Physics - Rotation Question with Solution | TestHub

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

$K_{\text{total}}=Mgh$

The total kinetic energy is the sum of translational and rotational kinetic energies:

$K_{\text{total}}=K_{\text{trans}}+K_{\text{rot}}$

For a solid cylinder, the moment of inertia is $I=\frac{1}{2}M{R}^2$. For pure rolling, $v=\omega R$.

$K_{\text{rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{4}M{v}^2$

$K_{\text{trans}}=\frac{1}{2}M{v}^2=2K_{\text{rot}}$

Substituting this into the total kinetic energy equation:

$K_{\text{total}}=2K_{\text{rot}}+K_{\text{rot}}=3K_{\text{rot}}$

Since $K_{\text{total}}=Mgh$, we have:

$3K_{\text{rot}}=Mgh$

$K_{\text{rot}}=\frac{Mgh}{3}$

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