Question

A uniform cylinder of radius $R$ is spinned about its axis to the angular velocity ${\omega }_0$ and then placed into a corner. The coefficient of friction between the corner walls and the cylinder is ${\mu }_k$. How many turns will the cylinder accomplish before it stops?

Answer 1

As the centre of mass of the cylinder does not accelerate hence $\sum F=0$

$\sum F_x=0,N_2={\mu }_k{N}_1=\mathrm{0...}\left(i\right)$

$\sum F_x=0,N_1={\mu }_k{N}_2-mg=\mathrm{0...}\left(ii\right)$

solving these equations $N_1=\frac{mg}{1+{\mu }_k^2},N_2=\frac{{\mu }_{k}mg}{1+{\mu }_k^2}$

The torque on the cylinder about the axis of rotation

${\tau }_{CM}=mg\times \times 0+N_1\times 0+N_2\times 0+{\mu }_k{N}_{2}R=\frac{{\mu }_k(1+{\mu }_k)}{(1+{\mu }_k)}mgR$

The moment of inertia about axis of rotation $I_{CM}=\frac{1}{2}m{R}^2$

The torque equation, $\tau =I\alpha$

$\frac{{\mu }_k(1+{\mu }_k)}{(1+{\mu }_k)}mgR=\left(\frac{1}{2}m{R}^2\right)\alpha \Rightarrow \alpha =\frac{{2\mu }_k(1+{\mu }_k)g}{(1+{\mu }_k^2)R}$

Using equation ${\omega }^2={\omega }_0^2-2\alpha \theta$

calculate the angular displacement $\theta$

$0^2={\omega }_0^2-2\left[\frac{{2\mu }_k(1+{\mu }_k)g}{(1+{\mu }_k^2)R}\right]\theta$

$\Rightarrow \theta =\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{4{\mu }_k(1+{\mu }_k)g}$

Revolution accomplished

$n=\frac{\theta }{2\pi }=\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{2\pi \times 4{\mu }_k(1+{\mu }_k)g}=\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{8\pi {\mu }_k(1+{\mu }_k)g}$