A small object of uniform density rolls up a curved surface with an initial velocity V- It reaches up to a maximum height of 3V24g w-r-t- the initial position -neglect initial potential energy- The object is a

Initial energy E_i=1/2mv^2+1/2Iω^2 Final energy E_f=mgh By using energy conservation, E_i=E_f 1/2mv^2+1/2I v^2/R^2=mg 3v^2/4g I=MR^2/2 and as we know a disc ha ...

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Byju's Answer

Standard XII

Physics

Rotational Work and Energy

A small objec...

Question

A small object of uniform density rolls up a curved surface with an initial velocity $V$ . It reaches up to a maximum height of $\frac{3 V^{2}}{4 g}$ w.r.t. the initial position [neglect initial potential energy]. The object is a

ring

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solid sphere

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hollow sphere

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disc

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Solution

The correct option is D disc
Initial energy $E_{i} = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
Final energy $E_{f} = m g h$

By using energy conservation,
$E_{i} = E_{f} \frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v^{2}}{R^{2}} = m g \frac{3 v^{2}}{4 g} I = \frac{M R^{2}}{2}$
and as we know a disc having moment of inertia as same.

Hence, the given body is a 'Disc'.

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A small object of uniform density rolls up a curved surface with an initial velocity V. It reaches up to a maximum height of 3V24g w.r.t. the initial position [neglect initial potential energy]. The object is a

The correct option is D discInitial energy Ei=12mv2+12Iω2 Final energy Ef=mgh By using energy conservation, Ei=Ef12mv2+12Iv2R2=mg3v24gI=MR22 and as we know a disc having moment of inertia as same. Hence, the given body is a 'Disc'.

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