Answer to Question #145701 in Mechanics | Relativity for abdullah noor

Question

A bowling ball that has an 11-cm radius and a 7.2-kg mass is rolling without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill. Model the bowling ball as a uniform sphere and find h.

Answer 1

According to the law of energy conservation kinetic energy of the rolling ball is equal to potential energy of the ball on height h:

$E_{k}=E_{p}$

$\frac{m\times V^2}{2}+\frac{I\times \omega^2}{2}=m\times g\times h$

I - moment of inertia of the ball

$I=\frac{2}{5}\times m \times r^2$

where m - mass, r - radius.

$\omega$ is anglular velocity

$\omega = \frac{V}{r}$

calculating

$\omega = \frac{2}{0.11}=18.18$

$I=\frac{2}{5}\times 7.2 \times 0.11^2=0.035$

$\frac{m\times V^2+I\times \omega^2}{2}=m\times g\times h$

$h=\frac{m\times V^2+I\times \omega^2}{2\times m \times g}$

$h=\frac{7.2\times 2^2+0.035\times 18.18^2}{2\times 7.2 \times 9.8}=\frac{28.8+11.57}{141.12}=0.29m$