A hollow sphere is released from the top of an inclined plane of inclination $\theta.$
- What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding?
- Find the kinetic energy of the ball as it moves down a length 1 on the incline if the friction coefficient is half the value calculated in part (a).
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- A hollow sphere is released from a top of an inclined plane of inclination $\theta.$ To prevent sliding, the body will make only perfect rolling. In this condition,
& torque about the centre
$\text{f}\times\text{R}=\frac{2}{3}\text{mR}^2\times\frac{\text{a}}{\text{R}}$
$\Rightarrow\text{f}=\frac{2}{3}\text{ma}\ \dots(2)$
Putting this value in equation (1) we get
$\Rightarrow\text{mg}\sin\theta-\frac{2}{3}\text{ma}=\text{ma}$
$\Rightarrow \text{a}=\frac{3}{5}\text{g}\sin\theta$
$\Rightarrow\text{mg}\sin\theta-\text{f}=\frac{3}{5}\text{mg}\sin\theta$
$\Rightarrow\text{f}=\frac{2}{5}\text{mg}\sin\theta$
$\Rightarrow\mu\text{mg}\cos\theta=\frac{2}{5}\text{mg}\sin\theta$
$\mu=\frac{2}{5}\tan\theta$
- $\frac{1}{5}\tan\theta(\text{mg}\cos\theta)\text{R}=\frac{2}{3}\text{mR}^2\alpha$
$\text{a}_{\text{c}}=\text{g}\sin\theta-\frac{\text{g}}{5}\sin\theta=\frac{4}{5}\sin\theta$
$\Rightarrow\text{t}^2=\frac{\text{2s}}{\text{a}_{\text{c}}}=\frac{\text{2l}}{\big(\frac{4\text{g}\sin\theta}{5}\big)}=\frac{5\text{l}}{2\text{g}\sin\theta}$
Again, $\omega=\alpha\text{t}$
$\text{K.E.}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2$
$=\Big(\frac{1}{2}\Big)\text{m}(\text{2as})+\Big(\frac{1}{2}\Big)\text{l}\big(\alpha^2\text{t}^2\big)$
$=\frac{1}{2}\text{m}\times\frac{4\text{g}\sin\theta}{5}\times2\times\text{l}+\frac{1}{2}\\\times\frac{2}{3}\text{mR}^2\times\frac{9}{100}\frac{\text{g}^2\sin^2\theta}{\text{R}}\times\frac{\text{5l}}{2\text{g}\sin\theta}$
$=\frac{\text{4mgl}\sin\theta}{5}+\frac{\text{3mgl}\sin\theta}{40}=\frac{7}{8}\text{mgl}\sin\theta$
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