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}} $.

Rotational Motion

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