A small sphere of radius $r$ is released from point A as shown in large hemispherical bowl of radius $R$ . Sphere purely rolls. If fraction of rotational energy when sphere reaches bottom of bowl is $\frac{2}{n}$ . Value of $n$ is

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Solution

$K_{t} = \frac{1}{2} m v^{2}$

$K_{R} = \frac{1}{2} I ω^{2} = \frac{1}{5} m v^{2}$ $∵ v = r ω$

$K = k_{t} + K_{R} = \frac{7}{10} m v^{2}$

$\frac{K_{t}}{K} = \frac{5}{7}$ and $\frac{K_{R}}{K} = \frac{2}{7}$

$K = k_{t} + K_{R} = \frac{7}{10} m v^{2}$

Now, $N = m g + \frac{m v^{2}}{R - r}$ Also, $\frac{7}{10} m v^{2} = m g (R - r)$ (By energy conservation)

$N = \frac{17}{7} m g$

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