Question

81. A uniform solid sphere of radius r is placed on the inside surface of a hemispherical bowl with radius R. The sphere is released from rest at an angle$\mathit{\theta }$to the vertical and rolls without slipping (Fig. P10.81). Determine the angular speed of the sphere when it reaches the bottom of the bowl.

Short answer

The sphere's angular speed is$\omega =\sqrt{\frac{10\left(Rg\right(cos\theta -1\left)\right)}{7r^2}}$

Step-by-step solution

Energy Conservation

As a result of energy conservation:

$\mathit{\Delta }\mathit{U}\mathbf{+}\mathit{\Delta }{\mathit{K}}_{\mathbf{r}\mathbf{o}\mathbf{t}}\mathbf{+}\mathit{\Delta }{\mathit{K}}_{\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{s}}\mathbf{=}\mathbf{0}$

$\Delta U$denotes potential energy change, $\Delta K_{rot}$denotes rotational kinetic energy change, and $\Delta K_{trans}$denotes translational kinetic energy change.

From the figure p10.81.

When there are treasures in the bottom of the bowl, use energy conservation,

$mg(R-r)(cos\theta -1)+[\frac{1}{2}m{v}^2-0]+\frac{1}{2}\left[\frac{2}{5}m{r}^2\right]{\omega }^2=0$

Substitute $\omega r$for.$v$

$mg(R-r)(cos\theta -1)+[\frac{1}{2}m{\left(\omega r\right)}^2-0]+\frac{1}{2}\left[\frac{2}{5}m{r}^2\right]{\omega }^2=0$

Calculating the angular speed

The angular speed of the solid sphere is calculated as,

$\begin{array}{l}\frac{1}{2}m{\omega }^2{r}^2+\frac{1}{5}m{\omega }^2{r}^2=mg(R-r)(cos\theta -1)\\ \frac{1}{2}{\omega }^2{r}^2+\frac{1}{5}{\omega }^2{r}^2=g(R-r)(cos\theta -1)\\ {\omega }^2\left(\frac{7r^2}{10}\right)=g(R-r)(cos\theta -1)\end{array}$

$\begin{array}{l}\omega =\sqrt{\frac{10\left(g\right(R-r\left)\right(cos\theta -1\left)\right)}{7r^2}}\\ \omega =\sqrt{\frac{10\left(Rg\right(cos\theta -1\left)\right)}{7r^2}}\end{array}$

As a result, the sphere's angular speed is$\omega =\sqrt{\frac{10\left(Rg\right(cos\theta -1\left)\right)}{7r^2}}$