Question

A table-tennis ball of radius 0.041mm rolls on the inside of a track of radius 0.448cm. If the ball starts from rest at the vertical edge of the track, what will it's speed be when it reaches the lowest point of the track, rolling without slipping?

Answer 1

Assume that $r=0.041 m$ and $R=0.448m$

$mgR=\frac{mv^2}{2}+\frac{I\omega^2}{2}+mgr=\frac{mv^2}{2}+\frac{2/5mr^2\cdot (v/r)^2}{2}+mgr=$

$=\frac{mv^2}{2}+\frac{mv^2}{5}+mgr=\frac{7}{10}mv^2+mgr$

$gR=\frac{7}{10}v^2+gr \to v=\sqrt{\frac{10}{7}g(R-r)}=\sqrt{\frac{10}{7}\cdot 9.81\cdot (0.448-0.041)}=2.39 m/s$