Question 657: Centre of Mass and Rotational Motion

Question

A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of $${{3{v^2}} \over {4g}}$$ with respect to the initial position. The object is

Accepted Answer

The kinetic energy of the rolling object is
converted into potential energy at height
$$h = \left( { {{3{v^2}} \over {4g}}} \right)$$
So by the law of conservation of 
mechanical energy, we have
$${1 \over 2}M{v^2} + {1 \over 2}I{\omega ^2} = Mgh$$  $$\left( \because{\omega  = {v \over R}} \right)$$
$${1 \over 2}M{v^2} + {1 \over 2}I{\left( {{v \over R}} \right)^2} = Mg\left( {{{3{v^2}} \over {4g}}} \right)$$
$${1 \over 2}I{{{v^2}} \over {{R^2}}} = {3 \over 4}M{v^2} - {1 \over 2}M{v^2}$$
$${1 \over 2}I{{{v^2}} \over {{R^2}}} = {1 \over 4}M{v^2}$$
$$ \Rightarrow I = {1 \over 2}M{R^2}$$
Hence, the object is disc.

Answer

hollow sphere

Answer

disc

Answer

ring

Answer

solid sphere

Centre of Mass and Rotational Motion

A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of ${{3{v^2}} \over {4g}}$ with respect to the initial position. The object is

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