / s
 2.2 rad / s
Angular Acceleration
We shall restrict ourselves to rotations about a fixed axis.
(Instantaneous) angular acceleration
at
ˆ
α  ω
Trajectory of point on rotating rigid body is a circle,
v / / θˆ
a
ar
ωˆ
 d 
d2 
  lim


 t 0 t
dt
d t2
i.e. r = const.
Its velocity v is always tangential:
v  r  θˆ
Its acceleration is in the plane of rotation (   ) :
d ˆ
d θˆ
dv
r
θ  r
a
 at  ar
dt
dt
dt
d ˆ
Tangential component:
at  r
θ  r  θˆ
dt
Radial component:
d θˆ
v2
2
ar  r 
  rˆ   r  rˆ
r
dt
d θˆ
d

rˆ
dt
dt
  rˆ
Angular vs Linear
Example 10.2. Spin Down
When wind dies, the wind turbine of Example 10.1 spins down with
constant acceleration of magnitude 0.12 rad / s2.
How many revolutions does the turbine make before coming to a stop?
 2  02  2 
# of rev.
  2  02

2
4 

0   2.2 rad / s 
2
4  3.14  0.12 rad / s
2

 3.2
10.2. Torque
Rotational analog of force
Torque :
τˆ
τ   τˆ
  r F sin 
 plane of r & F
[  ] = N-m ( not J )
  r
  sin 
Example 10.3. Changing a Tire
You’re tightening the wheel nuts after changing a flat tire of your car.
The manual specify a tightening torque of 95 N-m.
If your 45-cm-long wrench makes a 67 angle with the horizontal,
with what force must you pull horizontally to do the job?
  r F sin 
95 Nm   0.45 m F sin 180  67
F  230 N
Note:
sin      sin 
10.3. Rotational Inertia & 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   mi 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 ¼ of the
way from one end & perpendicular to it.
GOT IT? 10.2
Would I
(a)increase
 L  2  3L  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 & 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
 M R2
0

2
0
d

M
2 R
 R2  d m
Pipe of radius R & length L :
I 
L
0
 R3

2
0
R2  R d d z

M
2 R L
M
2 L  M R2  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 & mass M
about an axis through its center & perpendicular to it.
I   r2 d m
dm    2 r dr 
I
2M
R2

R
0


r3 d r 
2M
r dr
2
R
1
M R2
2
M
 R2
Table 10.2. Rotational Inertia
Parallel - Axis Theorem
Parallel - Axis Theorem:
I  Icm  M d 2
Ex. Prove the theorem
for a set of particles.
GOT IT? 10.3.
Explain why the rotational inertia for a solid sphere is
less than that of a spherical shell of the same M & R.
I sphere 
2
M R2
5
I shell 
2
M R2
3
Mass of shell is further away from the axis.
Example 10.8. De-Spinning a Satellite
A cylindrical satellite is 1.4 m in diameter, with its 940-kg mass distributed uniformly.
The satellite is spinning at 10 rpm but must be stopped for repair.
Two small gas jets, each with 20-N thrust, are mounted on opposite sides of it & fire
tangent to its rim.
How long must the jets be fired in order to stop the satellite’s rotation?
To stop the spin:
  0
Time required for a const ang accel
  r F sin   I 
t 

MR 
4F
10 rpm   2 rad / rev  
4  20 N 
1

min / s 
 60


t 


1

2R F   M R2  
2

 940 kg  0.7 m 
 8.6 s
Example 10.9. Into the Well
A solid cylinder of mass M & radius R is mounted on a frictionless horizontal axle over a well.
A rope of negligible mass is wrapped around the cylinder & 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
Example 10.10. Flywheel Storage
A flywheel has a 135-kg solid cylindrical rotor
with radius 30 cm and spins at 31,000 rpm.
How much energy does it store?
K rot 
1
1 1
I  2   M R 2   2
2
22

Flywheel for hybrid bus (30% fuel saving).
1
2 
 1

 135 kg  0.30 m   31, 000 rpm  2 rad / rev   min / s  
4
 60


2
 32 MJ
~ energy in 1 liter of gasoline
Modern flywheels 10s of kW of power for up to a min.
Carbon composite to withstand strain of 30,000 rpm.
Magnetic bearings to reduce friction.
supercondutor to reduce electrical losses.
Energy & Work in Rotational Motion
Work-energy theorem for rotations:
2
W    d  Krot 

1
1
1
I  2f  I i2
2
2
Example 10.11. Balancing a Tire
An automobile wheel with tire has rotational inertia 2.7 kg m2.
What constant torque does a tire-balancing machine need to apply in order to
spin this tire up from rest to 700 rpm in 25 revolutions?
W    
1
I  2f
2
 1

2 
2.7
kg
m
700
rpm
2

rad
/
rev
min
/
s








I  2f
 60




2  25 rev  2 rad / rev 
2 
2
 46 N m
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 & 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  Krot 0  U0  M g h
Finally:
E  Ktrans  Krot  U

1
1
M v2  I  2
2
2
2
1
12
 v 
 M v2   M R2     7 M v 2
2
25
 R
10
E  E0
 v
10
gh
7

2g h
sliding ball
Note: v is independent of M & R
GOT IT? 10.5.
A solid ball & a hollow ball roll without slipping down a ramp.
Which reaches the bottom first? Explain.
Solid ball.
Smaller I  smaller Krot
 larger v.