PHY 206 Spring 2003 HW #10 Solutions
5.
Find A  B :
 4iˆ    6iˆ  6 ˆj   24kˆ b)  4iˆ    6iˆ  6kˆ   24 ˆj
c)  2iˆ  3 ˆj    3iˆ  2 ˆj   4kˆ  9kˆ  13kˆ
a)
12.
20.
24.
A particle travels in a circular path. (a) If its linear momentum p is doubled, how is its
angular momentum affected? (b) If the radius is doubled at constant speed, how is L
affected?
(a) Since L  r  p , doubling p will double L as well. (b) Likewise, L will be doubled.
(a) True, the net external torque is parallel to the rate of change in angular momentum
(b) False; if the net torque is zero, the rate of change of L is zero, but not necessarily L itself.
**A 2-kg speaker is lowered over a pulley (mp = 0.6 kg, r = 0.08 m) and pulls a 4-kg
amplifier. The table is frictionless (see Fig. 10-24). Find (a) the net torque about the
center of the pulley, (b) the total angular momentum after 3.5 s., (c) the pulley’s angular
momentum after 3.5 s.
To find the torque, we set up Newton’s second law, as in Example 9-7. Calling the
speaker m1 and the amplifier m2, we have three equations,
m1 g  T1  m1a
T2  m2 a
 net  T1  T2  rp  I
Using the “non-slip” condition,  = a / r, we can solve for the acceleration to find a =
3.11 m/s2 . Substituting this into the equation for the net torque, we find
1
 a 
 net   m p rp2     0.0746N-m . (b) The total angular momentum is given by
2
  rp 
v
L  rp m1v  rp m2v  I p   . Substituting v  v0  at  3.11 sm2  3.5s.  10.9 ms we
r


find the total angular momentum to be L = 5.49 kg-m2/s. (c) The angular momentum of
the pulley is L = Ip = 0.26 kg-m2/s.
26.
**The rocket is rotating at 6 rev/min and we want to slow it to a stop using the jets affixed to
the radius, R = 3 m. The jets expel 10 g/s at a speed of 800 m/s, and the moment of inertia of
the ship is I = 4000 kg-m2. What time does it take?
The most important point to realize is that the expelled mass represents a force, since
mass/sec  velocity is equivalent to p/t = F. The initial angular momentum of the ship is
 rev 2 
L  I   4000kg-m2   6 min
 800
sec 
60 min


kg-m2
s
. To change the angular momentum,
we apply a torque,  = FR = L/t. The force of each jet is F = 0.01kg/s800 m/s = 8 N
and the total torque is then  = 48 N-m. The time needed to stop the ship’s rotation is then
t = (800/48)s. = 52 s.
29.
If the angular momentum of a system is constant, (e) zero net torque acts on the system.
36.
**A man has a moment of inertia of 6 kg-m2 and a rotation speed of 1.5 rev/s. Pulling in
his arms and a pair of weights, the new moment of inertia is 1.8 kg-m2. (a) What is the new
rotation speed? (b) What is the change in K? (c) Where did the (apparent) increase in
energy come from?
For part (a) we can use conservation of angular momentum, since no external forces are
acting. Thus, I11  I 2 2 and the new rotation speed is 5 rev/s. (b) To calculate the
change in Krot we need to change the rotation speed to rad/s. Initially K i 
The final rotational kinetic energy is K f 
1
I112  266J .
2
1
I 2 22  888J , so the change is 622 J. (c) The
2
additional kinetic energy came from work done by the man’s muscles, converting chemical
energy.
38.
Two disks, of radii r and 2r are spinning at an angular speed 0 in opposite directions.
They come together and stick, ending up with the same angular velocity. Calculate that
velocity in terms of the initial angular velocity.
The angular momentum vectors have opposite directions in this problem. The total initial angular
momentum is given by Ltot  I1 0  I 2 0 
1
1
3
m  r12  r22   0  m  r 2  4r 2   0   mr 2 0 ,
2
2
2
where the direction of the angular momentum vector for the smaller disk (see Fig. 10-36) is taken
as positive. Angular momentum is conserved in the inelastic collision, so the final angular
momentum is L f   I1  I 2   f 


