Question 514: Centre of Mass and Rotational Motion

Question

A body of mass $m$ slides down an incline and reaches the bottom with a velocity V . If the same mass were in the form of a disc which rolls down this incline, the velocity of the disc at bottom would have been

Accepted Answer

When a body of mass $ m $ slides down an incline without friction, it converts all its potential energy at the top into translational kinetic energy at the bottom. The velocity $ V $ at the bottom in this case is given by:
$$ V = \sqrt{2gh} $$
where $ g $ is the acceleration due to gravity, and $ h $ is the height of the incline.
For a disc that rolls without slipping, the energy at the bottom is distributed between translational kinetic energy and rotational kinetic energy. The total mechanical energy at the top is the same potential energy, but at the bottom, it will be divided as:
Translational kinetic energy: $ \frac{1}{2}mv^2 $
Rotational kinetic energy: $ \frac{1}{2}I\omega^2 $
For a disc, the moment of inertia $ I $ is $ \frac{1}{2}mr^2 $, and the relationship between linear velocity $ v $ and angular velocity $ \omega $ is $ \omega = \frac{v}{r} $.
Substituting these into the equations for kinetic energy, the total kinetic energy at the bottom becomes:
$$ \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 $$
The combined kinetic energy thus equals:
$$ \frac{3}{4}mv^2 $$
Equating the initial potential energy to the total kinetic energy at the bottom:
$$ mgh = \frac{3}{4}mv^2 $$
Solving for $ v $, we find:
$$ gh = \frac{3}{4}v^2 $$
$$ v^2 = \frac{4}{3}gh $$
$$ v = \sqrt{\frac{4}{3}gh} = \sqrt{\frac{4}{3}} \cdot \sqrt{gh} $$
Since $ \sqrt{gh} = \frac{V}{\sqrt{2}} $, the velocity $ v $ of the disc at the bottom becomes:
$$ v = \sqrt{\frac{4}{3}} \cdot \frac{V}{\sqrt{2}} $$
Further simplification gives the final expression:
$$ v = \frac{V}{\sqrt{3/2}} = V \cdot \sqrt{\frac{2}{3}} $$
Therefore, the correct option is
Option D: $ v \sqrt{\frac{2}{3}} $
.

Answer

$\mathrm{v} \sqrt{\frac{3}{4}}$

Answer

$\mathrm{v} \sqrt{\frac{3}{2}}$

Answer

$v \sqrt{\frac{1}{3}}$

Answer

$\mathrm{v} \sqrt{\frac{2}{3}}$

Centre of Mass and Rotational Motion

A body of mass m slides down an incline and reaches the bottom with a velocity V . If the same mass were in the form of a disc which rolls down this incline, the velocity of the disc at bottom would have been

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