Experiment 5: Rotational
Dynamics
and
Angular Momentum
8.01
W10D1
Young and Freedman: 10.5-10.6;
Announcements
Vote Tomorrow
Math Review Night Tuesday from 9-11 pm
Pset 9 Due Nov 8 at 9 pm
No Class Friday Nov 11
Exam 3 Tuesday Nov 22 7:30-9:30 pm
W010D2 Reading Assignment Young and Freedman:
10.5-10.6
Rotor Moment of Inertia
Review Table Problem: Moment
of Inertia Wheel
A steel washer is mounted on a
cylindrical rotor . The inner radius of the
washer is R. A massless string, with an
object of mass m attached to the other
end, is wrapped around the side of the
rotor and passes over a massless pulley.
Assume that there is a constant frictional
torque about the axis of the rotor. The
object is released and falls. As the mass
falls, the rotor undergoes an angular
acceleration of magnitude a1. After the
string detaches from the rotor, the rotor
coasts to a stop with an angular
acceleration of magnitude a2. Let g
denote the gravitational constant.
What is the moment of inertia of the rotor
assembly (including the washer) about
the rotation axis?
Review Solution: Moment of
Inertia of Rotor
Force and rotational equations
while weight is descending:
mg  T  ma
RT   f  I ra1
Constraint: a  Ra 1
Rotational equation while
slowing down
 f  I ra 2
Speeding up
Solve for moment of inertia:
Rm(g  Ra1 )  I ra 2  I ra1
 Ir 
Rm(g  Ra 1 )
(a 1  a 2 )
Slowing down
Review Worked Example:
Change in Rotational Energy and
Work
While the rotor is slowing down,
use work-energy techniques to find
frictional torque on the rotor.
t2
t2
1
1
1
K  0  I r 12     f  dt   f   dt
2
t
t
1
I r 12
 f  2t
2
  dt
t1
Experiment 5: Rotational
Dynamics
Review: Cross Product
Magnitude: equal to the area of the parallelogram defined by
the two vectors



