91Ó°ÊÓ

The angular velocity of object for rolling without slipping is$\mathrm{Ó\neg }=\frac{v_{CM}}{R}$

The mass of sphere or cylinder is $M$.

The radius of sphere or cylinder is $R$.

The moment of inertia of sphere or cylinder is $I$.

The height of hill is $h$.

The conservation of mechanical energy between top and bottom of the hill is used to find the speed of object at the bottom of hill.

Apply the conservation of mechanical energy between top and bottom of hill.

$\frac{1}{2}M{u}^2+Mgh=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I{\mathrm{Ó\neg }}^2$

Here, $u$ is the speed of the object at top of hill and its value is 0 because object is at rest at top of hill.

Substitute all the values in the above equation.

$\begin{array}{rcl}\frac{1}{2}M{\left(0\right)}^2+Mgh& =& \frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I{\left(\frac{v_{CM}}{R}\right)}^2\\ Mgh& =& \frac{1}{2}M{v}_{CM}^2+\frac{1}{2}I\left(\frac{v_{CM}^2}{R^2}\right)\\ v_{CM}^2& =& \frac{2Mgh{R}^2}{I+M{R}^2}\\ v_{CM}& =& \sqrt{\frac{2Mgh{R}^2}{I+M{R}^2}}\\ & & \end{array}$

Therefore, the speed of the sphere or cylinder at the bottom of hill is .$\sqrt{\frac{2Mgh{R}^2}{I+M{R}^2}}$