Question

A solid metal sphere starting from rest rolls down an incline If it rolls down from a helght of 8848 meters (Mt.Everest!), what is its speed at the bottom 2 yphere MR

Please explain step-by-step process THOROUGHLY.

Answer 1

When the ball was above mount everest it had potential energy which gets converted to rotational as well as transational kinetic energy on reaching the ground.

Iw mu mgh = 2. 2

We assume rolling without slipping as nothing is explicitely mentioned about the nature of rolling motion. For rolling without slipping , $v=R\omega$

$mgh=\frac{I(\frac{v}{R})^{2}}{2}+\frac{mv^{2}}{2}=\frac{Iv^{2}}{2R^{2}}+\frac{mv^{2}}{2}$

$mgh=\frac{2mR^{2}v^{2}}{5*2R^{2}}+\frac{mv^{2}}{2}=\frac{mv^{2}}{5}+\frac{mv^{2}}{2}=0.7mv^{2}$

$v=\sqrt{\frac{gh}{0.7}}=\sqrt{\frac{9.8*8848}{0.7}}= 351.95\ m/s$