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Standard VIII

Physics

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A solid spher...

Question

A solid sphere of mass $M$ and radius $r$ slips on a rough horizontal plane. At some instant it has translational velocity $v_{0}$ and rotational velocity about the centre $\frac{v_{o}}{2 r}$ . The translational velocity after the sphere starts pure rolling is $x v_{o} / y$ in forward direction. Find $(x + y)$

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Solution

Draw a diagram of given problem.

Find translational velocity after the sphere starts pure rolling.

As we know, angular momentum is conserved about lowest point.
Therefore,

$m v_{0} r + \frac{2}{5} m r^{2} \frac{v_{0}}{2 r} = m v^{'} r + \frac{2}{5} m r^{2} \frac{v^{'}}{r}$
$\frac{10 m v_{0} r + 2 m v_{0} r}{10} = \frac{7 m v^{'} r}{5}$
$v^{'} = \frac{6 v_{0}}{7}$
Hence, we get
$x = 6$ and $y = 7$
i.e., $(x + y) = 6 + 7$ $= 13$

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A solid sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about the centre vo2r. The translational velocity after the sphere starts pure rolling is xvo/y in forward direction. Find (x+y)

Draw a diagram of given problem. Find translational velocity after the sphere starts pure rolling. As we know, angular momentum is conserved about lowest point. Therefore, mv0r+25mr2v02r=mv′r+25mr2v′r 10mv0r+2mv0r10=7mv′r5 v′=6v07 Hence, we get x=6 and y=7 i.e., (x+y)=6+7=13