A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with vel

Q: Why does the center of mass of the ring stop completely?

The horizontal impulse imparted by the ball is exactly equal and opposite to the initial horizontal momentum of the ring. When you add them up, the final momentum becomes zero, bringing the CM to a halt.

Q: How do we find the exact point of collision?

We know the radius of the ring is $0.5\text{ m}$ and the collision height is $0.75\text{ m}$. The vertical distance from the CM is $0.25\text{ m}$. Using the equation of a circle $x^2 + y^2 = R^2$, we can find the horizontal coordinate $x$.

Q: Why do we take the angular impulse about the center of mass?

Taking the angular impulse about the CM allows us to decouple the translational and rotational motion. The equation $\vec{L}_J = \Delta \vec{L}_{cm}$ holds true even if the CM is accelerating during the impact.

Q: Why is the initial angular velocity $2\text{ rad/s}$?

The problem states the ring is rolling without slipping initially. For pure rolling, $v = \omega R$. Given $v = 1\text{ m/s}$ and $R = 0.5\text{ m}$, we get $\omega = 1 / 0.5 = 2\text{ rad/s}$.

Q: How do we determine the direction of friction after the collision?

Friction always opposes relative slipping between surfaces. We calculate the velocity of the bottom-most point of the ring. Since it moves to the right, kinetic friction must act to the left to oppose this motion.

Question

A thin ring of mass $2	ext{ kg}$ and radius $0.5	ext{ m}$ is rolling without slipping on a horizontal plane with velocity $1	ext{ m/s}$. A small ball of mass $0.1	ext{ kg}$, moving with velocity $20	ext{ m/s}$ in the opposite direction, hits the ring at a height of $0.75	ext{ m}$ and goes vertically up with velocity $10	ext{ m/s}$. Immediately after the collision,

Options:

(A) the ring has pure rotation about its stationary CM

(B) the ring comes to a complete stop

(C) friction between the ring and the ground is to the left

(D) there is no friction between the ring and the ground

Sigma Percentile · Accepted Answer

### A Tale of Collision: When a Ball Meets a Rolling Ring

Imagine a heavy ring rolling smoothly along a horizontal surface. Suddenly, a small, fast-moving ball strikes it. What happens next? Does the ring stop? Does it spin faster? Let's break down the physics of this fascinating collision step by step.

### The Setup

We have a thin ring of mass $M = 2	ext{ kg}$ and radius $R = 0.5	ext{ m}$. It's rolling without slipping to the left with a velocity of $v_r = 1	ext{ m/s}$. Because it's rolling without slipping, its initial angular velocity is $\omega_i = \frac{v_r}{R} = \frac{1}{0.5} = 2	ext{ rad/s}$ in the anti-clockwise direction.

Approaching from the opposite direction is a small ball of mass $m = 0.1	ext{ kg}$, moving to the right at a brisk $u = 20	ext{ m/s}$. The ball hits the ring at a height of $h = 0.75	ext{ m}$ from the ground.

### The Impact (Linear)

Immediately after the collision, the ball shoots vertically upwards with a velocity of $v = 10	ext{ m/s}$. To understand what happens to the ring, we first need to calculate the **impulse** experienced by the ball.

Impulse is simply the change in linear momentum: $\vec{J} = \Delta \vec{p}$.

The ball's initial momentum was purely horizontal: $\vec{p}_i = 0.1 	imes 20 \hat{i} = 2 \hat{i}	ext{ kg m/s}$.
Its final momentum is purely vertical: $\vec{p}_f = 0.1 	imes 10 \hat{j} = 1 \hat{j}	ext{ kg m/s}$.

Therefore, the impulse on the ball is $\vec{J}_{ball} = -2 \hat{i} + 1 \hat{j}	ext{ Ns}$.

By Newton's Third Law, the ring experiences an equal and opposite impulse: $\vec{J}_{ring} = 2 \hat{i} - 1 \hat{j}	ext{ Ns}$. This means the ring receives a horizontal kick of $2	ext{ Ns}$ to the right.

Let's apply the impulse-momentum theorem to the ring's horizontal motion. The ring's initial horizontal momentum was $P_{xi} = M(-v_r) = 2(-1) = -2	ext{ kg m/s}$. When we add the horizontal impulse of $+2	ext{ Ns}$, the final horizontal momentum becomes exactly zero!

This is a beautiful result: the collision perfectly cancels out the ring's translational motion. The center of mass of the ring comes to a complete stop ($v_f = 0$). Thus, the ring now has **pure rotation about its stationary CM**.

### The Impact (Angular)

But what about its rotation? To find the new angular velocity, we need to calculate the **angular impulse** about the center of mass.

First, we find the position vector $\vec{r}$ of the collision point relative to the CM. The vertical distance is $y = 0.75 - 0.5 = 0.25	ext{ m}$. Using the equation of a circle, the horizontal distance is $x = \sqrt{0.5^2 - 0.25^2} = \frac{\sqrt{3}}{4}	ext{ m}$. Since the ball hits the left side, $\vec{r} = -\frac{\sqrt{3}}{4} \hat{i} + 0.25 \hat{j}$.

The angular impulse is the cross product of this position vector and the linear impulse on the ring:

$\vec{L}_J = \vec{r} 	imes \vec{J}_{ring} = \left(-\frac{\sqrt{3}}{4} \hat{i} + 0.25 \hat{j}ight) 	imes (2 \hat{i} - 1 \hat{j})$

Evaluating this cross product gives us $\vec{L}_J = \left( \frac{\sqrt{3}}{4} - 0.5 ight) \hat{k}	ext{ Nms}$. This positive value indicates a net angular impulse in the anti-clockwise direction.

### The Aftermath

Now, we apply the angular impulse-momentum theorem: $I \omega_f - I \omega_i = L_J$.

The moment of inertia of the ring is $I = M R^2 = 2 	imes (0.5)^2 = 0.5	ext{ kg m}^2$. Plugging in our values:

$0.5 \omega_f - 0.5(2) = \frac{\sqrt{3}}{4} - 0.5$

Solving for $\omega_f$, we find $\omega_f = 1 + \frac{\sqrt{3}}{2}	ext{ rad/s}$. Since this is positive, the ring continues to spin anti-clockwise, but at a slightly slower rate.

Finally, let's determine the direction of friction. The velocity of the bottom-most point of the ring is given by $\vec{v}_{bottom} = \vec{v}_{cm} + \vec{\omega}_f 	imes \vec{r}_{bottom}$. Since the CM is stationary ($v_{cm} = 0$), the bottom point moves to the right with a speed of $\omega_f R$.

Because the bottom point is slipping to the right, **kinetic friction must act to the left** to oppose this motion. And there you have it—a perfect harmony of linear and angular mechanics!

Select Answer:

Visualized Solution

Visualizing the Setup

Impulse on the Ball

Impulse on the Ring

Linear Motion of the Ring

Geometry of the Collision Point

Angular Impulse on the Ring

Change in Angular Momentum

Final Angular Velocity & Friction

The Sigma Insight: Conservation of Angular Momentum

A Tale of Collision

The Setup

The Impact (Linear)

The Impact (Angular)

The Aftermath

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