1
2
m r 2   2r   f . Setting the two angular momenta equal
2
3
5
gives us  f    0 . The pair starts spinning together at a lower angular speed and in the same
direction as the initial angular velocity of the larger disk.
41.
The sun rotates with a period of 25.3 days and has a radius of 6.96108 m. If the sun
collapses to become a neutron star with a radius of 5 km, what is its new rotation speed?
(This scenario might very well take place – in several million years or so.) Since there is
no external torque acting during the collapse, angular momentum is conserved. Thus
I ii  I f  f . Using  = 2/T, where T is the period, we find, with I = 2/5 MR2, that
R12 R22

T1 T2
53.
 T2  1.3  109 days  0.11ms.
**A ball collides inelastically with a rod, which then swings up together with the ball.
The length of the rod is L = 1.2 m, the ball hits at a distance d = 0.8L below the pivot
point, the mass of the rod is M = 0.8 kg and that of the ball is m = 0.3 kg. If the final
angle is  = 60, what was the initial speed of the ball?
We must use conservation of momentum in the inelastic collision, along with
conservation of energy in the swinging of the rod + ball. Starting with the latter, we
determine that the gain in potential energy of the system in swinging is
U  mgd 1  cos   mg
L
1  cos  . This is what we found for the pendulum
2
problems we worked in previous chapters. We must use L/2 for the rod, since the
potential energy change is referred to the center of mass of the object, which lies halfway
along its length. Plugging in all the numbers, we find U = 3.78 J. This must be equal to
the initial rotational kinetic energy of the ball + rod as it began its swing upward,
K rot ,i 
1
11

I tot 2   ML2  md 2   2 . From this we can solve for the angular speed,
2
23

 = 3.38 rad/s. In the previous expression, the moment of inertia is that of a rod
suspended at one end (see Table 9-1) plus that contribution due to the ball of mass m at a
distance d from the pivot. Finally, we use conservation of angular momentum, using the
definition L  r  p  L  dmv . Setting this equal to L f  I tot , using the angular
speed found previously, we arrive at v = 7.75 m/s.
69.
71.
Answer e) is correct, but the reasoning is a bit subtle. The pole has a finite radius, which
means that there is a torque acting due to the cord tension, and that torque is pointed in
the opposite direction from the angular momentum vector. Thus the angular momentum
decreases.
**A particle of mass 3 kg moves with velocity v  3 ms iˆ along the line z = 0, y = 5.3 m.
Find L relative to the origin at x = 12 m, y = 5.3 m. (b) A force F  3Niˆ is applied.
Find the torque relative to the origin.
The angular momentum is L  r  p and p  mv . Thus we can write

 

2
ˆ
L  12miˆ  5.3mˆj  3kg  3 ms iˆ  47.7 kg-m
s k . The torque is
  r  F  12miˆ  5.3mˆj    3Niˆ   15.9N-mkˆ .
73.
**An ice skater starts her pirouette with arms outstretched, rotating at 1.5 rev/s. Estimate
her rotational speed when she brings her arms flat against her body.
We have to make some assumptions. Consider the body to be of mass 50 kg total and
radius R = 15 cm, with 10 kg of the mass in the arms of length L = 70 cm. The torso we
will consider to be a cylinder rotating about the axis, I b 
cylinders rotating about one end, I a 
1
M b R 2 . The arms are
2
1
M a L2 . Putting this together, the initial total
2
moment of inertia is
1
1
2
2
 40kg  0.15m   2   5kg  0.70m   2.08kg-m 2 . After
2
3
1
2
pulling in the arms, I tot   50kg  0.15m   0.56kg-m 2 . Conservation of angular
2
I tot
  5.6 revs .
momentum gives us   
I tot
I tot  I b  I a 
79.
**Referring to Fig. 10-45, the masses attached to the cords fly out to the ends of the
cylinder after the cord breaks. Find the initial and final angular speeds and the initial and
final energies. The moment of inertia of the cylinder is I = ML2/10 and the masses are m
= 0.4 kg and M = 0.8 kg. The relevant lengths are l = 0.6 m and L = 2.0 m. The breaking
tension of the cord is T = 108 N.
Using material from Chap. 5, we have T 
mv 2 mv 2 r
l
 2  m 2 . Solving for the
r
r
2
angular speed, we find  = 30 rad/s. After the cord breaks, angular momentum must be
2
  f where I tot  I cyl  I masses
conserved, so we have I toti  I tot
2
ML2
l

 2m   and
10
2
ML2
L
  I cyl  I masses 
I tot
 2m   . Substituting all the numbers we find f = 10.5
10
2
1
rad/s. The initial rotational kinetic energy is K i  I toti2  176J and the final kinetic
2
1
  2f  61.7J .
energy is K f  I tot
2