Question

A hollow cylinder with a very thin wall and a block are placed at rest at the top of a plane with inclination $\theta ={45}^{\circ }$ above the horizontal. The cylinder rolls down the plane without slipping and the block slides down the plane; it is found that both objects reach the bottom of the plane simultaneously. lf the coefficient of kinetic friction between the block and the plane is $\mu$, the value of $18\mu$ is

Answer 1

Block slides down the plane so kinetic friction will work on it equal to $\mu mgcos\theta$,

on applying Newton's second law on Block, let a is block acceleration along inclined surface- $mgsin\theta -mg\mu cos\theta =ma$

$a=g(sin\theta -\mu cos\theta )$

Applying Newton's second law on rolling cylinder along inclined surface:

$mgsin\theta -f=ma\to \left(1\right)$ ; (f = static frictional force)

and

$fr=I\alpha ;(I=m{r}^2,\alpha =\frac{a}{r})$

$f=m{r}^2\frac{a}{r^2}$

so $f=ma$

put this in (1) equation:

$mgsin\theta -ma=ma$

$2ma=mgsin\theta$

$a=\frac{g}{2}sin\theta$

because both objects reach the bottom of inclined at same time so according to Newton's motion equation; $S=ut+\frac{1}{2}a{t}^2$, because u = 0, a will be same for both

$\therefore$

$\frac{g}{2}sin\theta =g(sin\theta -\mu cos\theta )$

$\frac{1}{2}sin\theta =\mu cos\theta$

$\mu =\frac{1}{2}tan\theta$ and $\theta =45$

$\mu =\frac{1}{2}$

so, $18\mu =9$