acceleration, is not constant. Use conservation of energy instead. The potential energy at the release
point is mg (L /2) (L /2 because the rod’s center of mass is that far vertically above its lowest point).
This potential energy is converted entirely into rotational kinetic energy ^1 / 2 Iω 2 . The rod’s angular
velocity ω is equal to v /(L /2), where v is the linear speed of the center of mass that we’re solving
for. Plugging in, you get mgL /2 = ^1 / 2 I (v 2 /[L /2]^2 ). Solving for v , choice D emerges from the
mathematics.29 . D —Choose any point at all as the fulcrum; say, the center of mass. Rod B supports 500 N, and is
located 20 cm from the fulcrum, producing a total counterclockwise torque of 10,000 N·cm. Rod A
also supports 500 N; call its distance from the fulcrum “x ”. So 10,000 = 500x , and x = 20 cm. This
means Rod A is located 20 cm left of the center of mass, or 50 cm from the left edge.
30 . B —The period of a mass on a spring is
with the mass under the square root. So when the mass is quadrupled, the period is only multiplied
by two. The total mechanical energy is the sum of potential plus kinetic energy. At the greatest
displacement from equilibrium (i.e., at the amplitude), the mass’s speed is zero and all energy is
potential; potential energy of a spring is ^1 / 2 kx 2 and does not depend on mass. So, because the
amplitude of oscillation remains the same, the total mechanical energy does not change.31 . D —The maximum potential energy of the mass is at the amplitude, and equal to ^1 / 2 kA 2 . This is
entirely converted to kinetic energy at the equilibrium position, where the speed is maximum. So set . Solving for v (^) max , you get choice D. (Note: Only choices C and D have units ofvelocity! So guess between these if you have to!)
32 . B —The bottle’s lowest position is x = −A , and its highest position is x = +A . When t = 0, cos (0)
= 1 and the bottle is at x = +A . So, find the time when the cosine function goes to −1. This is when
ωt = π, so t = π/ω .
33 . B —Don’t try to calculate the answer by saying mg = GMm /r 2 ! Not only would you have had to
memorize the mass of the Earth, but you have no calculator and you only have a minute or so,
anyway. So think: the acceleration must be less than 9.8 m/s^2 , because that value is calculated at the
surface of the Earth, and the Shuttle is farther from Earth’s center than that. But the added height of
300 km is a small fraction ( 5%) of the Earth’s radius. So the gravitational acceleration will not be
THAT much less. The best choice is thus 8.9 m/s^2 . (By the way, acceleration is not zero—if it were,
the Shuttle would be moving in a straight line, and not orbiting.)
34 . B —The orbit can no longer be circular—circular orbits demand a specific velocity. Because the
satellite gains speed while at its original distance from the planet, the orbit is now elliptical.
Because the direction of the satellite’s motion is still tangent to the former circular path, in the next
instant the satellite will be farther from Earth than at point P , eliminating answer choice A. The
satellite will not “fly off into space” unless it reaches escape velocity, which cannot be 1% greater
than the speed necessary for a low circular orbit.