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 $v$, the height to which the disc will rise will be:

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

Answer 1

Answer: (B)

$\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2}= mgh$
$\frac{1}{2} m v^{2}+\frac{1}{2} \frac{M R^{2}}{2} \frac{v^{2}}{R^{2}}=m g h$
$\frac{3}{4} m v^{2}=mgh$
$h=\frac{3 v^{2}}{4g}$