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VITEEE 2019 Paper

Section: Physics

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Question : 28 of 125

Marks: +1, -0

A solid cylinder of mass m and radius R rolls down inclined plane without slipping. The speed of its C.M. when it reaches the bottom is

$\sqrt{2gh}$
$\sqrt{\frac{4}{3} g h}$
$\sqrt{\frac{3}{4} g h}$
$\sqrt{4gh}$

Solution:

By energy conservation

(\text{K.E})_i + (\text{P.E})_i = (\text{K.E})_f + (\text{P.E})_f

(\text{K.E})_i = 0, (\text{P.E})_i = mgh

,

(\text{P.E})_f = 0

(\text{K.E})_f = \frac{1}{2} I \omega^2 + \frac{1}{2} m v_{cm}^2

For solid cylinder, moment of inertia,

I = \frac{1}{2} m R^2

(\text{K.E})_i + (\text{P.E})_i = (\text{K.E})_f + (\text{P.E})_f

so,

mgh = \frac{1}{2} \left( \frac{1}{2} m R^2 \right) \left( \frac{v_{cm}^2}{R^2} \right) + \frac{1}{2} m v_{cm}^2

\Rightarrow v_{cm} = \sqrt{\frac{4}{3} g h}

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