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)0.70 N
B)1.4 N
C)2.8 N
D)4.2 N
E)5.6 N

Question

Quizplus · Accepted Answer

The force exerted by the bowl on the stone at the bottom can be found using the concept of circular motion. At the bottom, the stone's weight (mg) acts downward and the normal force (N) from the bowl acts upward. The net force is the centripetal force, which is $mv^2/r$, where $m$ is the mass of the stone, $v$ is its velocity at the bottom, and $r$ is the radius of the bowl. Using energy conservation, the potential energy at the top (mgh) is converted into kinetic energy ($1/2 mv^2$) at the bottom. Since $h = r$, $mgh = 1/2 mv^2$, and $v^2 = 2gh$. Substituting $v^2$ into the centripetal force equation gives $N - mg = m(2g)$, so $N = 3mg$. Given $mg = 1.4 N$, $N = 3 \times 1.4 N = 4.2 N$.

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