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:
$\tau=f R=I \alpha$
$ f =\mu n=\mu m g \cos \theta $
$ \mu =\frac{f}{m g \cos \theta}=\frac{I \alpha / R}{m g \cos \theta} $
$=\frac{\left(\frac{2}{3} g \sin \theta\right)\left(\frac{1}{2} m R^{2}\right)}{R^{2} m g \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