First slide

Rolling motion

Question

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h, from rest without sliding is

Moderate

A

$\sqrt{g h}$
B

$\sqrt{(\frac{g}{5}) g h}$
C

$\sqrt{(\frac{4}{3}) g h}$
D

$\sqrt{(\frac{10}{7}) g h}$

Solution

At bottom most point, total kinetic will be mgh.
Total kinetic energy $= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) {(\frac{v}{R})}^{2} = \frac{7}{10} m v^{2}$
$Now, \frac{7}{10} m v^{2} = m g h \Rightarrow v = \sqrt{\frac{10}{7} g h}$

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