Question

A disc of mass M and radius R rolls in a horizontal surface and ten rolls up an inclined plane in the fig. If the velocity of the disc is v, the height to which the disc will rise be then

• A. $\frac{3v^2}{2g}$
• B. $\frac{3v^2}{4g}$
• C. $\frac{v^2}{4g}$
• D. $\frac{v^2}{2g}$

Answer 1

The correct option is B $\frac{3v^2}{4g}$

Given,

Initial velocity, $v$

Radius of gyration of disc, $k=\frac{R}{\sqrt{2}}$

Potential Energy = Translational Kinetic Energy + Rotational Kinetic Energy

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$mgh=\frac{1}{2}m{v}^2(1+\frac{k^2}{R^2})$

$h=\frac{3v^2}{4g}$

Hence, the height to which the disc will rise be $\frac{3v^2}{4g}$