A disc of mass M and radius R rolls in a horizontal surface and then rolls up an inclined plane as s

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6.System of Particles and Rotational Motion

hard

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

A

$\frac{{3{v^2}}}{{2g}}$

B

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

C

$\frac{{{v^2}}}{{4g}}$

D

$\frac{{{v^2}}}{{2g}}$

Solution

Initial velocity, $v$

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

Potential Energy $=$ Translational Kinetic Energy $+$ Rotational Kinetic Energy

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

$m g h=\frac{1}{2} m v^{2}\left(1+\frac{k^{2}}{R^{2}}\right)$

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

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

Standard 11

Physics

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