Physics - Rotation Question with Solution | TestHub

Let $h$ be the vertical height of the inclined plane.

For a body sliding down a smooth inclined plane, the potential energy is converted entirely into translational kinetic energy. By conservation of energy:

$mgh=\frac{1}{2}m{V}^2$

$V=\sqrt{2gh}$

For a ring rolling down the same inclined plane, the potential energy is converted into both translational and rotational kinetic energy. By conservation of energy:

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

For a ring, the moment of inertia is $I=m{R}^2$ and for pure rolling, the angular velocity is $\omega =\frac{v^{\prime }}{R}$. Substituting these values:

$mgh=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}(m{R}^2){\left(\frac{v^{\prime }}{R}\right)}^2$

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

$mgh=m{v}^{\prime 2}$

$v^{\prime }=\sqrt{gh}$

Comparing the two velocities:

$v^{\prime }=\frac{\sqrt{2gh}}{\sqrt{2}}=\frac{V}{\sqrt{2}}$

Answer: $\frac{\mathrm{V}}{\sqrt{2}}$