Question

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

• A. $\frac{{(1+{\mu }_k)}^2{\omega }_0^{2}R}{8\pi {\mu }_k(1+{\mu }_k)g}$
• B. $\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{8\pi {\mu }_k(1+{\mu }_k)g}$
• C. $\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{4\pi {\mu }_k(1+{\mu }_k)g}$
• D. $\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{8\pi {\mu }_k(1+{\mu }_k^2)6g}$

Answer 1

The correct option is B $\frac{(1+{\mu }_k^2){\omega }_0^{2}R}{8\pi {\mu }_k(1+{\mu }_k)g}$

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

$\sum F_x=0,N_2-{\mu }_k{N}_1=0$ ...(1)
$\sum F_x=0,N_1-{\mu }_k{N}_2-mg=0$ ...(2)
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

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

The torque equation $T=1\alpha$

Using equation ${\omega }^2={\omega }_0^2+2\alpha \theta$, Calculate the angular displacement $\theta$,

Revolution accomplished,