Rankers Physics
Topic: Rotational Motion
Subtopic: Rolling on Inclined Plane

A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be
\[ \sqrt[]{2gh} \]
\[ \sqrt[]{\frac{7}{10}gh} \]
\[ \sqrt[]{\frac{10}{7}gh} \]
\[ \left( \sqrt[]{\frac{10}{7}} \right)gh \]

Solution:

\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \]

Solving we get,

\[ v=  \sqrt[]{\frac{10}{7}gh} \]

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