A small 1.4-N stone slides down a frictionless bowl, starting from rest at the rim. The bowl itself is a hemisphere of radius 75 cm. Just as the stone reaches the bottom of the bowl, how hard is the
Bowl pushing on it?
A) 2.8 N
B) 5.6 N
C) 4.2 N
D) 0.70 N
E) 1.4 N

Question

Quizplus · Accepted Answer

The force exerted by the bowl on the stone at the bottom can be found by considering the centripetal force required to keep the stone moving in a circular path. At the bottom, all the gravitational potential energy (mgh) has been converted into kinetic energy ($\frac{1}{2}mv^2$), where $h = 2r$ for a hemisphere and $r = 0.75m$. The centripetal force, which is provided by the normal force at the bottom, is equal to $mv^2/r$. Since $mg = 1.4N$ and $h = 1.5m$, we can find that the normal force (centripetal force) at the bottom is $mg + \frac{mv^2}{r} = mg + 2mg = 3mg = 3 \times 1.4N = 4.2N$, because the gravitational force still acts on the stone in addition to the centripetal force required to keep it in circular motion.

A Small 14-N Stone Slides Down a Frictionless Bowl, Starting from Rest