Question

A uniform solid cylinder of mass m and radius R is released from the top of a fixed rough inclined plane of inclination a The coefficient of friction between the cylinder and the inclined plant is $\mu =\left(1/4\right)tan\theta$ Find
(i) the time taken by the cylinder to reach the bottom.
(ii) total kinetic energy of the cylinder at the bottom
(iii) total work done by friction.

Answer 1

for pure rolling of disc minimum

$\mu =\frac{1}{3}\tan \theta$

here $\mu =\frac{1}{4}\tan \theta$

therefore there will be slipping and full frictin will be be present

$f=\frac{1}{4}\tan \theta \times mg\sin \theta =\frac{mg\sin \theta }{4}$

$a=\frac{mg\sin \theta -f}{m}=\frac{3g\sin \theta }{4}\frac{1}{2}a{t}^2=l\Rightarrow t=\sqrt{\frac{2l}{a}}=\sqrt{\frac{2l\times 4}{3g\sin \theta }}=\sqrt{\frac{8l}{3g\sin \theta }}f=\frac{mg\sin \theta }{4}$

displacement is against f

$\therefore workdonebyfriction=-\frac{lmg\sin \theta }{4}KE=\Delta u-f=mgl\sin \theta -\frac{mgl\sin \theta }{4}=\frac{3mgl\sin \theta }{4}$