Question

A hollow cylinder with a very thin wall and a block are placed at the top of a plane with inclination $\theta ={45}^0$ 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. If the coefficient of kinetic friction between the block and the plane is $\mu$, find the value of 18$\mu$

Answer 1

$mgsin\theta -f=ma=m(gsin\theta -\mu gcos\theta )$

$fR=m{R}^2\alpha$

$=>f=ma$

Acceleration of the cylinder:

$a_{cylinder}=\frac{g\sin \theta }{1+\frac{I}{m{R}^2}}=\frac{g\sin \theta }{1+\frac{m{R}^2}{m{R}^2}}=\frac{g\sin \theta }{2}$

$a_{block}=g\sin \theta -\mu g\cos \theta$

Given,

$a_{cylinder}=a_{block}$

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

$sin\theta =cos\theta$ since $\theta ={45}^{\circ }$

$\Rightarrow \frac{1}{2}=1-\mu$ since $sin\theta =cos\theta$ due to $\theta ={45}^{\circ }$

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

$\Rightarrow 18\mu =18\times \frac{1}{2}=9$