Physics - Rotation Question with Solution | TestHub

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

For the sphere rolling down without slipping, the conservation of mechanical energy gives:

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

Using the radius of gyration $I=m{k}^2$ and the rolling condition $\omega =\frac{v}{R}$:

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

When the sphere slides down the frictionless inclined plane without rolling, there is no rotational kinetic energy. The speed at the bottom is given as $v^{\prime }=\frac{5v}{4}$. Conservation of energy gives:

$mgh=\frac{1}{2}m(v^{\prime }{)}^2=\frac{1}{2}m{\left(\frac{5v}{4}\right)}^2=\frac{1}{2}m{v}^2\left(\frac{25}{16}\right)$

Equating the two expressions for $mgh$:

$\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)=\frac{1}{2}m{v}^2\left(\frac{25}{16}\right)$

$1+\frac{k^2}{R^2}=\frac{25}{16}$

$\frac{k^2}{R^2}=\frac{25}{16}-1=\frac{9}{16}$

$k=\frac{3R}{4}$

Answer: $\frac{3R}{4}$