Download Kinematics Multiples

Angular Momentum
From Princeton Review Book
1. An ice skater is rotating with an angular velocity . If she extends her arms and leg
outward to increase her rotational inertia by a factor of 2, what will her new angular
velocity be?
a. /4
b. /2
c.
d.

2
2
e. 2 
*B. Conserve angular momentum (L)
Io o = If f
Io o = (2Io) f
Solving for f yields: /2
2. Which of the following conditions will ensure that angular momentum is conserved?
I. Conservation of linear momentum.
II. Zero net external force.
III. Zero net external torque.
a. I and II only.
b. I and III only.
c. II and III only.
d. II only.
e. III only.
*E. These two concepts are linked.
Zero net external force guarantees that linear momentum is conserved.
Here is an example of a system for which linear
momentum would be conserved, but NOT angular
momentum because there are external forces which
cause torque, but add to zero. If you strike a rod as
shown (bird’s eye view), it will rotate but not
translate. Thus, it has not gained linear momentum,
but it has gained angular momentum.
From Old AP’s
3. (1984) A particle of mass m moves with a constant speed v along the dashed line y = a.
When the x-coordinate of the particle is xo, the magnitude of the angular momentum of
the particle with respect to the origin of the system is:
a. zero.
b. mva.
c. mvxo
d. mv x02  a 2
e.
mva
x02  a 2
*B. Remember that L = rmvsin.
Do not be confused by the fact that the object is not directly above the origin as it would
be if it were a force vector and we were calculating torque. The “lever arm” is still “a”
because when the particle IS above the origin, it will be a distance “a” away from the
origin and V will be perpendicular
SO: L = rmvsin = (a)(m)(v)sin90=mva
4. (1984) A cylinder rotates with constant angular acceleration about a fixed axis. The
cylinder’s moment of inertia about the axis is 4 kg m2. At time t = 0 the cylinder is at
rest. At time t = 2 seconds its angular velocity is 1 radian per second. What is the
angular momentum of the cylinder at time t = 2 seconds?
a. 1 kgm2/s
b. 2 kgm2/s
c. 3 kgm2/s
d. 4 kgm2s


e. It cannot be determined without knowing the radius of the cylinder
DDon’t forget that there are TWO formulas for angular momentum. The cylinder is not
a “blob” whose angular momentum is rmvsin. It is a solid object.
L = I  = ( 4 kg m2) (1 r/s) = 4 kgm2/s
5. A disk rotates freely with angular velocity on frictionless bearings, as shown above.
A second identical disk Y, initially not rotating, is placed on X so that both disks rotate
together without slipping. When the disks are rotating together, which of the following is
half of what it was before?
a. Moment of inertia of X.
b. Moment of inertia of Y.
c. Angular velocity of X.
d. Angular velocity of Y.
e. Angular momentum of both disks.
* C. This is similar to a linear inelastic collision of two bodies with the same mass.
Because the forces (kinetic friction) which speeds disk Y up and slows disk X down until
they have the SAME speed (so there is no more kinetic friction) are internal, angular
momentum is conserved. Since the rotational inertia (what the problem is calling moment
of inertia) has doubled, the rotational velocity has halved, in order to keep the total
angular momentum a constant.
6. (1974) A mass M on the end of a string moves in a circle on a horizontal frictionless
table as shown above. If the tension in the string increases, the:
a. angular momentum of M remains constant.
b. angular momentum of M increases.
c. angular momentum of M decreases.
d. kinetic energy of M remains constant.
e. kinetic energy of M decreases.
* A. Even though the string is an external force, it is acting radially inward, and so it
causes no net torque which means that the angular momentum must remain constant. The
radius will decrease and the speed will increase and the total angular momentum = RMV
will remain constant. Because the force of the string and radially inward movement of the
mass are in the same direction, the force does positive work, increasing the KE of the
mass.
7. (1993) The rigid body shown in the diagram above consists of a vertical support post
and two horizontal cross-bars with spheres attached. The masses of the spheres and the
lengths of the crossbars are indicated in the diagram. The body rotates about a vertical
axis along the support post with constant angular speed w. If the masses of the support
post and the crossbars are negligible, what is the ratio of the angular momentum of the
two upper spheres to that of the two lower spheres?
a. 2/1
b. 1/1
c. ½
d. ¼
e. 1/8
*E.
Ltop  I  (mL 2  mL 2 )  2mL 2


LBottom  I  (2m)(2L) 2  (2m)( 2 L) 2   (16mL2 )
Ratio :
2mL2 1

16mL2 8
8. (1984) An ice skater is spinning about a vertical axis with arms fully extended. If the
arms are pulled in closer to the body, in which of the following ways are the angular
momentum and kinetic energy of the skater affected?
a.
b.
c.
d.
e.
Angular Momentum
Kinetic Energy
Increases
Increases
Remains constant
Remains constant
Decreases
Increases.
Remains constant
Increases.
Remains constant
Remains constant
* C
Because the skater’s muscles are internal forces, the angular momentum must remain
constant.
However, the skater’s muscles are nonconservative forces. The force of the muscles is in
the same direction as the radially inward motion of the arms. This means that the KE of
the skater increases. This is why skaters, diver, and gymnasts have such strong upper
bodies—it takes work to make themselves rotate faster.
9. (1974) An ice skater starts a spin with kinetic energy of 12 I 0 02 . As she pulls her
arms in, her moment of inertia decreases to 13 I 0 . Her angular speed then becomes:
a.
b.
0
3
0
3
c.  0
d.
3 0
e. 3  0
* E. It is a trick that they tell you the KE of the ice skater. Since her mass distribution is
changing, it is a conservation of angular momentum problem.
I 0 0  I f  f
I 0 0 
13 I 0  f
 f  3 0
10.
(2004, 38%)
A long board is free to slide on a sheet of frictionless ice. As shown in the top view
above, a skater skates to the board and hops onto the end, causing the board to slide and
rotate. In this situation, which of the following occurs?
a. Linear momentum is converted to angular momentum.
b. Kinetic energy is converted to angular momentum.
c. Rotational kinetic energy is conserved.
d. Translational kinetic energy is conserved.
e. Linear momentum and angular momentum are both conserved.
*E
A is not true--linear momentum is conserved--so the linear momentum of the skater
equals the linear momentum of the skater/board combination.
B is not true--Kinetic energy is not conserved. Some is lost. Energy can never turn into
momentum
C is not true--The skater had no rotational energy to begin with.
D is not true--Since some of the initial translational energy becomes rotational energy, it
is not conserved.
E is true. Linear momentum is conserved--no outside forces. Angular momentum is
conserved--no outside torques.