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 $\left(3v^2/4g\right)$. If the moment of inertia of the body is $\frac{M{r}^2}{k}$. Then $k$ is :

Answer 1

As the surface are smooth and the mass rolls on the horizontal surface. So, using conservation of energy,

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

where I is moment of inertia

$w=\frac{v}{r}$

$\frac{1}{2}M{v}^2+\frac{1}{2}\times \frac{M{r}^2}{k}{\left(\frac{v}{r}\right)}^2=mg\times \frac{3v^2}{4g}$

$\Rightarrow \frac{1}{2}+\frac{1}{2k}=\frac{3}{4}$

$\Rightarrow \frac{1}{2k}=\frac{3}{4}-\frac{1}{2}$

$k=2$