Transcript File - Phy 2048-0002

Chapter 11
Angular Momentum
Rolling, Torque, and Angular Momentum
I. Rolling
- Kinetic energy
- Forces
II. Torque
III. Angular momentum
- Definition
IV. Newton’s second law in angular form
V. Angular momentum
- System of particles
- Rigid body
- Conservation
I. Rolling
- Rotation + Translation combined.
Example: bicycle’s wheel.
ds d
s  R  
R    R  vCOM
dt dt
Smooth rolling motion
The motion of any round body rolling smoothly over a surface can be
separated into purely rotational and purely translational motions.
- Pure rotation.
Rotation axis  through point where wheel contacts ground.
Angular speed about P = Angular speed about O for
stationary observer.
vtop  ( )( 2 R)  2(R)  2vCOM
Instantaneous velocity vectors = sum of
translational and rotational motions.
- Kinetic energy of rolling.
K
I p  I COM  MR 2
1
1
1
1
1
2
I p 2  I COM  2  MR 2 2  I COM  2  MvCOM
2
2
2
2
2
A rolling object has two types of kinetic
energy 
Rotational:
(about its COM).
(translation of its COM).
Translational:
1
2
I COM 
2
1
2
MvCOM
2
- Forces of rolling.
(a)Rolling at constant speed  no sliding
at P
 no friction.
(b) Rolling with acceleration  sliding
at P  friction force opposed to sliding.
Static friction  wheel does not slide 
smooth rolling motion  aCOM = α R
Sliding
Increasing acceleration
Example1: wheels of a car moving forward while its tires are
spinning madly, leaving behind black stripes on the road 
rolling with slipping = skidding Icy pavements.
Antiblock braking systems are designed to ensure that tires
roll without slipping during braking.
Example2: ball rolling smoothly down a ramp. (No slipping).
1. Frictional force causes the rotation.
Without friction the ball will not roll
down the ramp, will just slide.
Sliding
tendency
2. Rolling without sliding  the point of
contact between the sphere and the
surface is at rest  the frictional force
is the static frictional force.
3. Work done by frictional force = 0 
the point of contact is at rest (static
friction).
Example: ball rolling smoothly down a ramp.
Fnet , x  max  f s  Mg sin   MaCOM , x
Note: Do not assume fs = fs,max . The only fs requirement is that
its magnitude is just right for the body to roll smoothly down the
ramp, without sliding.
Newton’s second law in angular form
 Rotation about center of mass
  r F   f  R  f s
s
F N  0
g
 net  I  R  f s  I COM   I COM
 f s   I COM
aCOM , x
R2
 aCOM , x
R
f s  Mg sin   MaCOM , x
f s   I COM
aCOM , x
aCOM , x
 Mg sin   MaCOM , x  ( M 
R2
g sin 

1  I com / MR 2
I COM
)aCOM , x  Mg sin 
2
R
Linear acceleration of a body rolling
along a incline plane
- Yo-yo
Potential energy (mgh) kinetic energy:
translational (0.5mv2COM) and rotational
(0.5 ICOMω2)
Analogous to body rolling down a ramp:
- Yo-yo rolls down a string at an angle
θ = 90º with the horizontal.
- Yo-yo rolls on an axle of radius R0.
- Yo-yo is slowed by the tension on it from
the string.
aCOM , x
 g sin 
g


2
1  I com / MR
1  I com / MR0 2
II. Torque
- Vector quantity.
 
  r F

Direction: right hand rule.
Magnitude:
  r  F sin   r  F  (r sin  ) F  r F
Torque is calculated with respect to (about) a point. Changing
the point can change the torque’s magnitude and direction.
III. Angular momentum
- Vector quantity:
  
 
l  r  p  m(r  v )
Units: kg m2/s
Magnitude:
l  r  p sin   r  m  v sin   r  m  v  r  p  (r sin  ) p  r p  r m  v
Direction: right hand rule.
l positive  counterclockwise
l negative  clockwise
Direction of l is always perpendicular to plane formed
by r and p.
IV. Newton’s second law in angular form
Linear

dp
Fnet 
dt

Angular
 net
dl

dt
Single particle
The vector sum of all torques acting on a particle is equal to the
time rate of change of the angular momentum of that particle.




