1. Chapter 11: Rotational Vectors and Angular Momentum
Vector (cross) products
Definition of vector product and its properties
Axis of rotation
unit vector normal to the plane
defined by
and

2. Vector (cross) products (cont’d)
Properties of vector product z
: unit vector in x,y,z direction y
Consider three vectors: x
Then

3. Vector (cross) products (cont’d)
Properties of vector product (cont’d) z y x

4. Torque Case for 1 point-like object of mass m with 1 force
Torque is a quantitative measure of the tendency of a force to cause
or change the rotational motion.
Case for 1 point-like object of mass m with 1 force
r is constant
only tangential component
of causes rotation y x
massless rigid rod
torque
moment of inertia
unit Nm

5. Torque Case for 1 point-like object of mass m with 1 force Definey
Define x
is aligned with the rotation axis
lever arm

6. Torque (cont’d) Case for 2 point-like objects with 2 forces increasesy
increases
decreases
Example: A see-saw in balance x M m R r mg Mg

7. Work & energy (I) W=DK A massive body on massless rigid rod
Work done by the force:
Also x
W=DK

8. Work & energy (I) (cont’d)
Power in rotational motion
Work done by the force:
Power :

9. Correspondence between linear & angular quantities
displacement
velocity
acceleration
mass
force
Newton’s law
kinetic energy
work

10. Work & energy (II) A massive body in rotational & translational motion
Kinetic energy:
Now
where
is the velocity with respect to
the center of mass (COM) and
the velocity of COM. Then
com

11. Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
kinetic energy due to
translational motion
kinetic energy due
to rotational motion
about a rotation axis
through COM

12. Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
Example: A rolling cylinder (without sliding due to friction) O O s P s P
View point 1: +
translation =
rotation O O O
This view point was used in the last few slides

13. Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
Example: A rolling cylinder (without sliding) (cont’d)
View point 2: Any point in the cylinder rotates around P
From the parallel axis theorem P
rotation axis
rotational
translational

14. Angular momentum & torque
Angular momentum of a particle y
Define angular momentum as:
angular momentum x
Since
and
lever arm
net torque

15. Angular momentum & torque
Angular momentum of a multi-particle system
As before let’s break down vectors into two components
velocity of com
Then
position vector for com
ang. mom. about COM + ang. mom. of COM

16. Conservation of angular momentum
If the net torque is zero, the total angular momentum
of the system is conserved.
Conservation of angular momentum

17. Angular momentum & torque
Angular momentum vector and axis of rotation z
plane A
plane B x
plane A
plane B
In general,
because of non-zero x/y component of
angular momentum
unless the object is symmetric about the
axis of rotation.
If the object is symmetric about the axis of rotation, then

18. Gyroscope Gyroscopic motion remove support pivot  mg precession pivot

19. Gyroscope
Principle of gyroscopic z y
pivot x 

20. Gyroscope y Assumption: The angular momentum vector
is associated only with the spin
of the flywheel and is purely
vertical.
The procession is much slower
than the rotation, x
Precession angular speed:

21. Gyroscope

22. Gyroscope

23. Spinning Top

24. Example
A disk of mass M and radius R rotates around the axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f.
1) wf = wi 2) wf = ½ wi 3) wf = ¼ wi z f z i

25. First realize that there are no external torques acting on
the two-disk system. Angular momentum will be conserved! f z i z 2 1

26. Example
You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

27. What happens to the angular momentum as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
CORRECT L1 L2

28. What happens to your angular velocity as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
CORRECT w1 w2 I2 I1 L

29. What happens to your kinetic energy as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
CORRECT
(using L = I ) w1 w2 I2 I1 L

30. Example
A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?
1. She starts to spin CCW. 2. She starts to spin CW. 3. Nothing
CORRECT

31. Since there is no net external torque acting on the student
stool system, angular momentum is conserved.
Remember, L has a direction as well as a magnitude!
Initially: LINI = LW,I
Finally: LFIN = LW,F + LS LS
LW,I
LW,I = LW,F + LS
LW,F

32. Example
A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase?
(a) (b) (c) 8 w

33. Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved.
L1 = I1w1 = mR2w1 w1 m R
mR2w1 = m R2w2
w1 = w2
w2 = 4w1

34. Example
A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed wF of the stick after the collision? (Ignore gravity) M F D m
D/4 v1 v2
initial
final

35. Set Li = Lf using M F D m
D/4
initial
final v1 v2

36. Exercises Problem 1 Find the acceleration of an object of mass m. n y
Solution R
For the object, Newton’s 2nd law gives:
ay>0 in –y direction
(1) T
For the cylinder, the total torque is: M
(2) Mg T m
The tangential acceleration of the cylinder
is equal to that of the object: mg
(3) h
(2)+(3): x
(4)

37. Exercises Problem 1 (cont’d) Solution (cont’d) n (1)+(4): y R (5) T
(4)+(5): M Mg T
The final velocity of the object when it
was at rest initially: m mg h x

