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Chapters 10, 11
Rotation and angular momentum
Rotation of a rigid body
• We consider rotational motion of a rigid body about
a fixed axis
• Rigid body rotates with all its parts locked together
and without any change in its shape
• Fixed axis: it does not move during the rotation
• This axis is called axis of rotation
• Reference line is introduced
Angular position
• Reference line is fixed in the body, is perpendicular
to the rotation axis, intersects the rotation axis, and
rotates with the body
• Angular position – the angle (in radians or degrees)
of the reference line relative to a fixed direction (zero
angular position)
Angular displacement
• Angular displacement – the change in angular
position.
• Angular displacement is considered positive in the
CCW direction and holds for the rigid body as a
whole and every part within that body
   f   i
Angular velocity
• Average angular velocity
avg
 f  i



t f  ti
t
• Instantaneous angular velocity – the rate of change
in angular position
 d
  lim

t 0 t
dt
Angular acceleration
• Average angular acceleration
 avg
 f  i



t f  ti
t
• Instantaneous angular acceleration – the rate of
change in angular velocity
 d
  lim

t 0 t
dt
Rotation with constant angular
acceleration
• Similarly to the case of 1D motion with a constant
acceleration we can derive a set of formulas:
Chapter 10
Problem 6
A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular
speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant
angular acceleration of the wheel?
Relating the linear and angular
variables: position
• For a point on a reference line at a distance r from
the rotation axis:
s  r
• θ is measured in radians
Relating the linear and angular
variables: speed
ds d (r )
d

r
s  r v 
 r
dt
dt
dt
• ω is measured in rad/s
• Period (recall Ch. 4)
2r 2
T

v

Relating the linear and angular
variables: acceleration
dv d (r )
d
at 

r
 r
dt
dt
dt
• α is measured in rad/s2
• Centripetal acceleration (Ch. 4)
v
(r )
2
 r
ac 

r
r
2
2
Rotational kinetic energy
• We consider a system of particles participating in
rotational motion
• Kinetic energy of this system is
2
i i
mv
K 
2
i
• Then
mv
mi (i ri )
K 

2
2
i
i
2
i i
2


2
2
 m (r )
i
i
i
2
Moment of inertia
• From the previous slide
K

2
2
 m (r )
i
i
i
• Defining moment of inertia (rotational inertia) as
I   mi (ri )
2
i
• We obtain for rotational kinetic energy
I
K
2
2
2
Moment of inertia: rigid body
• For a rigid body with volume V and density ρ(V) we
generalize the definition of a rotational inertia:
I

r
dm
r

dV


2
2
volume
• This integral can be calculated for different shapes
and density distributions
• For a constant density and the rotation axis going
through the center of mass the rotational inertia for 9
common body shapes is given in Table 10-2 (next
slide)
Moment of inertia: rigid body
Moment of inertia: rigid body
• The rotational inertia of a rigid body depends on the
position and orientation of the axis of rotation relative
to the body
Parallel-axis theorem
• Rotational inertia of a rigid body
with the rotation axis, which is
perpendicular to the xy plane and
going through point P:
I
r

dV


2
volume
r
dm

2
volume
• Let us choose a reference
frame, in which the center of
mass coincides with the origin
Parallel-axis theorem
I   r dm   [( x  a )  ( y  b) ]dm
2
2
2
  ( x 2  y 2 )dm   (a 2  b 2 )dm
 2a  xdm  2b  ydm

rcom 




 r dm / M

 iˆ  xdm  ˆj  ydm / M  0
Parallel-axis theorem
I   r dm   [( x  a )  ( y  b) ]dm
2
2
2
  ( x 2  y 2 )dm   (a 2  b 2 )dm
  ( R 2 )dm
R
  (h 2 )dm
 I CM  Mh 2
I  I CM  Mh
2
Parallel-axis theorem
I  I CM  Mh
2
Chapter 10
Problem 22
Rigid rods of negligible mass lying along the y axis connect three particles.
The system rotates about the x axis with an angular speed of 2.00 rad/s. Find
(a) the moment of inertia about the x axis and the total rotational kinetic energy
and (b) the tangential speed of each particle and the total kinetic energy. (c)
Compare the answers for kinetic energy in parts (a) and (b).
Torque
• We apply a force at point P to a rigid body that is
free to rotate about an axis passing through O
• Only the tangential component Ft = F sin φ of the
force will be able to cause rotation
Torque
• The ability to rotate will also depend on how far from
the rotation axis the force is applied
• Torque (turning action of a force):
  ( Ft )(r )  ( F sin  )(r )
• SI unit: N*m (don’t confuse with J)
Torque
• Torque:
  ( Ft )(r )  ( F sin  )(r )  ( F )( r sin  )
• Moment arm: r┴=
r sinφ
• Torque can be redefined as:
force times moment arm
τ = F r┴
Newton’s Second Law for rotation
• Consider a particle rotating under the influence of a
force
• For tangential components
  Ft r  mat r  m(r )r  (mr 2 )  I
  I
• Similar derivation for rigid body
Newton’s Second Law for rotation
  I
   i
i
Chapter 10
Problem 39
An electric motor turns a flywheel through a drive belt that joins a pulley on the
motor and a pulley that is rigidly attached to the flywheel. The flywheel is a
solid disk with a mass of 80.0 kg and a diameter of 1.25 m. It turns on a
frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. The
tension in the upper (taut) segment of the belt is 135 N, and the flywheel has a
clockwise angular acceleration of 1.67 rad/s2. Find the tension in the lower
(slack) segment of the belt.
Rotational work
• Work
dW  Ft ds  Ft rd  d
f
W   d
i
• Power
dW d
 
