Question 3

When the cylinder reaches the corner, it begins to rotate about the point of contact at the corner $P$. The angular momentum about point $P$ is conserved during the instantaneous transition from pure rolling to pivoting around the corner.

Angular momentum just before the corner: $L_P = I_{cm}\omega_0 + Mv_0R = \frac{1}{2}MR^2(\frac{v_0}{R}) + Mv_0R = \frac{3}{2}Mv_0R$.

Angular momentum just after the transition: $L_P = I_P\omega = (\frac{1}{2}MR^2 + MR^2)\omega = \frac{3}{2}MR^2\omega$.

Equating these, we find the angular velocity at the start of the pivot is $\omega = \frac{v_0}{R}$.

As the cylinder rotates by an angle $\theta$ (from the vertical), we apply conservation of energy:

$\frac{1}{2} I_P \omega_{initial}^2 + MgR = \frac{1}{2} I_P \omega^2 + MgR \cos \theta$

Substituting $I_P = \frac{3}{2}MR^2$ and $\omega_{initial} = \frac{v_0}{R}$:

$\frac{3}{4} M v_0^2 + MgR = \frac{3}{4} MR^2 \omega^2 + MgR \cos \theta$

The cylinder loses contact when the normal force $N$ becomes zero. The radial equation is:

$Mg \cos \theta - N = M \omega^2 R \implies g \cos \theta = \omega^2 R$

Substitute $\omega^2 = \frac{g \cos \theta}{R}$ and $v_0^2 = \frac{gR}{3}$ into the energy equation:

$\frac{3}{4} M \left( \frac{gR}{3} \right) + MgR = \frac{3}{4} M gR \cos \theta + MgR \cos \theta$

$\frac{1}{4} MgR + MgR = \frac{7}{4} MgR \cos \theta \implies \frac{5}{4} = \frac{7}{4} \cos \theta \implies \cos \theta = \frac{5}{7}$

The speed of the center of mass at this moment is $v = \omega R = \sqrt{gR \cos \theta} = \sqrt{\frac{5gR}{7}}$.