Question

A disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v_1$ the height to which the disc will rise will be:

Answer 1

From figure

$\frac{1}{2}M{v}_1^2+\frac{1}{2}I{\omega }^2=Mgh$

$\frac{1}{2}M{v}_1^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\left(\frac{v_1^2}{R^2}\right)=Mgh$

$\frac{1}{2}M{v}_1^2+\frac{1}{4}IM{v}_1^2=Mgh$

$\frac{3}{4}M{v}_1^2=Mgh$

$h=\frac{3v_1^2}{4g}$