A uniform cylinder of mass $m$ and radius $R$ is spun about its axis to an initial angular velocity

Question

A uniform cylinder of mass $m$ and radius $R$ is spun about its axis to an initial angular velocity $\omega_0$ and then placed into a corner as shown in the figure. The coefficient of kinetic friction between both the corner walls and the cylinder is equal to $k$. Which of the following statements is/are correct about the motion of the cylinder before it completely stops?

Professor PC Thomas Classes · Answer

Based on the diagram, the cylinder rotates counter-clockwise. The point of contact with the floor moves right, so the kinetic friction from the floor $f_1$ acts to the left. The point of contact with the wall moves down, so the kinetic friction from the wall $f_2$ acts upwards. Let $N_1$ be the normal force from the floor (upwards) and $N_2$ be the normal force from the wall (rightwards).

Applying translational equilibrium conditions since the cylinder's center of mass doesn't move:
Horizontal ($\Sigma F_x = 0$): $N_2 - f_1 = 0 \implies N_2 = f_1$
Vertical ($\Sigma F_y = 0$): $N_1 + f_2 - mg = 0 \implies N_1 + f_2 = mg$

Using the law of kinetic friction $f_1 = k N_1$ and $f_2 = k N_2$, the horizontal equation becomes $N_2 = k N_1$. Substituting this into the vertical equation gives:
$N_1 + k(k N_1) = mg \implies N_1(1+k^2) = mg \implies N_1 = \frac{mg}{1+k^2}$ (Option A is correct).
Consequently, $N_2 = k N_1 = \frac{k mg}{1+k^2}$ (Option B is incorrect).

The total frictional torque opposing the rotation is $ au = f_1 R + f_2 R = k(N_1 + N_2)R = k\left(\frac{mg}{1+k^2} + \frac{k mg}{1+k^2} ight)R = \frac{k(1+k)mgR}{1+k^2}$ (Option C is correct).

The moment of inertia of the cylinder is $I = \frac{1}{2}mR^2$. The angular deceleration is $\alpha = \frac{ au}{I} = \frac{2k(1+k)g}{R(1+k^2)}$.
Using the kinematic equation $\omega^2 = \omega_0^2 - 2\alpha heta$ with final $\omega = 0$, the total angle rotated is $ heta = \frac{\omega_0^2}{2\alpha} = \frac{\omega_0^2 R(1+k^2)}{4k(1+k)g}$.
The number of turns is $n = \frac{ heta}{2\pi} = \frac{\omega_0^2 R (1+k^2)}{8\pi k(1+k) g}$ (Option D is correct).