Solve the following :
Question:

A small spherical ball is released from a point at a height $h$ on a rough track shown in figure. Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.

Solution:

By energy conservation,

$m q h=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$

$\mathrm{mgh}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right)\left(\frac{v^{2}}{R^{2}}\right)+\frac{1}{2} m v^{2}$

$v=\sqrt{\frac{10 h}{7}}$

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