 

   
 
dl
 dv dr  
l  m( r  v ) 
 m r    v   mr  a  v  v   m(r  a ) 
dt
dt dt


Proof: 
  
  
dl 
 r  ma  r  Fnet   r  F   net
dt


V. Angular momentum
- System of particles:
n 
  

L  l1  l2  l3  ...  ln   li
i 1
 n 

n


dli
dL
dL

  net ,i   net 
dt i 1 dt i 1
dt
Includes internal torques (due to forces between particles
within system) and external torques (due to forces on the
particles from bodies outside system).
Forces inside system  third law force pairs  torqueint sum =0
 The only torques that can change the angular momentum of
a system are the external torques acting on a system.
The net external torque acting on a system of particles is
equal to the time rate of change of the system’s total angular
momentum L.
- Rigid body (rotating about a fixed axis with constant angular speed ω):
Magnitude
li  (ri )( pi )(sin 90 )  (ri )( mi vi )
Direction: li  perpendicular to ri
and pi
liz  (li ) sin   (ri sin  )(mi vi )  ri mi vi
v    r
 n
2 
Lz   liz   mi vi ri   mi (  ri )  ri     mi ri 
i 1
i 1
i 1
 i 1

Lz  I z
n
L  I
n
n
Rotational inertia of a rigid body about a fixed axis
- Conservation of angular momentum:
Newton’s second law

 net

dL

dt
If no net external torque acts on the
system  (isolated system)
Law of conservation of angular momentum:


dL
 0  L  cte
dt
 
Li  L f
(isolated system)
Net angular momentum at time ti = Net angular momentum at later time tf
If the net external torque acting on a system is zero, the
angular momentum of the system remains constant, no matter
what changes take place within the system.
If the component of the net external torque on a system along a certain
axis is zero, the component of the angular momentum of the system
along that axis cannot change, no matter what changes take place within
the system.
This conservation law holds not only within the frame of Newton’s
mechanics but also for relativistic particles (speeds close to light) and
subatomic particles.
I ii  I f  f
( Ii,f, ωi,f refer to rotational inertia and angular speed before and after
the redistribution of mass about the rotational axis ).
Examples:
Spinning volunteer
If < Ii (mass closer to rotation axis)
Torque ext =0  Iiωi = If ωf
ωf > ωi
Springboard diver
- Center of mass follows parabolic path.
- When in air, no net external torque about COM
 Diver’s angular momentum L constant
throughout dive (magnitude and direction).
- L is perpendicular to the plane of the figure
(inward).
- Beginning of dive  She pulls arms/legs closer
Intention: I is reduced  ω increases
- End of dive  layout position
Purpose: I increases  slow rotation rate 
less “water-splash”
Rotation
Translation
Force

F
Linear momentum
 
Torque   r  F
  
Angular momentum l  r  p


p
Linear
momentum
Angular
momentum
(system of particles,
rigid body)



P   pi  MvCOM
i
Newton’s
second law
Conservation law

 dP
F
dt

p  cte
(Closed isolated system)

L

i
L  I
Newton’s
second law

li
System of particles
Rigid body, fixed
axis L=component
along that axis.

 net
Conservation law

dL

dt

L  cte
(Closed isolated system)
Motion of a Top
• The only external forces
acting on the top are the
normal force N and the
gravitational force Mg
• The direction of the angular
momentum L is along the
axis of symmetry
• The right-hand rule
indicates that
=rF=rMg
is in the xy plane
Motion of a Top
• The direction of ΔL is
parallel to that of  .
• The fact that
Lf = ΔL + Li
indicates that the top
precesses about the z
axis.
– The precessional motion
is the motion of the
symmetry axis about the
vertical
– The precession is
usually slow relative to
the spinning motion of
the top
Gyroscope
• A gyroscope can be used
to illustrate precessional
motion
• The gravitational force Mg
produces a torque about
the pivot, and this torque
is perpendicular to the
axle
• The normal force
produces no torque
Gyroscope
• The torque results in a
change in angular
momentum dL in a direction
perpendicular to the axle.
The axle sweeps out an
angle df in a time interval dt.
• The direction, not the
magnitude, of L is changing
• The gyroscope experiences
precessional motion
Gyroscope
• To simplify, assume the angular
momentum due to the motion of the center
of mass about the pivot is zero
– Therefore, the total angular momentum is
L = I due to its spin
– This is a good approximation when  is large
Precessional Frequency
• The vector diagram
shows that in the time
interval dt, the angular
momentum vector rotates
through an angle dφ,
which is also the angle
through which the axle
rotates.
• From the vector triangle
formed by the vectors Li,
Lf, and dL:
dL dt ( Mgh)dt
sin( df )  df 


