Question

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an incined plane of inclination $\theta$?

• A. $\frac{2}{7}$ $\tan \theta$
• B. $\frac{1}{3}g$ $\tan \theta$
• C. $\frac{1}{2}$ $\tan \theta$
• D. $\frac{2}{5}$ $\tan \theta$

Answer 1

The correct option is A $\frac{2}{7}$ $\tan \theta$

Here we are talking about minimum coefficient of friction ,this suggests that

friction acting would be $\mu mg\cos \theta$

taking moment about centre,

$\mu mg\cos \theta \times r=I\times \alpha$

$⟹\alpha r=\frac{\mu mg\cos \theta r^2}{I}$

now from force balance,

$mg\sin \theta -\mu mg\cos \theta =ma,$

$⟹a=g\sin \theta -\mu g\cos \theta$

from condition of rolling,we know that

$a=r\alpha$

$\frac{\mu mg\cos \theta r^2}{I}=g\sin \theta -\mu mg\cos \theta ,$

simplifying this,we get

$\mu =\frac{\tan \theta }{1+\frac{m{r}^2}{I}}$

$I=\frac{2}{5}m{r}^2$

putting the value of moment of inertia in exprssion of coefficient of friction,we get

$\mu =\frac{2}{7}\tan \theta$