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A mass m of radius r is rolling horizontally without any slip with a linear speed v. It then rolls upto a height given by $\frac{3}{4} \frac{v^{2}}{g}$

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the body is identified to be a disc or a solid cylinder

the body is a solid sphere

moment of inertia of the body about instantaneous axis of rotation is $\frac{3}{2} m r^{2}$

moment of inertia of the body about instantaneous axis of rotation is $\frac{7}{5} m r^{2}$

answer is A, C.

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Detailed Solution

$\frac{K_{R}}{K_{T}} = \frac{I}{m r^{2}} ∴ K_{T} = (\frac{m r^{2}}{I + m r^{2}}) K_{T o t a l} o r K_{T o t a l} = (\frac{I + m r^{2}}{m r^{2}}) (\frac{1}{2} m v^{2}) = m g h = m g (\frac{3}{4} \frac{v^{2}}{g})$

Solving, we get $I = \frac{1}{2} m r^{2}$

So, it is either solid cylinder or disc.

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