L
L
L
Precessional Frequency
• Analyzing the previous vector triangle, the
rate at which the axle rotates about the
vertical axis can be found
df Mgh
p 

dt
I
p is the precessional frequency
Gyroscope in a Spacecraft
• The angular momentum of
the spacecraft about its
center of mass is zero
• A gyroscope is set into
rotation, giving it a nonzero
angular momentum
• The spacecraft rotates in
the direction opposite to
that of the gyroscope
• So the total momentum of
the system remains zero
V. Precession of a gyroscope
Gyroscope: wheel fixed to shaft and
free to spin about shaft’s axis.
Non-spinning gyroscope

 dL
 
dt
If one end of shaft is placed on a support
and released  Gyroscope falls by
rotating downward about the tip of the
support.
The torque causing the downward rotation (fall)
changes angular momentum of gyroscope.
Torque  caused by gravitational force acting on COM.
  Mgr sin 90  Mgr

Rapidly spinning gyroscope
If released with shaft’s angle slightly upward 
first rotates downward, then spins horizontally
about vertical axis z  precession due to nonzero initial angular momentum
Simplification: i) L due to rapid spin >> L due to
precession
ii) shaft horizontal when precession
starts
L  I
I = rotational moment of gyroscope
about shaft
ω = angular speed of wheel about
shaft
Vector L along shaft, parallel to r
Torque perpendicular to L  can only change the
Direction of L, not its magnitude.
 
dL   dt  dL  dt  Mgrdt
dL Mgrdt
d 

L
I
Rapidly spinning gyroscope
 
dL   dt  dL  dt  Mgrdt
dL Mgrdt
d 

L
I
Precession rate:
d Mgr


dt
I
A solid cylinder of radius 15 cm and mass 3.0 kg rolls
without slipping at a constant speed of 1.6 m/s. (a) What is
its angular momentum about its symmetry axis? (b) What is
its rotational kinetic energy? (c) What is its total kinetic
1
energy? ( I cylinder= MR 2 )
2
A solid cylinder of radius 15 cm and mass 3.0 kg rolls
without slipping at a constant speed of 1.6 m/s. (a) What is
its angular momentum about its symmetry axis? (b) What is
its rotational kinetic energy? (c) What is its total kinetic
energy? ( I cylinder= 1 MR 2 )
2
v

R
(a) The angular speed of the cylinder is
, the
1
2
I

MR
rotational inertia for cylinder
. The angular
2
momentum about the symmetry axis is
1
2  v 
L  I   MR    MRv 
2
 R 
0.5  (3.0kg)(0.15m)(1.6m / s)  0.36kg  m2 / s
A solid cylinder of radius 15 cm and mass 3.0 kg rolls
without slipping at a constant speed of 1.6 m/s. (a)
What is its angular momentum about its symmetry axis?
(b) What is its rotational kinetic energy? (c) What is its
total kinetic energy? ( I cylinder= 1 MR 2 )
2
(b)
2
K Rot
1 2 11
1
1
2  v 
2
 I   MR    Mv  (3kg)(1.6m / s) 2  1.9 J
2
22
4
4
 R 
(c)
K tot  K lin  K Rot
1 2 1 2
 mv  mv  3.8 J  1.9 J  5.7 J
2
4
A light rigid rod 1.00 m in
length joins two particles, with
masses 4.00 kg and 3.00 kg,
at its ends. The combination
rotates in the xy plane about a
pivot through the center of the
rod. Determine the angular
momentum of the system
about the origin when the
speed of each particle is 5.00
m/s.
A light rigid rod 1.00 m in
length joins two particles, with
masses 4.00 kg and 3.00 kg,
at its ends. The combination
rotates in the xy plane about a
pivot through the center of the
rod. Determine the angular
momentum of the system
about the origin when the
speed of each particle is 5.00
m/s.
L   m iviri
  4.00 kg 5.00 m s  0.500 m    3.00 kg 5.00 m s  0.500 m 
L  17.5 kg  m 2 s


L  17.5 kg  m 2 s kˆ
A conical pendulum consists of a
bob of mass m in motion in a
circular path in a horizontal plane
as shown. During the motion, the
supporting wire of length 
maintains the constant angle
with the vertical. Show that the
magnitude of the angular
momentum of the bob about the
center of the circle is
 m g sin f 
L

cosf


2
3
4
1/ 2
 Fx  m ax
m v2
T sin  
r
 Fy  m ay
T cos  m g
sin 
v2

cos
rg
sin 
v  rg
cos
L  rm v sin 90.0
L  rm
sin 
rg
cos
sin 
L  m gr
cos
r  sin  ,so
2
L
3
m 2g
3
sin 4 
cos
m
The position vector of a particle of mass 2.00 kg is given
as a function of time by r  6.00ˆi  5.00t ˆj m . Determine
the angular momentum of the particle about the origin, as a
function of time.


