Why Doesn't the Ball Rotate When Hit by a Pivoting Rod?

[alkaspeltzar]

Okay, I asked this post(below) the other day on how ang. momentum was conserved by the ball. The answer was the ball receives an equal torque back from the door.
https://www.physicsforums.com/threa...lar-momentum-of-the-ball.993221/#post-6388435

But what I still don't get, is why doesn't the ball rotate if there is a torque on it? Or even more so, why don't we include this torque in free body diagram?

I think of an example of a winch rotated by a falling mass attached via string. There is torque on the winch and force of gravity on the mass. Wouldnt there be a torque on the mass too?

 

[A.T.]

But what I still don't get, is why doesn't the ball rotate if there is a torque on it?

Torque doesn't imply rotation of some rigid body.

 

[Halc]

Angular momentum is not necessarily spin.

Think of a system of a rod (free floating say, so we can say it is an isolated system) and a ball. The rod is oriented east-west and not rotating. The two are moving towards each other north-south, but not with aligned centers of gravity. So neither part is rotating, but the system has angular momentum since there is motion around the center of gravity of the system which is between the CoG of the rod and ball respectively.

Now the ball hits the rod square on (not at an angle) but well to the side. For simplicity, let's assume the ball is halted (relative to the system CoG) by this collision. By conservation of momentum, the rod is also halted in its southward trajectory, but the torque of the impact begins to spin it. It absorbs all of the angular momentum of the system, leaving none to the ball. The thrust on the ball, tangential to the center of mass of the system, is torque applied to/by the ball, despite the ball never spinning in this scenario.

The rod rotates 180 degrees and smacks the ball from the other side, halting the rod's spin and sending both items on their original course, but with the rod facing the other way now.

Linear and angular momentum was conserved throughout.

 

[kuruman]

Remember @PeroK's admonition in the thread you mentioned?

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

Please specify torque about what point. In your previous question the ball had angular momentum about the pivot before it collided with the rod. As a result of the collision, the ball lost angular momentum about the pivot which means that during the collision there was a torque about the pivot acting on the ball. In other words, the ball changed its angular speed about the pivot.

We do not normally include torques in FBDs as a matter of clarity. We include all the forces acting on the system with the understanding that once we do so correctly, we can calculate torques about any point we wish and write them down. There is an infinity of points about which we can draw an infinity of different torques in a FBD so what's the point?

 

[Lnewqban]

Summary:: Do equal and opposite torques exist, and if so, how come we don't account for them or include them in many FBD's
...
But what I still don't get, is why doesn't the ball rotate if there is a torque on it? Or even more so, why don't we include this torque in free body diagram?

I think of an example of a winch rotated by a falling mass attached via string. There is torque on the winch and force of gravity on the mass. Wouldnt there be a torque on the mass too?

Could you show us a free body diagram of your question?
I am sorry, I don't see how is the initial rotation of the ball (is the thing simultaneously sliding and rotating?) and what torque is acting on it after the collision.

 

[Dale]

But what I still don't get, is why doesn't the ball rotate if there is a torque on it?

Please clarify:
Why doesn't the ball rotate about what axis?
If there is a torque on it about what axis?

One of the more difficult things to learn in physics is when you have actually said a complete thought. Torque depends on the axis and rotation is always about an axis. So if you haven't specified the axis then your thought is incomplete.

EDIT: I see @kuruman and @PeroK already made this point to you. This is now three knowledgeable people making the same point to you. Perhaps it is an important point.

 

[A.T.]

EDIT: I see @kuruman and @PeroK already made this point to you. This is now three knowledgeable people making the same point to you. Perhaps it is an important point.

The entire previous thread was people making this point to the OP:
https://www.physicsforums.com/threa...lar-momentum-of-the-ball.993221/#post-6388435

 

[alkaspeltzar]

Here was my initial problem. A ball striking the door. The door swings, ball slow/stops. How does ball conserved angular momentum?

What i am having a hard time getting my head around is the torque on the ball. In many cases, we would only look at the forces on the ball which is what i did. But if my understanding is correct, there is a torque about the pivot acting on the ball. But my question is why doesn't the ball then rotate? OR is it such that yes, there is a torque about the pivot on the ball, but all the ball itself see is the force, so it travels straight.

 

[alkaspeltzar]

Remember @PeroK's admonition in the thread you mentioned?

Please specify torque about what point. In your previous question the ball had angular momentum about the pivot before it collided with the rod. As a result of the collision, the ball lost angular momentum about the pivot which means that during the collision there was a torque about the pivot acting on the ball. In other words, the ball changed its angular speed about the pivot.

We do not normally include torques in FBDs as a matter of clarity. We include all the forces acting on the system with the understanding that once we do so correctly, we can calculate torques about any point we wish and write them down. There is an infinity of points about which we can draw an infinity of different torques in a FBD so what's the point?

Okay, so there is a torque acting about the pivot on the ball, so in relation to the pivot the ball loses its ang. momentum.

So then is it wrong to say the ball itself, only sees the reaction force hence why it travels straight and it doesn't rotate? Trying to understand why the ball doesn't rotate, but i assume it is because the torque is ONLY about the pivot in regards to the ball.

 

[sophiecentaur]

@alkaspeltzar are you sure you are not mixing up ideas of equilibrium and Newton's Third Law? It happens all the time for people considering linear motion. This is a similar possibility.

 

[etotheipi]

@alkaspeltzar I think you misunderstand what a torque is. I have the feeling this could go on forever, so I thought it might help if I spell out some things (but you should really read a textbook on intro classical mechanics).

You first have to choose a coordinate system; it doesn't matter if it's inertial or non-inertial, you just need to bear in mind that if it's non-inertial then you need to take into account inertial forces. Secondly, the torque of a force with respect to this coordinate system is by definition $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is to the point of application. