A thin spherical shell lying on a rough horizontal surface is hit… - Vidyadip

Question

A thin spherical shell lying on a rough horizontal surface is hit by a cue in such a way that the line of action passes through the centre of the shell. As a result, the shell starts moving with a linear speed v without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface.

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Answer

The shell will move with a velocity nearly equal to v due to this motion a frictional force well act in the background direction, for which after some time the shell attains a pure rolling. If we consider moment about A, then it will be zero.

Therefore, Net angular momentum about A before pure rolling = net angular momentum after pure rolling.
Now, angular momentum before pure rolling about A = M(V × R) and angular momentum after pure rolling:
$\Big(\frac{2}{3}\Big)\text{MR}^2\times\Big(\frac{\text{V}_0}{\text{R}}\Big)+\text{MV}_0\text{R}$
($V_0$ = velocity after pure rolling)
$\Rightarrow\text{MVR}=\frac{2}3{}\text{MV}_0\text{R}+\text{MV}_0\text{R}$
$\Rightarrow\Big(\frac{5}{3}\Big)\text{V}_0=\text{V}$
$\Rightarrow\text{V}_0=\frac{3\text{V}}{5}$

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