Question

A uniform cylinder of radius $R(=3m)$ is spin about its axis at an angular velocity ${\omega }_0(=40\sqrt{\pi }$ rad/sec) and placed between two perpendicular wall. The coefficient of friction between the walls and cylinder is $\mu$ ( =2). Cylinder makes $25K$ turns before it stops. Find the value of $K$.

Answer 1

$\mu N_1+N_2=mg$ ........ (1)
$N_1=\mu N_2$ ........ (2)
$N_2=\frac{mg}{(1+{\mu }^2)}$
$N_1=\frac{\mu mg}{(1+{\mu }^2)}$
Torque about centre of mass
$\mu (N_1+N_2)R=\frac{1}{2}m{R}^2\alpha$
$\alpha =\frac{2\mu (N_1+N_2)}{mR}=\frac{2\mu (1+\mu )g}{(1+{\mu }^2)R}$
${\omega }^2={\omega }_0^2-2\alpha \theta$
$\theta =\frac{{\omega }_0^2}{2\alpha }$
$\theta =\frac{{\omega }_0^2(1+{\mu }^2)R}{2\times 2\mu (1+\mu )g}$
Hence number of turns
$N=\frac{\theta }{2\pi }=\frac{{\omega }_0^{2}R(1+{\mu }^2)}{4\mu (1+\mu )g.2\pi }$
$N=\frac{{\mu }_0^{2}R(1+{\mu }^2)}{8\mu \pi g(1+\mu )}$