Download = ∫ ( ) = ∫ ( )

PES 1110 Fall 2013, Spendier
Lecture 32/Page 1
Today:
- Work and Rotational Kinetic Energy (10.10)
- Angular Momentum (11.7-1.10)
- HW 8 assigned, due next Wednesday November 20th
Work and Rotational Kinetic Energy
When you pedal a bicycle, you apply force to a rotating body and do work on it. Similar
things happen in many other real-life situations, such as a rotating motor shaft driving a
power tool or a car engine propelling the vehicle. We can express this work in terms of
torque and angular displacement.
Translational kinetic energy and work:
 
Work done by constant force
W  F d
Work done by variable force
Work = area under force versus position plot
xf
W
 F ( x )dx
xi
Wtotal  K ...... work energy theorem
Equivalent work equations for angular motion:
Work done by constant torque: W       f   i 
[Joules]
f
Work done by variable torque: W    ( )d
i
Again work could be (+ , speeding up) or (- , slowing down)
This work can change the rotational kinetic energy:
1
1
Wtotal  K  I  2f  I i2
2
2
For example, the rotational kinetic energy of a wind turbine is equal to the total work
done to set the turbine spinning.
Power: The rate at which work is done
dW
d
P
   f   i    [Watt]
dt
dt
W
Pavg 
t
This is analog of the relationship P = Fv we developed earlier for linear motion.
PES 1110 Fall 2013, Spendier
Lecture 32/Page 2
Example 1:
A 1.50 kg grinding wheel is in the form of a solid cylinder of radius 0.100 m
a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in
2.5s?
b) Through what angle has it turned during that time?
c) Calculate the work done by the torque.
d) What is the grinding wheel’s kinetic energy when it is rotating at 1200 rev/min?
Compare your answer to the result in part c)
e) What average power is needed to stop it rotating in 2.5 s?
PES 1110 Fall 2013, Spendier
Lecture 32/Page 3
What we are still missing is the counterpart for linear momentum!
Linear Momentum, p:





dp
 dv  d
 F  ma  m  dt   dt  mv   dt


linear momentum: p  mv
Angular moment:
Rotating rigid bodies have an angular momentum
In rotation, torque plays the role of force.




dL
 d  d 
  I   I  dt   dt  I    dt
Newton's Second Law for Rotation:
 dL

  dt  I 


angular moment for a rigid object: L  I 
How hard it is to torque an object.
[kg m2/s] (dropped radians)
For point particles
I   mi ri 2
i
Consider a point a particle (an object with a single value of v) going around a circle of
radius r, then I is
I = mr2
PES 1110 Fall 2013, Spendier
Lecture 32/Page 4
  
and v = ωr ( v  w  r ) ==> ω=v/r


and p  mv


   
L  I  m  r  v   r  p
  
 
angular moment for point particles: L  r  p  m  r  v 
(used in Bohr model for angular momentum of electrons: depicts the atom as small,
positively charged nucleus surrounded by electrons that travel in circular orbits around
the nucleus)
magnitude: L = r m v sin(ϕ) = r p sin(ϕ)


- If motion is purely rotational, then r is perpendicular to p and L = r m v


- If motion is NOT purely rotational, then there may be some angle ϕ between r and p
and L = r m v sin(ϕ)


- If r is parallel to p there is no angular momentum
  
 
angular moment for point particles: L  r  p  m  r  v 


angular moment for a rigid object: L  I  [kg m2/s] (dropped radians)
Example 2:
A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.50
rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m.
Calculate the magnitude of the total angular momentum of the woman-plus-disk system.
(Assume that you can treat the woman as a point.)
PES 1110 Fall 2013, Spendier
Lecture 32/Page 5
Example 3:
A 2.00 kg rock has a horizontal velocity of magnitude 12.0
m/s when it is at point P.
a) At this instant what are the magnitude and direction of its
angular momentum relative to point O?
b) If the only force acting on the rock is its weight, what is
the rate of change (magnitude and direction) of its angular
momentum at this instant?
Example 3:
Consider a point particle at position x = 2.0 m and y = 0 m with instantaneous velocity
vx = 0 m/s and vy = 3.0 m/s and mass 2.0 kg. What is its’ angular momentum about the
origin (magnitude and direction)?
PES 1110 Fall 2013, Spendier
Lecture 32/Page 6
Example 4:
Find the torque exerted by the 25 N and 20 N forces shown. If the torques remain
constant and the 3 kg mass has an initial speed of 3.4 m/s, how fast will it be going after
0.25 s? Assume the mass continues on its 2-m circle.
PES 1110 Fall 2013, Spendier
Example 5: Proof Newton's second law by starting with
a) the point particle definition of angular momentum and by,
b) the solid body definition of angular momentum
Lecture 32/Page 7