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.

Answer

A small spherical ball is released from a point at a height on a rough track & the sphere does not slip. Therefore potential energy it has gained w.r.t the surface will be converted to angular kinetic energy about the centre & linear kinetic energy. Therefore $\text{mgh}=\Big(\frac{1}{2}\Big)\text{l}\omega^2=\Big(\frac{1}{2}\Big)\text{mv}^2$$\Rightarrow\text{mgh}=\frac{1}{2}\times\frac{2}{5} \text{mR}^2\omega^2+\Big(\frac{1}{2}\Big)\text{mv}^2$
$\Rightarrow\text{gh}=\frac{1}{5}\text{v}^2+\frac{1}{2}\text{v}^2$
$\Rightarrow\text{v}^2=\frac{10}{7}\text{gh}$
$\Rightarrow\text{v}=\sqrt{\frac{10}{7}\text{gh}}$

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