A uniform hollow hemi-sphere of mass m and radius r is released from rest in the position shown. Knowing that the hemi-sphere rolls without sliding. Determine the reaction at the horizontal surface at the instant when it has rolled through $90 ° .$

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$\frac{5 m g}{2}$

$\frac{5 m g}{3}$

$\frac{7 m g}{4}$

$\frac{8 m g}{3}$

answer is A.

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Detailed Solution

$r_{C M} = \frac{R}{2}$

$I_{1} = I_{C M} + m {(\frac{R}{2})}^{2} = \frac{2}{3} m R^{2}$

$I_{2} = I_{C M} + m {(\frac{R}{2})}^{2} = \frac{2}{3} m R^{2}$

$W = Δ K$

$m g \frac{R}{2} = \frac{1}{2} \frac{2}{3} m R^{2} ω^{2} = \frac{3 g}{2 R}$

Using $F = m a_{C M}$
$N - m g = m \frac{ω^{2} R}{2}$
$N = m g + \frac{3 m g}{4} = \frac{7 m g}{4}$

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