Physics 1A
Week 5 Questions
Fall 1998
1. A flywheel rotates with constant angular velocity. Does a point on its rim have a tangential acceleration? A
radial acceleration? Are these accelerations constant in magnitude? In direction? In each case, give the
reasoning behind the answer.
2. A cylindrical body has mass M and radius R. Can the mass be distributed within the body in such a way that its
moment of inertia about its axis of symmetry is greater than MR2? Which mass distribution will give the
maximum possible value of moment of inertia?
3. If you stop a spinning raw egg for the shortest instant and then release it, the egg will start spinning again. If
you do the same to a hard-boiled egg, it will remain stopped. Try it. Explain it.
4. If the entire population of the world (5  109 persons) moved to Antarctica, would it affect the length of the
day? If yes, estimate the change in length (Me = 6  1024 kg, Re = 6.4  106 m, average human body mass =
70kg).
5. The earth moves in a circular orbit of radius 1.495  1011 m and has a mass of 5.975  1024 kg. What is the
linear speed, angular speed, and angular momentum of the earth associated with its orbital motion?
6. A uniform disk of mass 2.0 kg and radius 4.0 m is spinning at 3.0 rad/s. It falls flat onto another disk of mass
4.0 kg and radius 6.0 m that is initially not rotating. The center of the fist disk is directly above the center of
the second. After 0.2 s have passed, the disks are turning together at the same angular speed. What is that
speed? What is the average torque between the disks during the 0.2 s?
Physics 1A
Week 5 Solutions
Fall 1998
1. A point in the rim of the flywheel is moving in a circle of radius r at a constant speed v = r. As the angular
velocity is constant,  = 0 and then tan = r = 0. The radial acceleration is not zero: rad = v2/r = 2r  0.
Both components of the acceleration vector are constant. The acceleration vector is constant in magnitude, bit
not in direction.
2. No. Imaging dividing the flywheel into small masses m1, m2, m3, … at distances r1, r2, r3, … etc. from the axis.
The moment of inertia is :
I  m1 r12  m2 r22  m3 r32  
In general, each of the r's in the above equation is smaller than R, then:
I  m1 R 2  m2 R 2  m3 R 2    (m1  m2  m3  ) R 2  MR 2
The biggest possible value for I is MR2, and it can only be obtained if every piece of mass is at a distance R
from the axis. This corresponds to a hollow, thin cylinder of radius R.
3. A hard boiled egg is a rigid body. When you stop it, the angular velocity at any point is zero, the rotational
K.E. of the egg is zero, and it will not change after the egg is released. For a raw egg the situation is different.
When you "stop" it you only set the angular velocity of the white of the egg to zero, but the yolk is still
rotating, and the remaining rotational K.E. will be shared with the white once you release it. As a result, you
will see the egg spinning again.
4. Call I the moment of inertia of earth  people before the move, I' the m.o.i. after the move,  and ' the initial
and final values of the rotational angular velocity. Note that I  52 M e R 2  23 M p R 2 (people are uniformly
distributed on the Earth's surface), whereas I '  52 MRe2 (everyone lives near the pole at a nearly null distance
from the axis of rotation). By angular momentum conservation:
I ' '  I 0   ' 
I
I'
0  T ' T
,
I'
I
where T is the length of one day. Now


I'
1
 1  5M p /(3M e )  1  5M p /(3M e )
I
(as M p / M e  5 109  70 /(6 10 24 )  5.8 10 14  tiny number),
then, T '  (24  3600s)(1  5M p /(3M e )) ,
T  T '  (24  3600s)5M p /(3M e )  (24  3600s)(9.7 10 14 )  8 10 9 s ,
i.e., days would be 8  10 9 s shorter than what they currently are.
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Physics 1A
Week 5 Solutions
Fall 1998
5. The angular speed of the earth is given by  = 2/T, where T is the period. With T measured in seconds, we
get  = 1.99  10-7 rad/s. Since v = r, we can multiply our result for  times the radius to obtain v = 29,800
m/s. Note that we could also get this by dividing the circumference of the earth's orbit, 2r, by the orbital
period. We can then get the angular momentum from L = mrv = 2.66  1040 kg m2/s.
6. Angular momentum is conserved. The initial angular momentum is that of the rotating disk, (1/2)mr2, where
m = 2.0 kg, r = 4.0 m, and  = 3.0 rad/s. The final angular momentum is given by
1 2
1
1
mr  f  MR 2 f  (mr 2  MR 2 ) f
2
2
2
where M = 4.0 kg, R = 6.0 m, and f is the common angular velocity of the disks after the 0.2 s have passed.
We can therefore equate initial and final angular momentum and solve for f :
mr 2 i
1 2
1
mr  f  (mr 2  MR 2 ) f   f 
 0.55 rad/s
2
2
mr 2  MR 2
2
L 12 MR  f (0.5)(4.0)(6.0) 2 (0.55)
Then the average torque is: T 


 200 Nm .
T
T
0.2
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