A solid cylinder of radius R rolls without slipping with a center of mass speed v_0 = √(gR/3) on a h

Q: Why doesn't the angular velocity change when it hits the corner?

During the collision with the corner, the impulsive forces (normal and friction) act exactly at the pivot point. Since the distance from the pivot is zero, they create no torque about the corner. Therefore, angular momentum about the corner is conserved, keeping the angular velocity constant right after the impact.

Q: Why do we use the moment of inertia about the corner instead of the center of mass?

Because the cylinder is purely rotating around the corner! Using the parallel axis theorem, $I_{\text{corner}} = I_{cm} + mR^2$. This makes our kinetic energy calculation much simpler: $K = \frac{1}{2} I_{\text{corner}} \omega^2$.

Q: What exactly does 'losing contact' mean in terms of forces?

Losing contact means the surface is no longer pushing against the cylinder. In physics terms, the normal force $N$ becomes exactly zero.

Q: Why is the centripetal force equal to $mg \cos\theta$?

For an object moving in a circle, the net force towards the center must equal $m\omega^2 R$. The forces acting on the cylinder are gravity and the normal force. The component of gravity pointing towards the pivot is $mg \cos\theta$. When contact is lost ($N=0$), this gravity component alone provides the centripetal force.

Q: What happens to the cylinder after it loses contact?

It becomes a projectile! It will follow a parabolic path under the influence of gravity while continuing to spin at the constant angular velocity it had at the moment it lost contact.

Question

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 the 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:

Options:

(A) $0$

(B) $\sqrt{\frac{5gR}{7}}$

(C) $\sqrt{\frac{gR}{15}}$

(D) $\sqrt{\frac{3gR}{7}}$

Sigma Percentile · Accepted Answer

### The Setup: Rolling to the Edge

Imagine a solid cylinder rolling smoothly along a horizontal surface, heading straight for a sharp vertical drop.

We are given its initial center of mass speed, $v_0 = \sqrt{\frac{gR}{3}}$.

Because the cylinder is in **pure rolling** without slipping, its initial angular velocity is directly tied to its linear speed.

We can easily find this by dividing the linear speed by the radius: $\omega_0 = \frac{v_0}{R} = \sqrt{\frac{g}{3R}}$.

### The Pivot: A Sudden Shift

When the cylinder reaches the edge, it doesn't just fall straight down; it begins to pivot around the sharp corner.

You might wonder, does its angular velocity change abruptly upon impact?

The answer is no!

During the collision, the impulsive normal and frictional forces act exactly at the corner itself.

Because the distance from the pivot is zero, these forces create **zero torque** about the corner.

Therefore, **angular momentum about the corner is conserved**, meaning the cylinder begins its downward swing with the exact same angular velocity, $\omega_0$.

### The Drop: Work and Energy

Now, let's visualize the cylinder rotating around this corner by some angle $	heta$.

As its center of mass swings downward, gravity does positive work, which speeds up the cylinder's rotation.

To find the speed at any angle, we apply the **Work-Energy Theorem**.

The work done by gravity as the center of mass drops by a vertical distance of $R(1 - \cos	heta)$ is equal to the change in rotational kinetic energy.

$$mgR(1 - \cos	heta) = \frac{1}{2} I_{	ext{corner}} \omega^2 - \frac{1}{2} I_{	ext{corner}} \omega_0^2$$

Remember, since the cylinder is pivoting around its edge, we must use the **parallel axis theorem** to find the moment of inertia.

$$I_{	ext{corner}} = I_{cm} + mR^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$$

### The Critical Moment: Losing Contact

At what exact moment does the cylinder finally lose contact with the corner and become a projectile?

This happens when the surface is no longer pushing against the cylinder, meaning the **normal force drops to zero**.

At this precise instant, the radial component of gravity must entirely provide the necessary **centripetal force** to keep the center of mass moving in its circular arc.

$$mg \cos	heta = m \omega^2 R$$

This elegant constraint gives us a direct relationship: $g \cos	heta = \omega^2 R$.

### The Final Calculation

We now have two powerful equations.

Let's substitute our contact-loss condition into the Work-Energy equation to eliminate $	heta$ entirely.

$$mgR - m(\omega^2 R)R = \frac{3}{4} mR^2 \omega^2 - \frac{3}{4} mR^2 \omega_0^2$$

Notice how beautifully the mass $m$ cancels out from every term.

Rearranging the equation to group the $\omega$ terms on one side, we get:

$$gR + \frac{3}{4} R^2 \omega_0^2 = \frac{7}{4} R^2 \omega^2$$

We are almost at the finish line!

Let's plug in our initial angular velocity squared, $\omega_0^2 = \frac{g}{3R}$.

$$gR + \frac{3}{4} R^2 \left(\frac{g}{3R}ight) = \frac{7}{4} R^2 \omega^2$$

The $3$ cancels out, leaving us with $gR + \frac{1}{4} gR$, which simplifies to $\frac{5}{4} gR$.

$$\frac{5}{4} gR = \frac{7}{4} R^2 \omega^2$$

Solving for $\omega^2$, we find that $\omega^2 = \frac{5g}{7R}$.

Finally, the question asks for the linear speed of the center of mass.

We simply multiply our final angular velocity by the radius $R$.

$$v = \omega R = \sqrt{\frac{5g}{7R}} \cdot R = \sqrt{\frac{5gR}{7}}$$

And there we have it!

The final speed of the center of mass is **$\sqrt{\frac{5gR}{7}}$**.

Select Answer:

Visualized Solution

Initial State

Conservation of Angular Momentum

Pivoting by Angle $θ$

Condition for Losing Contact

Work-Energy Theorem

Substituting the Variables

Simplifying the Equation

Solving for $ω$

Final Center of Mass Speed

Beyond the Edge

The Sigma Insight: Dynamics of Rigid Body Rotation

The Setup

The Pivot

The Drop

The Critical Moment

The Final Calculation

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