P

dt
dt
• Work – kinetic energy theorem
I
I
K 

W
2
2
2
f
2
i
Corresponding relations for
translational and rotational motion
Smooth rolling
• Smooth rolling – object is rolling without slipping or
bouncing on the surface
• Center of mass is moving at speed vCM
• Point P (point of momentary contact between two
surfaces) is moving at speed vCM
s = θR
ds/dt = d(θR)/dt = R dθ/dt
vCM = 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 vCM
Rolling: translation and rotation
combined
I CM 
Mv
K

2
2
2
2
CM
Chapter 10
Problem 53
A cylinder of mass 10.0 kg rolls without slipping on a horizontal surface. At a
certain instant its center of mass has a speed of 10.0 m/s. Determine (a) the
translational kinetic energy of its center of mass, (b) the rotational kinetic
energy about its center of mass, and (c) its total energy.
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) = 2vCM
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
aCM = α R
fs opposes tendency to slide
Rolling down a ramp
Fnet,x = M aCM,x
fs – M g sin θ = M aCM,x
R fs = ICM α
α = – aCM,x / R
fs = – ICM aCM,x / R2
aCM,x
 g sin 

2
1  I CM / MR
Rolling down a ramp
aCM,x
 g sin 

2
1  I CM / MR
Vector product of two vectors
• The result of the vector (cross) multiplication of two
vectors is a vector
  
a b  c
• The magnitude of this vector is
c  ab sin 

• Angle φ is the smaller of the two angles between b

and a
Vector product of two vectors

• Vector c is perpendicular
to the plane that contains

vectors a and b and its direction is determined by
the right-hand rule
• Because of the right-hand rule, the order of
multiplication is important (commutative law does not
apply)
 
 
b  a  ( a  b )
• For unit vectors
iˆ  iˆ  0 
iˆ  ˆj  k̂
ˆj  ˆj  kˆ  kˆ
ˆj  kˆ  iˆ kˆ  iˆ  ˆj
Vector product in unit vector notation
 
a  b  (axiˆ  a y ˆj  az kˆ)  (bxiˆ  by ˆj  bz kˆ)
axiˆ  bxiˆ  axbx (iˆ  iˆ)  0
a x iˆ  by ˆj  axby (iˆ  ˆj )  axby kˆ
 
a  b  (a y bz  by az )iˆ 
 (az bx  bz ax ) ˆj  (axby  bx a y )kˆ
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
Chapter 11
Problem 15
A particle of mass m moves in a circle of radius R at a constant speed. The
motion begins at point Q at time t = 0. Determine the angular momentum of the
particle about point P as a function of time.
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
dt
n
n

dL 
  net
dt
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
Chapter 11
Problem 50
A projectile of mass m moves to the right with a speed v. The projectile strikes
and sticks to the end of a stationary rod of mass M, length d, pivoted about a
frictionless axle through its center. (a) Find the angular speed of the system
right after the collision. (b) Determine the fractional loss in mechanical energy
due to the collision.
Answers to the even-numbered problems
Chapter 10
Problem 4
− 226 rad/s2
Answers to the even-numbered problems
Chapter 10
Problem 16
(a) 54.3 rev;
(b) 12.1 rev/s
Answers to the even-numbered problems
Chapter 10
Problem 26
11mL2/12
Answers to the even-numbered problems
Chapter 10
Problem 32
168 N⋅m clockwise
Answers to the even-numbered problems
Chapter 10
Problem 34
(a) 1.03 s;
(b) 10.3 rev
Answers to the even-numbered problems
Chapter 10
Problem 48
276 J
Answers to the even-numbered problems
Chapter 11
Problem 4
(a) 168°;
(b) 11.9° principal value;
(c) Only the first is unambiguous.
Answers to the even-numbered problems
Chapter 11
Problem 12
(− 22.0 kg⋅m2/s)ˆk