First Examination
Physics 104
21 September 1982

#1.  A uniform disk (mass 0.60 kg, radius 0.15 m) is mounted at the center of a (massless) axle which is 0.30 m long, to make a top as shown in the sketch.  A pair of equal and opposite forces, F = 0.2 N, act at the ends of the axle (along the +x and - x directions) while the top is spinning at 90 rev/s, as shown.  the axle of the top lies along the y-axis.  Referring to the coordinate system shown in the sketch, the angular speed of precession has the direction given by   


Answer:
a) +j     b) -k    c) +i     d) -j     e) +k    

#2.  In problem #1 above, the angular speed of precession of the top (in rad/s) is most  nearly equal to
Answer:

a) 0.0988     b) 0.0157     c) 0.00250     d) 0.621     e) zero

#3. A simple pendulum takes exactly 0.800 seconds to complete one oscillation.  Its frequency is most nearly equal to
Answer:

a) 0.800 seconds      b) 1.25 rad/s      c) 1.25 Hz       d) 7.85 Hz      e) 0.199 rad/s     

#4.   The differential equation of motion for a particular simple harmonic oscillator is

 d2x / dt2 + [ 0.5625 / sec2 ] x = 0  ,

where x is measured in cm and t in seconds.  The period (in seconds) of the oscillator is most nearly equal to
Answer:

a) 4.7      b) 9.4      c) 0.75      d) 1.5      e) 8.4

#5. A mass M undergoes SHM according to the equation

x = A  cos [ w t + j  ]

The total mechanical energy of the system is equal to

Answer:

a) M w2 A     b) M A2 / 2    c)  M(wA)2 / 2   d) MwA2 / 2    e)  Mw2 / 2

#6. A mass M is on a frictionless horizontal surface, connected by three springs having restoring force constants as shown in the sketch.  When set into SHM, the period of the system is.

Answer:
a) 2p Ö(M/k)    b) 2p Ö(2M/5k)     c) 2p Ö(2M/k)    d) 2p Ö(M/2k)  e) 2p Ö(2M/3k)

#7.  The Lissajous Figure shown in the sketch is the result of combining the two SHM's

x = Ax cos (wx t )           and             y = Ay cos ( wy t + d )

From the figure, the ratio  wx/wy  is equal to

Answer:
a) 1 / 2     b) 2 / 1       c)1 / 6         d)   1 / 3     e)  3 / 1

#8. A solid sphere is suspended at the end of a vertical wire attached to the ceiling.  The sphere is twisted through an angle of  p/2 radian and then released so that it undergoes torsional SHM.  If its period is 2p seconds, then its maximum angular speed (in rad/s) is most nearly equal to


Answer:
a) p/2       b) 2p     c) p       d) p2     e) 0.25

#9. It is true that, in the strictest sense,
Answer:

a) inertial mass is that used in Newton's law of universal gravitation.
b) Newton's second law involves inertial mass.
c) Newton's third law involves gravitational mass.
d) Newton's first law involves gravitational mass.
e) none of the above.

#10. The mean distance of Venus from the sun is 1.08 ´ 109 m, while that of earth is 1.50 ´  109 m. The number of earth years required for Venus to complete one revolution around the sun is most nearly equal to
Answer:
a) 0.61    b) 0.72   c)  0.52   d) 1.64   e) 1.35

#11.  A simple pendulum on earth's surface has a period of 1.80 sec.  The same pendulum is found to have a period of 2.00 sec when operated on a strange planet.  The gravitational field at the strange planet's surface has a magnitude most nearly equal to
Answer
a) 8.82 m/sec2    b) 12.1 N / kg    c) 10.9 N / kg   d) 7.93 N / kg  e) 9.8 m/sec2

#12. A projectile (mass m) leaves earth's surface with an initial speed of v = Ö(gR), straight up, where R is earth's radius and g is the gravitational field at earth's surface, as usual.  the maximum distance that the projectile reaches from earth's center is equal to
Answer:

a) 3R / 2    b) 4 R  c) 2 R    d) 3 R    e) Ö2 R

#13. A planet has  mass M distributed uniformly throughout its volume, a sphere of radius R.  A small tunnel is bored through the planet along a diameter, through its center.  Assume the planet does not spin.  A ball of mass m is released from rest at one end of the tunnel so that it falls toward the planet's center.  At the instant it is at radius r from the center, the force acting on the ball is equal to
Answer:
a) -GMm / r2    b) -GMm r / R3    c) -GMm / R2   d) -GMm /(R- r)2   e) -GMm r / R2

#14. A spaceship (mass m) is on the surface of a planet (mass M, radius R) which moves in a circular orbit (radius Rs) about a star (mass Ms).  The planet does not spin.  In order for the spaceship to escape the gravitational fields of both the planet and the star, it must use an amount of energy equal to
Note:  Assume m << M << Ms, and take into account of the kinetic energy of the spaceship as it moves with the planet around the star.


Answer:

a) Gm M/R  b) Gm[M/R + Ms/Rs]   c) Gm [M/R + Ms/(2R)s]  d)Gm Ms/Rs   e) Gm/2 [M/R + Ms/Rs] 

#15. A mass of 0.20 kg is connected to the bottom end of a vertically held, unstretched light spring with a force constant of 5.0 N/m.  The mass is then released from rest.   Take the X-direction to be positive vertically up, and the initial position of the mass to be at X = 0.  then the position (in m) of the mass exactly 2.500 seconds after its release from rest is most nearly equal to
Answer:
a) + 0.392    b) - 0.784    c) - 0.400    d) - 0.392    e) zero