r r
r r
r r
r
r
A  B  A B sin  A B sin  A sin B
Direction: determined by
the Right-Hand-Rule
(0     )
Angular Momentum of a Point
Particle
Point particle of mass m moving with
a velocity v
p  mv
Momentum
Fix a point S
Vector rS from the point S to the
location of the object
Angular momentum about the point
S
[kg  m 2  s-1 ]
LS  rS  p
Cross Product: Angular
Momentum of a Point Particle
Magnitude:
LS  rS  p
L S  rS p sin 
a)
moment arm
rS ,  rS sin 
L S  rS , p
b)
perpendicular
momentum
pS ,  p sin 
L S  rS p
Angular Momentum of a Point
Particle: Direction
Direction: Right Hand Rule
Concept Question:
In the above situation where a particle is moving in the x-y plane
with a constant velocity, the magnitude of the angular momentum
about the point S (the origin)
1)decreases then increases
2)increases then decreases
3)is constant
4)is zero because this is not circular motion
Table Problem: Angular
Momentum and Cross Product
A particle of mass m = 2 kg moves
with a uniform velocity
v  3.0 m  s î  3.0 m  s ĵ
-1
-1
At time t, the position vector of the
particle with respect ot the point S is
rS  2.0 m î  3.0 m ĵ
Find the direction and the magnitude
of the angular momentum about the
origin, (the point S) at time t.
Angular Momentum and Circular
Motion of a Point Particle:
Fixed axis of rotation: z-axis
r
Angular velocity    z kφ
Velocity
r r r
v    r   z kφ  R rφ R z φ
Angular momentum about the
point S
r
r r r
r
LS  rS  p  rS  mv  Rmv kφ  RmR z kφ  mR2 z kφ
Concept Question
A particle of mass m moves in a circle of
radius R at an angular speed  about the z
axis in a plane parallel to but above the x-y
plane. Relative to the origin
1. L0 is constant.
2. L 0
is constant but L 0 / L 0 is not.
3. L0 / L0 is constant but
L0
4. L0 has no z-component. .
is not.
Worked Example : Changing
Direction of Angular Momentum
A particle of mass m moves in a circle of
radius R at an angular speed  about the z
axis in a plane parallel to but a distance h
above the x-y plane.
Find the magnitude and the direction of the
angular momentum L0 relative to the
origin.
Also find the z component of L0 .
Solution: Changing Direction of
Angular Momentum
Solution: Changing Direction of
Angular Momentum
Table Problem: Angular
Momentum of a Two Particles
About Different Points
Two point like particles 1 and 2, each of mass m, are
rotating at a constant angular speed about point A. How
does the angular momentum about the point B compare to
the angular momentum about point A? What about at a
later time when the particles have rotated by 90 degrees?
Table Problem: Angular
Momentum of Two Particles
Two identical particles of mass m move in a
circle of radius R, 180º out of phase at an
angular speed  about the z axis in a plane
parallel to but a distance h above the x-y
plane.
a) Find the magnitude and the direction of
the angular momentum L0 relative to the
origin.
b) Is this angular momentum relative to the
origin constant? If yes, why? If no, why is it
not constant?
Table Problem: Angular
Momentum of a Ring
A circular ring of radius R and mass dm
rotates at an angular speed  about the zaxis in a plane parallel to but a distance h
above the x-y plane.
a) Find the magnitude and the direction of
the angular momentum L0 relative to the
origin.
b) Is this angular momentum relative to the
origin constant? If yes, why? If no, why is it
not constant?
Concept Question
A non-symmetric body rotates with constant
angular speed  about the z axis. Relative
to the origin
1. L0 is constant.
2. L 0
is constant but L 0 / L 0
is not.
3. L 0 / L 0 is constant but L 0 is not.
4. L0 has no z-component.
Concept Question
A rigid body with rotational symmetry body rotates at
a constant angular speed  about it symmetry (z)
axis. In this case
1. L is constant.
0
2. L 0 is constant but L / L is not.
0
0
3. L / L is constant but L is not.
0
0
0
4. L0 has no z-component.
5. Two of the above are true.
Angular Momentum of
Cylindrically Symmetric Body
A cylindrically symmetric rigid body rotating
about its symmetry axis at a constant angular
velocity r   z kφ
with
has
z  0
angular momentum about any point on its axis
r
Lz 
r
dLz 

body

dmr2 z kφ
body


2
   dmr   z kφ
 body

 I  kφ
z
z
Angular Momentum
for Fixed Axis Rotation
Angular Momentum about the point S
L S ,i  rS ,i  pi  (r,i rˆ  zi kˆ )  ptan,i ˆ
L  r p kˆ  z p rˆ
 ,i
S ,i
tan,i
i
tan,i
Tangential component of momentum
ptan,
 mi vtan,i  mi r,i z
i
z-component of angular momentum
about S:
LS ,z,i  r ,i ptan,i  r ,i mi r ,i  z  mi r2,i  z
i N
i N
i1
i1
2
LS ,z   LS ,z,i   mi r,i
 z  I S z
Concept Question: Angular
Momentum of Disk
A disk with mass M and radius R is spinning with angular
speed ω about an axis that passes through the rim of the disk
perpendicular to its plane. The magnitude of its angular
momentum is:
1
1.
M R2 2
4
2.
1
M R 2 2
2
3
3.
M R 2 2
2
1
4.
M R2
4
5.
1
M R2
2
3
6.
M R2
2
Kinetic Energy of Cylindrically
Symmetric Body
A cylindrically symmetric rigid body with
moment of inertia Iz rotating about its
symmetry axis at a constant angular velocity
r
   z kφ
z  0
Angular Momentum
Kinetic energy
K rot
Lz  I z z
2
L
1
 I z z2  z
2
2I z
Concept Question: Figure Skater
A figure skater stands on one spot on the ice (assumed
frictionless) and spins around with her arms extended.
When she pulls in her arms, she reduces her rotational
moment of inertia and her angular speed increases.
Assume that her angular momentum is constant.
Compared to her initial rotational kinetic energy, her
rotational kinetic energy after she has pulled in her arms
must be
1.
2.
3.
4.
the same.
larger.
smaller.
not enough information is given to decide.