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A non uniform ball of mass M and radius R, rolls from rest down a ramp and onto a circular loop of radius r. The ball is initially at height h above the bottom of loop. At the bottom of loop the normal force on ball is twice its weight. Moment of inertia of ball is given by $I = β M R^{2}$ . Find $β$ (Assume h>>R and r>>R) (centre of mass of ball will lie at it’s centre)

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$\frac{h}{R}$

$\frac{2 h}{r}$

$\frac{2 h}{r} - 1$

$\frac{2 h}{r} + 1$

answer is C.

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Detailed Solution

$2 M g - M g = \frac{M V^{2}}{r} a n d M g h = \frac{1}{2} M V^{2} + \frac{1}{2} I ω^{2}$

$V = R ω \Rightarrow β = \frac{2 h}{r} - 1$

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