Question

A body of mass $M$ and radius $r$, rolling on a smooth horizontal floor with velocity $v$, rolls up an irregular inclined plane up to a vertical height $\left(3v^2/4g\right)$. It's moment of inertia is $M{r}^2/x$. Find the value of $x$.

Answer 1

By conservation of energy, KE at bottom is equal to PE at top.
$KE=PE$

we get $M{v}^2/2+I{\omega }^2/2=Mgh$, ..(1)

where $I=M{r}^2/x$ ,$h=3v^2/\left(4g\right)$ ,$v=\omega r$

putting the value in equation 1 , we get

$\frac{M{v}^2}{2}+\frac{M{r}^2{\omega }^2}{2x}=\frac{Mg\times 3v^2}{4g}$

we get $x=2$.