Question

A uniform solid disk rolling down an incline making angle $\theta$ with the horizontal. The minimum coefficient of friction required to maintain pure rolling motion for the disk is

• A. $\left(\frac{2}{3}\tan \theta \right)$
• B. $\left(\frac{1}{3}\tan \theta \right)$
• C. $\left(\frac{2}{5}\tan \theta \right)$
• D. $\left(\frac{1}{2}\tan \theta \right)$

Answer 1

Answer: (D)

Torque about the CM is caused by friction because the lever arm of the weight force is zero:
$\tau =fR=I\alpha$
$f=\mu n=\mu mg\cos \theta$
$\mu =\frac{f}{mg\cos \theta }=\frac{I\alpha /R}{mg\cos \theta }$
$=\frac{\left(\frac{2}{3}g\sin \theta \right)\left(\frac{1}{2}m{R}^2\right)}{R^{2}mg\cos \theta }=\left(\frac{1}{2}\tan \theta \right)$
As there is no slipping between any point of contact hence distance moved by the man is 2L