Question

A uniform sphere of mass $m$ and radius $R$ rolls without slipping down an inclined plane set at an angle $\alpha$ to the horizontal. Find the magnitudes of the friction coefficient ($k$) at which slipping is absent.

• A. $k\ge \frac{2}{7}\tan 2\alpha$
• B. $k\ge \frac{1}{7}\tan 2\alpha$
• C. $k\ge \frac{2}{7}\tan \alpha$
• D. $k\ge \frac{1}{7}\tan \alpha$

Answer 1

The correct option is C $k\ge \frac{2}{7}\tan \alpha$

We have moment of inertia about instantaneous axis as

$I=\frac{2}{5}m{R}^2+m{R}^2=\frac{7}{5}m{R}^2$

Thus angular momentum

$L=I\omega =\frac{7}{5}m{R}^2\omega$

Now moment of the weight about instantaneous axis

$\tau =mgl=mgRsin\alpha$

Now using the relation

$\frac{dL}{dt}=\tau$

$\frac{d\left(\frac{7}{5}m{R}^2\omega \right)}{dt}=mgRsin\alpha$

$\frac{7}{5}m{R}^{2}d\omega =mgRsin\alpha dt$

Integrating both sides

$\frac{7}{5}R\omega =gtsin\alpha$

$\omega =\frac{5gtsin\alpha }{7R}$

Now angular acceleration

$\beta =\frac{\omega }{t}=\frac{5gsin\alpha }{7R}$

Now kinetic energy of the sphere

$E_k=\frac{1}{2}m{v}_c^2+\frac{1}{2}{I}_c{\omega }^2$

$E_k=\frac{1}{2}m\frac{25g^2{t}^{2}si{n}^2\alpha }{49}+\frac{1}{2}.\frac{2}{5}m{R}^2\frac{25g^2{t}^{2}si{n}^2\alpha }{49R^2}$

$E_k=\frac{5}{14}m{g}^2{t}^{2}si{n}^2\alpha$

Now as $\tau =I\beta$

$F_{r}R=\frac{2}{5}m{R}^2\times \frac{5gsin\alpha }{7R}$

$kmgcos\alpha R=\frac{2}{7}mgRsin\alpha$

$k=\frac{2}{7}tan\alpha$

Thus slipping would be absent for

$k\ge \frac{2}{7}tan\alpha$