The position vector of a particle of mass 2.00 kg is given
as a function of time by r  6.00ˆi  5.00t ˆj m . Determine
the angular momentum of the particle about the origin, as a
function of time.


ˆ 5.00tˆ
r  6.00i
jm


dr
v
 5.00ˆ
jm s
dt


p  m v  2.00 kg 5.00ˆ
j m s  10.0ˆ
j kg  m s
ˆ
i
ˆ
j
L  r p  6.00 5.00t
0 10.0
kˆ
0
0

60.0 kg  m
2

s kˆ
A particle of mass m is shot with an initial velocity vi making
an angle with the horizontal as shown. The particle moves
in the gravitational field of the Earth. Find the angular
momentum of the particle about the origin when the particle
is (a) at the origin, (b) at the highest point of its trajectory,
and (c) just before it hits the ground. (d) What torque
causes its angular momentum to change?
(a)
zero
(b) At the highest point of the trajectory,
vi2 sin 2
1
x R 
2
2g
y  hm ax 
 visin  
2
2g
L1  r1  m v1
2 
 v2 sin 2
visin   ˆ

i
ˆ

i
j  m vxiˆ
i
2g
 2g



 m  visin   vicos ˆ

k
2g
2
(c)
2
v
i sin 2
ˆ
L 2  R i m v2 ,w here R 
g


 m Rˆ
i vicos ˆ
i visin  ˆ
j
3

m
v
i sin 2 sin  ˆ
ˆ
  m Rvisin  k 
k
g
A wad of sticky clay with mass m and velocity vi is fired at a
solid cylinder of mass M and radius R. The cylinder is initially
at rest, and is mounted on a fixed horizontal axle that runs
through its center of mass. The line of motion of the projectile
is perpendicular to the axle and at a distance d < R from the
center. (a) Find the angular speed of the system just after the
clay strikes and sticks to the surface of the cylinder. (b) Is
mechanical energy of the clay-cylinder system conserved in this
process? Explain your answer.
(a) Consider the system to consist of the wad of clay and the
cylinder. No external forces acting on this system have a
torque about the center of the cylinder. Thus, angular
momentum of the system is conserved about the axis of the
cylinder.
L f  Li
I  m vid
1
2
2
 2 M R  m R    m vid
No

M
2m vid
 2m  R 2
(b)
. Some mechanical energy changes to internal
energy in this perfectly inelastic collision
Two astronauts, each having a mass of 75.0 kg, are connected by a
10.0-m rope of negligible mass. They are isolated in space, orbiting their
center of mass at speeds of 5.00 m/s. Treating the astronauts as
particles, calculate (a) the magnitude of the angular momentum of the
system and (b) the rotational energy of the system. By pulling on the
rope, one of the astronauts shortens the distance between them to 5.00
m. (c) What is the new angular momentum of the system? (d) What are
the astronauts’ new speeds? (e) What is the new rotational energy of the
system? (f) How much work does the astronaut do in shortening the
rope?
(a)
 d
Li  m 1v1ir1i  m 2v2ir2i  2m v 
 2
Li  2 75.0 kg 5.00 m s  5.00 m
Li  3 750 kg  m
2
s

(b)
1
1
2
K i  m 1v1i  m 2v22i
2
2
2
 1
K i  2   75.0 kg 5.00 m s  1.88 kJ
 2
(c) Angular momentum is conserved:
L f  Li  3750 kg  m
2
s
(d)
3750 kg  m 2 s
vf 

 10.0 m s
2 75.0 kg  2.50 m 
2 m rf
(e)
2
 1
K f  2   75.0 kg10.0 m s  7.50 kJ
 2
(f)
Lf
 
W  K f  K i  5.62 kJ
A uniform solid disk is set into rotation with an angular speed
ωi about an axis through its center. While still rotating at this
speed, the disk is placed into contact with a horizontal
surface and released as in Figure. (a) What is the angular
speed of the disk once pure rolling takes place? (b) Find the
fractional loss in kinetic energy from the time the disk is
released until pure rolling occurs. (Hint: Consider torques
about the center of mass.)
(a)
The net torque is
zero at the point of contact,
so the angular momentum
before and after the collision
must be equal.


 21 M
 
1 1
2 2M
R 2  i2
1
1
2
2
2
M
R


M
R


M
R


 i 

2
2
(b)
E

E

1 1
2 2M
R
2
 
i 2
3

R i 2
3

 21


1
2M

i
R 2  i2
3
2
 
3