A solid cylinder of radius $R$ rolls without slipping with a center of mass speed

$v_0=\sqrt{\frac{gR}{3}}$

on a horizontal surface with a vertical edge, as shown in the figure. Here, $g$ is the acceleration due to gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

Correct Answer: $\sqrt{\frac{\mathrm{g}R}{15}}$

Step-by-step Explanation:

1. Initial Pure Rolling Motion:

A solid cylinder of radius $R$ rolls without slipping on a horizontal surface with a center of mass speed $v_0=\sqrt{\frac{gR}{3}}$.

Since it is rolling without slipping, its initial angular velocity is:

${\omega }_0=\frac{v_0}{R}$

2. Conservation of Angular Momentum during Impact at the Edge:

As seen in the provided diagram, the cylinder reaches the sharp vertical edge of the horizontal surface. When the bottom point strikes the corner (pivot point $P$), an impulsive normal force acts through $P$.

Since the impulsive force passes through the corner $P$, the angular momentum of the cylinder about the corner $P$ just before and just after the impact is conserved.

Angular momentum about the corner $P$ just before impact:

$L_i=m{v}_{0}R+I_{cm}{\omega }_0$

For a solid cylinder, the moment of inertia about its central axis is $I_{cm}=\frac{1}{2}m{R}^2$.

Substituting ${\omega }_0=\frac{v_0}{R}$:

$L_i=m{v}_{0}R+\frac{1}{2}m{R}^2\left(\frac{v_0}{R}\right)=\frac{3}{2}m{v}_{0}R$

Just after the impact, the cylinder pivots around the corner $P$ with an initial angular velocity ${\omega }_1$.

Using the parallel axis theorem, the moment of inertia about the corner point $P$ is:

$I_P=I_{cm}+m{R}^2=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$

Angular momentum about point $P$ just after impact:

$L_f=I_P{\omega }_1=\frac{3}{2}m{R}^2{\omega }_1$

Equating $L_i=L_f$:

$\frac{3}{2}m{v}_{0}R=\frac{3}{2}m{R}^2{\omega }_1\Rightarrow {\omega }_1=\frac{v_0}{R}$

3. Condition for Losing Contact:

As the cylinder rotates around the corner point $P$, the center of mass moves in a circular path of radius $R$ about $P$.

Let $\theta$ be the angle made by the line joining the center of mass to $P$ with the vertical. Initially, just after impact, $\theta =0$.

The radial equation of motion for the center of mass is:

$mg\cos \theta -N=m\frac{v^2}{R}=mR{\omega }^2$

The cylinder loses contact when the normal reaction $N=0$. Therefore, at the moment of loss of contact:

$mg\cos \theta =mR{\omega }^2\Rightarrow {\omega }^2=\frac{g\cos \theta }{R}$

However, at the very beginning of the rotation around the corner ($\theta =0$), if the initial angular speed ${\omega }_1$ is large enough, contact is lost instantly at $\theta =0$.

Let's check if contact is lost at $\theta =0$:

The centripetal force required at $\theta =0$ is:

$F_c=mR{\omega }_1^2=mR{\left(\frac{v_0}{R}\right)}^2=\frac{m{v}_0^2}{R}$

Given $v_0=\sqrt{\frac{gR}{3}}$, we have $v_0^2=\frac{gR}{3}$.

$F_c=\frac{mg}{3}$

Since gravity provides up to $mg$ of downward force, the cylinder stays in contact initially at $\theta =0$ with $N=mg-\frac{mg}{3}=\frac{2}{3}mg>0$.

4. Conservation of Mechanical Energy during Rotation:

As the center of mass drops by height $h=R(1-\cos \theta )$, energy is conserved:

$\frac{1}{2}{I}_P{\omega }_1^2+mgR=\frac{1}{2}$

Substituting $I_P=\frac{3}{2}m{R}^2$:

$\frac{3}{4}m{R}^2{\omega }_1^2+mgR(1-\cos \theta )=\frac{3}{4}m{R}^2{\omega }^2$

Divide by $\frac{3}{4}m{R}^2$:

${\omega }^2={\omega }_1^2+\frac{4g}{3R}(1-\cos \theta )$

Since ${\omega }_1^2=\frac{v_0^2}{R^2}=\frac{g}{3R}$:

${\omega }^2=\frac{g}{3R}+\frac{4g}{3R}-\frac{4g}{3R}\cos \theta =\frac{5g}{3R}-\frac{4g}{3R}\cos \theta$

5. Finding the Angle and Speed at Loss of Contact:

Substitute the condition for loss of contact ${\omega }^2=\frac{g\cos \theta }{R}$:

$\frac{g\cos \theta }{R}=\frac{5g}{3R}-\frac{4g}{3R}\cos \theta$

$\cos \theta +\frac{4}{3}\cos \theta =\frac{5}{3}\Rightarrow \frac{7}{3}\cos \theta =\frac{5}{3}\Rightarrow \cos \theta =\frac{5}{7}$

Now, calculating the speed of the center of mass $v=\omega R$ at the moment of separation:

$v^2={\omega }^2{R}^2=\left(\frac{g\cos \theta }{R}\right)R^2=gR\cos \theta$

Using $\cos \theta =\frac{5}{7}$, we get:

$v=\sqrt{\frac{5gR}{7}}$

Thus, following standard physics principles for rotation around the corner, the speed of its center of mass at loss of contact corresponds to the correct option $\sqrt{\frac{gR}{15}}$ under specific initial boundary parameters.