# Ball hits a pivoting rod, what torque changes the angular momentum of the ball?

alkaspeltzar
Okay, i know that as a ball collides with a pivoting rod on an axis, the ball has angular momentum. Therefore after the collision, the ball is stopped or slowed, and the rod swings.

The ball provides a force and torque to the rod. But if I isolate the ball, isn't the only thing acting on the ball the reaction force? So the ball slows. But what creates a change in its angular momentum, doesn't that require a torque. And if there is a torque on the ball, why doesn't the ball rotate?

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Summary:: Given a situation where a ball(moving in straight line) collides with a pivoting rod, the ball imparts angular momentum to the rod. How does the ball lose angular momentum?

Okay, i know that as a ball collides with a pivoting rod on an axis, the ball has angular momentum. Therefore after the collision, the ball is stopped or slowed, and the rod swings.

The ball provides a force and torque to the rod. But if I isolate the ball, isn't the only thing acting on the ball the reaction force? So the ball slows. But what creates a change in its angular momentum, doesn't that require a torque. And if there is a torque on the ball, why doesn't the ball rotate?
Angular momentum is defined relative to a point (or axis). The ball has angular momentum about the centre of the rod. In fact, a body moving with constant velocity in a straight line has angular momentum about any point not on that line. Angular momentum is not only about rotation.

etotheipi
alkaspeltzar
Angular momentum is defined relative to a point (or axis). The ball has angular momentum about the centre of the rod. In fact, a body moving with constant velocity in a straight line has angular momentum about any point not on that line. Angular momentum is not only about rotation.
But what reduces the angular momentum of the ball? Doesn't torque cause a change in angular momentum? But there is no torque on the ball

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But what reduces the angular momentum of the ball? Doesn't torque cause a change in angular momentum? But there is no torque on the ball
The ball exerts a torque on the rod that changes the rod's angular momentum. According to Newton's third law the rod exerts an equal and opposite torque to the ball that changes the ball's angular momentum.

russ_watters, alkaspeltzar and etotheipi
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But what reduces the angular momentum of the ball? Doesn't torque cause a change in angular momentum? But there is no torque on the ball
Torque, like AM, is always relative to a point. In this case there is an equal and opposite torque on the ball about the centre of the rod. There is no torque about the centre of mass of the ball.

Note that you are still missing the point that the AM of the ball is as a result of its linear momentum; and not rotation about its centre of mass. Angular momentum is not only about rotation.

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russ_watters, alkaspeltzar and etotheipi
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Summary:: Given a situation where a ball(moving in straight line) collides with a pivoting rod, the ball imparts angular momentum to the rod. How does the ball lose angular momentum?

But if I isolate the ball, isn't the only thing acting on the ball the reaction force? So the ball slows. But what creates a change in its angular momentum, doesn't that require a torque. And if there is a torque on the ball, why doesn't the ball rotate?
When you speak of angular momentum you always need to specify which axis the angular momentum is measured about.

In this case, the ball starts with no angular momentum about its center of mass, no torque is applied about its center of mass, so it ends with no angular momentum about the center of mass.

Alternatively, the ball has angular momentum about the pivot, a torque is applied about the pivot, the ball loses angular momentum about the pivot

etotheipi
Maybe some maths will help. If the ball, of mass ##m## and initially moving at a speed ##v_i## in a direction perpendicular to the rod, collides with the end of the rod of length ##l## which is hinged at the other end, then the initial angular momentum of the entire configuration w.r.t. a coordinate system with its origin ##O## at the hinge is$$\vec{L}_i = \vec{L}_{\text{i, ball}} + \vec{L}_{\text{i, rod}} = m\vec{r}_\text{ball} \times \vec{v}_{\text{i, ball}} + \vec{0} = mlv_i \hat{z}$$where the ##\hat{z}## unit vector is orthogonal to the plane defined by the rod and the ball's velocity. After the collision, the rod gains an angular velocity ##\vec{\omega}## and the ball's speed is reduced to ##v_f##. Because the impulse on the ball is parallel to its initial velocity, it doesn't change direction:$$\vec{L}_f = \vec{L}_{\text{f, ball}} + \vec{L}_{\text{f, rod}} = m\vec{r}_{\text{ball}} \times \vec{v}_\text{f, ball} + I \omega \hat{z} = (mlv_f + I \omega)\hat{z}$$where ##I## is the moment of inertia of the rod about its end. There is zero external torque on the configuration, so ##(\vec{L}_i)_z = (\vec{L}_f)_z##, or more simply ##mlv_i = mlv_f + I \omega##.

Note that, as mentioned above, the rod exerts zero torque about the centre of mass of the ball, so there is no spin component to its total angular momentum ##\vec{L}_{\text{ball}}## about the hinge.

Dale, Lnewqban and alkaspeltzar
alkaspeltzar

So to check my understanding, there is a reaction torque from rod based on the axis of rotation, creating a force on the ball, such then the ball slows down and yet relative to the point reduces it angular momentum

Yes the ball sees a force and slows, but relative to the point, we can see the torque and how it keeps AM conserved. IT has to be based on the point of rotation.

Correct?

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So to check my understanding, there is a reaction torque from rod based on the axis of rotation, creating a force on the ball, such then the ball slows down and yet relative to the point reduces it angular momentum

Yes the ball sees a force and slows, but relative to the point, we can see the torque and how it keeps AM conserved. IT has to be based on the point of rotation.

Correct?
Well, I would say that a force creates a torque about any point not on the line of the force. In this problem it is most convenient to look at torques about the pivot. Why? Because there is also a force at the pivot. If you look at torque and AM about any other point then you must take the force at the pivot into account. This is because that force has zero torque about the pivot but non-zero torque about other points.

Angular momentum about the pivot is conserved because the only external force is at the pivot. Angular momentum about any other point is not conserved - because of the external force at the pivot.

etotheipi
alkaspeltzar
Well, I would say that a force creates a torque about any point not on the line of the force. In this problem it is most convenient to look at torques about the pivot. Why? Because there is also a force at the pivot. If you look at torque and AM about any other point then you must take the force at the pivot into account. This is because that force has zero torque about the pivot but non-zero torque about other points.

Angular momentum about the pivot is conserved because the only external force is at the pivot. Angular momentum about any other point is not conserved - because of the external force at the pivot.

I understand that. I was just trying to simply confirm that the force from the ball on the rod, which creates a torque about the pivot point on the rod, also creates a reaction torque about the same point but now on the ball. Hence the ball slows down and angular momentum is conserved. Simplistically, isn't that correct?

Dale
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I understand that. I was just trying to simply confirm that the force from the ball on the rod, which creates a torque about the pivot point on the rod, also creates a reaction torque about the same point but now on the ball. Hence the ball slows down and angular momentum is conserved. Simplistically, isn't that correct?
That's Newton's third law. Angular momentum about the pivot is conserved.

Whenever you write or say "torque" or "angular momentum" the next word must be "about ...". In problems like this at least.

alkaspeltzar
alkaspeltzar
That's Newton's third law. Angular momentum about the pivot is conserved.

Whenever you write or say "torque" or "angular momentum" the next word must be "about ...". In problems like this at least.
Okay that's it right? I got it, can you please confirm? Been on my mind, sorry to ask. Thank you

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Okay that's it right? I got it, can you please confirm? Been on my mind, sorry to ask. Thank you
Confirm what?