Question

A sphere or cylinder of mass M, radius R and moment of inertia I rolls without slipping down a hill of height h, starting from rest. As explained in problem P.33, if there is no slipping $\mathbf{\omega }\mathbf{=}{\mathbf{v}}_{\mathbf{CM}}\mathbf{/}\mathbf{R}$. (a) In terms of given variables (M,R,I and h), what is ${\mathit{V}}_{\mathbf{C}\mathbf{M}}$ at the bottom of hill? (b) If the object is a thin hollow cylinder, what is ${\mathit{V}}_{\mathbf{C}\mathbf{M}}$ at the bottom of hill? (c) If the object is a uniform density hollow cylinder, ), what is${\mathit{V}}_{\mathbf{C}\mathbf{M}}$ at the bottom of hill? (d) If the object is a uniform density sphere what is ${\mathit{V}}_{\mathbf{C}\mathbf{M}}$ at the bottom of hill? An interesting experiment that you can perform that is to roll various objects down an inclined board and see how much time each one takes to reach the bottom.

Short answer

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

Step-by-step solution

Identification of given data

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

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 $h$.

Conceptual Explanation

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.

Determination of velocity of object at the bottom of hill

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

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

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 .$\sqrt{\frac{2Mgh{R}^2}{I+M{R}^2}}$