Question

A uniform cylinder of radius $R$ is spinned about its axis to the angular velocity ${\omega }_0$ and then placed into a corner (figure shown above).The coefficient of friction between the corner walls and the cylinder is equal to $k$. The number of turns the cylinder accomplish before it stops is given by $n=\frac{{\omega }_0^2(1+k^2)R}{x\pi k(1+k)g}$. Find the value of $x$.

Answer 1

In the problem, the rigid body is in translation equilibrium but there is an angular retardation. We first sketch the free body diagram of the cylinder. Obviously the friction forces, acting on the cylinder, are kinetic. From the condition of translational equilibrium for the cylinder,
$mg=N_1+k{N}_2$; $N_2=k{N}_1$
Hence, $N_1=\frac{mg}{1+k^2}$; $N_2=k\frac{mg}{1+k^2}$
For pure rotation of the cylinder about its rotation axis, $N_z=I{\beta }_z$
or, $-k{N}_{1}R-k{N}_{2}R=\frac{m{R}^2}{2}{\beta }_z$
or, $-\frac{kmgR(1+k)}{1+k^2}=\frac{m{R}^2}{2}{\beta }_z$
or, ${\beta }_z=-\frac{2k(1+k)g}{(1+k^2)R}$
Now, from the kinematical equation,
${\omega }^2={\omega }_0^2+2{\beta }_z\Delta \phi$ we have,
$\Delta \phi =\frac{{\omega }_0^2(1+k^2)R}{4k(1+k)g}$, because $\omega =0$
Hence, the sought number of turns,
$n=\frac{\Delta \phi }{2\pi }=\frac{{\omega }_0^2(1+k^2)R}{8\pi k(1+k)g}$