38. Exercises Problem 2 Solution n
(a) The normal force on the cylinder is: y R T
(b) Compare n with (m+M)g: M
As the suspended mass is accelerating down
the tension is less than mg. Therefore
n is less than the total weight (m+M)g. Mg T m
(c) If the cylinder is initially rotating clockwise
and so that the object gets initial velocity
upward, what effect does this have on T
and n?
As long as the cable remains taut, T and n remain the same. mg h x

39. Exercises pulley Problem 3 m1 T1 M
A glider of mass m1 slides without friction on
a horizontal air track. It is attached to an
object of mass m2 by a massless string.
The pulley is a thin cylindrical shell with
mass M and radius R, and the string turns
the pulley without slipping or stretching.
Find the acceleration of each body, the
angular acceleration of the pulley, and the
tension in each part of the string. R T2
glider m2
hanging
object
Solution y x x y
glider:
(1) + x
(2) n1
object: T1 T1 y T2
pulley: m1 m2 T2
m1g
m2g
(3) Mg
no stretching and slipping: n2
(4)
glider
hanging obj.
pulley

40. Exercises
Problem 3 (cont’d)
Solution cont’d
(1)-(4):

41. Exercises Problem 4 A primitive yo-yo is made by wrapping a string
several times around a solid cylinder with mass
M and radius R. You hold the end of the string
stationary while releasing the cylinder with no
initial motion. The string unwinds but does not
slip or stretch as the cylinder drops and rotates.
Find the speed vcm of the center of mass of the
solid cylinder after it has dropped a distance h. 1 h 2
Solution
Energy conservation

42. Exercises Problem 5 (Atwood’s machine) R I
Find the linear accelerations of blocks A and B,
the angular acceleration of the wheel C, and
the tension in each side of the cord if there is
no slipping between the cord and the surface of
the wheel. C mA mB A B
Solution
The accelerations of blocks A and B will have the same magnitude .
As the cord does not slip, the angular acceleration of the pulley will be
. If we denote the tensions in cord as and , the equations
of motion are:

43. Exercises Problem 6 A solid uniform spherical boulder
starts from rest and rolls down a
50.0 m high hill. The top half of the
hill is rough enough to cause the
boulder to roll without slipping, but
the lower half is covered with ice and
there is no friction. What is the
translational speed of the boulder
when it reaches the bottom of the hill?
rough
h=50.0 m
smooth
Solution
1st half (rough):
2nd half (smooth):

44. Exercises
Problem 7
Occasionally, a rotating neutron star undergoes a sudden and
unexpected speedup called a glitch. One explanation is that a
glitch occurs when the crust of the neutron star settles slightly,
decreasing the moment of inertia about the rotation axis. A
neutron star with angular speed w0=70.4 rad/s underwent such
a glitch in October 1975 that increased its angular speed to
w=w0+Dw, where Dw/w0=2.01x10-6. If the radius of the neutron
star before the glitch was 11 km, by now how much did its
radius decrease in the starquake? Assume that the neutron
star is a uniform sphere.
Solution
Conservation of angular momentum:

45. Exercises
Problem 8
A small block with mass kg is attached to a string passing
through a hole in a frictionless, horizontal surface. The block is
originally revolving in a circle with a radius of m about the
hole with a tangential speed of 4.00 m/s. The string is then pulled
slowly from below, shortening the radius
of the circle in which the block revolves.
The breaking strength of the string is 30.0
N. What is the radius of the circle when the
string breaks.
Solution
The tension on the string:
The radius at which the string breaks is obtained from:

46. Exercises Problem 9 pivot A ball catcher whose mass is M and moment
of inertia is I is hung by a frictionless pivot.
A ball with a speed v and mass m is caught by the
catcher. The distance between the pivot point and
the ball is r which is much greater that the radius of
the ball. r v
Find the angular speed of the catcher
right after it catches the ball.
(b) After the ball is caught, the catcher –
ball system swings up as high as h.
Find the angular speed at the maximum
height h.
pivot
catcher
M,I

47. Exercises Problem 10 pivot Solution The initial angular momentum with
respect to the pivot is:
The final total moment of inertia is r v h
Since
pivot
catcher
M,I
(b) The kinetic energy after the collision
is:
potential energy at height h

48. Exercises d= 42.0 cm Problem 11
A 42.0 cm diameter wheel, consisting of
a rim and six spokes, is constructed from
a thin rigid plastic material having a linear
mass density of 25.0 g/cm. This wheel is
released from rest at the top of a hill 58.0 m
high.
How fast is it rolling when it reaches the
bottom of the hill?
(b) How would your answer change if the
linear mass density and the diameter of
the wheel were each doubled?
l= 25.0 g/cm
No sliding
h= 58.0 m

49. Exercises d= 42.0 cm Problem 11 (cont’d) l= 25.0 g/cm Solution
(a) Conservation of energy:
No sliding
h= 58.0 m
(b) Doubling the density would have no effect! As , doubling the
diameter would reduce the angular velocity by half. But would
be unchanged.