(a) In which case would the speed of the penny be greater, if it were placed at point A , or if it were
placed at point B ? Explain.
(b) At which point would the penny require the larger centripetal force to remain in place? Justify your
answer.
(c) Point B is 0.25 m from the center of rotation. If the coefficient of friction between the penny and the
turntable is μ = 0.30, calculate the maximum linear speed the penny can have there and still remain
in circular motion.Solutions to Practice Problems
1 . D— In Newton’s law of gravitation,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 = 4R 2 in the denominator here.
2 . E— In the centripetal acceleration equationthe 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 .
3 . 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.)
4 . 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.)
5 . 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.6 . (a) 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 .
(b) The coin needs more centripetal force at point B . The centripetal force is equal to mv 2 /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 = 4mπ^2 r /T 2 . Because both positions
take equal times to make a rotation, the coin with the larger distance from the center needs more