Question

A uniform solid cylinder of mass $m$ and radius $R$ is released from the top of a fixed rough inclined plane of inclination $\theta$. The coefficient of friction between the cylinder and the inclined plane 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

As the cylinder is sliding down the inclined plane, force of friction F acts up the plane. If a is acceleration produced in the block, then net force on the cylinder down the plane is,

$f=ma=mg\sin \theta -F$

$=mg\sin \theta -\mu R$

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

$=mg(\sin \theta -\mu \cos \theta )$

or $a=g(\sin \theta -\frac{1}{4}\times \tan \theta \times \cos \theta )$

$a=g(\sin \theta -\frac{1}{4}\times \sin \theta )=\frac{3}{4}g\sin \theta =7.5\sin \theta$

Now,

$u=0$

(i) Using, $s=ut+\frac{1}{2}a{t}^2$

$l=0+\frac{1}{2}\times 7.5\sin \theta \times t^2$

$t^2=\frac{2l}{7.5\sin \theta }$

$t=\sqrt{\frac{l}{3.75\sin \theta }}$

Again ,

$v^2-u^2=2as$

$v^2-0=2\times 7.5\sin \theta \times l$

$v=3.8\sqrt{l\sin \theta }$

(ii) Kinetic energy $=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times m\times 2\times 7.5\sin \theta \times l$

$=7.5ml\sin \theta$

Now again,

$N-mg\cos \theta =0$

(iii) The work done by the friction force is then

$W={\int }_0^1{F}_{x}dx=-\mu \times N\times l$

$=-\frac{1}{4}\tan \theta \times mg\cos \theta \times l$

$=-\frac{1}{4}mgl\sin \theta$