Download Kinetic energy of rolling.

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 (ω) = Angular speed (ω)about O for stationary observer.
vtop  ( )( 2 R)  2(R)  2vCOM
Instantaneous velocity vectors = sum of translational
and rotational motions.
- Kinetic energy of rolling.
I p  I COM  MR 2
1
1
1
1
1
2
2
2 2
2
2
K  I p  I COM   MR   I COM   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
I COM  2
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.
Anti-block 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 (1)
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
R
2
(2)
 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
2
1  I com / MR
1  I com / MR0
II. Torque
 
  r F

- Vector quantity.
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
Angular
Linear

dp
Fnet 
dt

 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
d
v
d
r

 m r    v   mr  a  v  v   m(r  a ) 
Proof: l  m(r  v ) 
dt
dt dt



  
  
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
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
(a) The angular speed of the cylinder is  
R
1
rotational inertia for cylinder I 
, the
MR 2. 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.
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  
L

cos


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.


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
L  r p  6.00
0
ˆ
j
5.00t
10.0
kˆ
0
0

60.0 kg  m
2

s kˆ
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 the 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