Animated Solution for Physics - Rotational Motion: A hoop of radius $r$ and mass $m$ rotating with an angular velocity ${\omega }_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

A Spinning Hoop Meets a Rough Surface

Imagine you are holding a bicycle wheel (which is essentially a hoop) off the ground and giving it a really fast spin. It has an initial angular velocity ${\omega }_0$, but its center of mass isn't moving anywhere, so its initial linear velocity is $v=0$.

Now, you gently drop this spinning wheel onto a rough horizontal road. What happens the moment it touches the ground? Because the wheel is spinning, the bottom-most point of the wheel is moving backward relative to the ground. The rough surface doesn't like this relative motion, so it immediately exerts a forward force to stop the slipping.

Kinetic Friction

This forward force is kinetic friction ($f_k$). Kinetic friction acts exactly at the point of contact between the hoop and the ground. It has two simultaneous effects on the hoop: 1. Translational Acceleration: It pushes the center of mass forward, causing the hoop to gain linear velocity ($v$). 2. Rotational Deceleration: It creates a torque that opposes the spin, causing the angular velocity ($\omega$) to decrease.

You could solve this problem using Newton's Second Law ($f_k=ma$) and the rotational equivalent ($\tau =I\alpha$), finding the time it takes for the slipping to stop. But there is a much more elegant, "Jedi-level" trick we can use to bypass time entirely!

Choosing the Right Reference Point

In physics, choosing the right reference point can turn a nightmare calculation into a one-line solution. Let's choose our reference point to be any point on the ground along the line of motion.

Why is this brilliant? Because the force of kinetic friction acts exactly along the ground! The line of action of the frictional force passes directly through our chosen reference point.

Since Torque ($\tau$) is the cross product of the position vector and the force ($\vec{\tau }=\vec{r}\times \vec{F}$), and the perpendicular distance from our reference point to the force is zero, the torque due to friction is exactly zero.

If the net external torque about a point is zero, the Angular Momentum ($L$) about that point is conserved.

Conservation of Angular Momentum

Let's calculate the initial angular momentum ($L_i$) about our point on the ground. The hoop is only spinning, not translating. The angular momentum of a rigid body about an arbitrary point is the sum of its spin angular momentum and its orbital angular momentum:

Initially, $v_{cm}=0$, so the orbital part is zero. The moment of inertia of a hoop about its center is $I_{cm}=m{r}^2$.

Pure Rolling

As friction does its work, the hoop speeds up linearly and slows down rotationally. Eventually, the backward velocity of the bottom point due to rotation perfectly cancels the forward velocity of the center of mass. The bottom point comes to rest relative to the ground.

This is the magical state of pure rolling. The slipping ceases, and the strict condition for pure rolling is met:

Now, let's calculate the final angular momentum ($L_f$) about the same point on the ground. The hoop is now both translating with velocity $v$ and rotating with angular velocity $\omega$.

Substitute $I_{cm}=m{r}^2$ and $\omega =\frac{v}{r}$:

Solving for Velocity

Since angular momentum is conserved, we simply equate the initial and final states:

Notice how beautifully the mass $m$ and one radius $r$ cancel out from both sides!

And there we have it! The final velocity of the center of the hoop is exactly half of its initial rotational velocity times the radius. This elegant result was achieved without calculating a single force or tracking the time, all thanks to the power of conservation laws!