Download Slide 1

Chapter 11
Rolling, torque, and angular momentum
Smooth rolling
• Smooth rolling – object is rolling without slipping or
bouncing on the surface
• Center of mass is moving at speed vcom
• Point P (point of momentary contact between two
surfaces) is moving at speed vcom
s = θR
ds/dt = d(θR)/dt = R dθ/dt
vcom = ds/dt = ωR
Rolling: translation and rotation
combined
• Rotation – all points on the wheel move with the
same angular speed ω
• Translation – all point on the wheel move with the
same linear speed vcom
Rolling: pure rotation
• Rolling can be viewed as a pure rotation around the
axis P moving with the linear speed vcom
• The speed of the top of the rolling wheel will be
vtop = (ω)(2R)
= 2(ωR) = 2vcom
Chapter 11
Problem 2
Friction and rolling
• Smooth rolling is an idealized mathematical
description of a complicated process
• In a uniform smooth rolling, P is at rest, so there’s
no tendency to slide and hence no friction force
• In case of an accelerated smooth rolling
acom = α R
fs opposes tendency to slide
Rolling down a ramp
Fnet,x = M acom,x
fs – M g sin θ = M acom,x
R fs = Icom α
α = – acom,x / R
fs = – Icom acom,x / R2
acom,x
 g sin 

2
1  I com / MR
Torque revisited
• Using vector product, we can redefine torque
(vector) as:
     
  r  F  r  F  r  F

  rF sin   r sin F
Angular momentum
• Angular momentum of a particle of mass m and

velocity v with respect to the origin O is defined as
  
 
l  r  p  m( r  v )
• SI unit: kg*m2/s
  
l  r  p
  
l  r  p
Newton’s Second Law in angular form
  
 
l  r  p  m( r  v )



   
dl
  dv dr  
 m r    v   mr  a  v  v 
dt
dt dt


 
 

  
 mr  a   r  ma  r  Fnet   r  Fi
i


  i   net

i
dl 
  net
dt
Angular momentum of a system of
particles


L   ln
 n



dln
dL
  net,n   net

dt
n
n dt

dL 
  net
dt
Chapter 11
Problem 33
Angular momentum of a rigid body
• A rigid body (a collection of elementary masses
Δmi) rotates about a fixed axis with constant angular
speed ω
• Δmi is described by
mi

ri

pi
Angular momentum of a rigid body
liz  (ri )( mi vi )
Lz   liz   (ri )( mi vi )
i
i
  ri mi (ri )
i
   mi (ri )  I z
2
i
Lz  I z
Conservation of angular momentum
• From the Newton’s Second Law

dL 
  net
dt
• If the net torque acting on a system is zero, then

dL
0
dt

L  const
• If no net external torque acts on a system of
particles, the total angular momentum of the system
is conserved (constant)
• This rule applies independently to all components
 net , x  0  Lx  const
Conservation of angular momentum
L  I  const
I ii  I f  f
Conservation of angular momentum

L  const
More corresponding relations for
translational and rotational motion
(Table 11-1)
Chapter 11
Problem 51
Answers to the even-numbered problems
Chapter 11:
Problem 4
(a) 8.0º;
(b) more
Answers to the even-numbered problems
Chapter 11:
Problem 18
(a) (6.0 N · m)ˆj + (8.0 N · m) ˆk;
(b) (− 22 N · m)ˆi
Answers to the even-numbered problems
Chapter 11:
Problem 26
(a) (6.0 × 102 kg · m2/s) ˆk;
(b) (7.2 × 102 kg · m2/s)ˆk
Answers to the even-numbered problems
Chapter 11:
Problem 32
(a) 0;
(b) (−8.0N · m/s)tˆk;
(c) − 2.0/√t ˆk in newton·meters for t in
seconds;
(d) 8.0 t−3 ˆk in newton·meters for t in
seconds
Answers to the even-numbered problems
Chapter 11:
Problem 42
(a) 750 rev/min;
(b) 450 rev/min;
(c) clockwise