Gravitation and Circular Motion Practice Problems for AP Physics B & C (page 2)

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By — McGraw-Hill Professional
Updated on Feb 10, 2011


  1. D—In Newton's law of gravitation,
  2. the distance used is the distance between the centers of the planets; here that distance is 2R. But the denominator is squared, so (2R)2 = 4R2 in the denominator here.

  3. E—In the centripetal acceleration equation
  4. the distance used is the radius of the circular motion. Here, because the planets orbit around a point right in between them, this distance is simply R.

  5. B—Consider the vertical forces acting on the rock. The rock has weight, so mg acts down. However, because the rock isn't falling down, something must counteract the weight. That something is the vertical component of the rope's tension. The rope must not be perfectly horizontal, then. Because the circle is horizontal, the centripetal force must be horizontal as well. The only horizontal force here is the horizontal component of the tension. (Gravity acts down, last we checked, and so cannot have a horizontal component.)
  6. A—Once the Ewok lets go, no forces (other than gravity) act on the rock. So, by Newton's first law, the rock continues in a straight line. Of course, the rock still must fall because of gravity. (The Ewok in the movie who got hit in the head forgot to let go of the string.)
  7. E—A circular orbit is allowed at any distance from a planet, as long as the satellite moves fast enough. At 300 km above the surface Earth's atmosphere is practically nonexistent. At 10 km, though, the atmospheric friction would quickly cause the shuttle to slow down.
    1. Both positions take the same time to make a full revolution. But point B must go farther in that same time, so the penny must have bigger speed at point B.
    2. The coin needs more centripetal force at point B. The centripetal force is equal to mv2/r. However, the speed itself depends on the radius of the motion, as shown in part (a). The speed of a point at radius r is the circumference divided by the time for one rotation T, v = 2πr /T. So the net force equation becomes, after algebraic simplification, Fnet = 4mπ2r/T2. Because both positions take equal times to make a rotation, the coin with the larger distance from the center needs more centripetal force.
    3. The force of friction provides the centripetal force here, and is equal to μ times the normal force. Because the only forces acting vertically are FN and mg, FN = mg. The net force is equal to mv2/r, and also to the friction force μmg. Setting these equal and solving for v,
    4. Plug in the values given (r = 0.25 m, μ = 0.30) to get v = 0.87 m/s. If the speed is faster than this, then the centripetal force necessary to keep the penny rotating increases, and friction can no longer provide that force.

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