Physics - Rotation Question with Solution | TestHub

By conservation of energy, the total kinetic energy at the bottom of the inclined plane is equal to the potential energy at the top.

$E_{total}=mgh$

For a rolling body, the total kinetic energy is the sum of translational and rotational kinetic energies.

$E_{total}=E_{trans}+E_{rot}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

For a disc, the moment of inertia is $I=\frac{1}{2}m{r}^2$ and for pure rolling, $v=r\omega$.

$E_{rot}=\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{v}{r}\right)}^2=\frac{1}{4}m{v}^2$

$E_{total}=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2$

Equating the total kinetic energy to the potential energy:

$\frac{3}{4}m{v}^2=mgh\Rightarrow m{v}^2=\frac{4}{3}mgh$

The rotational kinetic energy is:

$E_{rot}=\frac{1}{4}m{v}^2=\frac{1}{4}\left(\frac{4}{3}mgh\right)=\frac{mgh}{3}$

Answer: $\frac{\mathrm{mgh}}{3}$