Question

A hollow sphere is released from the top of an inclined plane of inclination θ. (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding? (b) Find the kinetic energy of the ball as it moves down a length l on the incline if the friction coefficient is half the value calculated in part (a).

Answer 1

It is given that a hollow sphere is released from the top of an inclined plane of inclination θ.
(a) To prevent sliding, the body will make only perfect rolling. In this condition, we have:

$$\begin{gathered} mgl\sin \mathrm{\theta }-f=ma...\left(1\right) \\ f\times R=\left(\frac{2}{3}\right)m{R}^2\times \left(\frac{a}{R}\right) \\ \Rightarrow f=\frac{2}{3}ma...\left(2\right) \end{gathered}$$

On putting this value in the equation (1), we get:
$mg\sin \mathrm{\theta }-\frac{2}{3}ma=ma$
$\Rightarrow a=\frac{3}{5}g\sin \mathrm{\theta }$
From equation (1), we have:

$$\begin{gathered} \mathit{}mg\mathit{}\sin \mathit{}\mathrm{\theta }\mathit{-}f\mathit{=}\frac{3}{5}\mathit{}mg\mathit{}\sin \mathit{}\mathrm{\theta } \\ \mathit{\Rightarrow }\mathit{}f\mathit{=}\frac{2}{5}\mathit{}mg\mathit{}\sin \mathit{}\mathrm{\theta } \\ \mathit{\Rightarrow }\mathit{}\mu mg\mathit{}\cos \mathit{}\mathrm{\theta }\mathit{=}\frac{\mathit{2}}{\mathit{5}}\mathit{}mg\mathit{}\mathit{sin}\mathrm{\theta } \\ \mathit{\Rightarrow }\mathit{}\mu \mathit{=}\frac{2}{5}\mathit{}\tan \mathit{}\mathrm{\theta } \end{gathered}$$

(b)
$$\begin{gathered} \left(\frac{1}{5}\right)\tan \mathrm{\theta }\left(mg\cos \mathrm{\theta }\right)R=\frac{2}{3}m{R}^2\alpha \\ \Rightarrow \alpha =\frac{3}{10}\left(\frac{g\sin \mathrm{\theta }}{R}\right) \end{gathered}$$
$$\begin{gathered} a_c=g\sin \mathrm{\theta }-\left(\frac{g}{5}\right)\sin \mathrm{\theta } \\ =\left(\frac{4}{5}\right)g\sin \mathrm{\theta } \\ \Rightarrow t^2=\frac{2l}{a_c} \\ =2l\left(4g\frac{\sin \mathrm{\theta }}{5}\right)\left(\frac{5}{2g\sin \mathrm{\theta }}\right) \end{gathered}$$

$$\begin{gathered} \therefore \omega =at \\ \mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\nu }^2+\frac{1}{2}I{\omega }^2 \\ =\frac{1}{2}m\left(2al\right)+\frac{1}{2}l\left(a^2{t}^2\right) \\ =\frac{1}{2}m\left(4g\frac{\sin \mathrm{\theta }}{5}\right)\times 2\times l+\frac{1}{2}\times \frac{2}{3}m{R}^2\times \frac{9}{100} \\ =\left(\frac{{\sin }^2\mathrm{\theta }}{R}\right)\times \left(\frac{5L}{2g\sin \mathrm{\theta }}\right) \\ =4mgl\frac{\sin \mathrm{\theta }}{5}+3mgl\frac{\sin \mathrm{\theta }}{40} \\ =\frac{7}{8}mgl\sin \mathrm{\theta } \end{gathered}$$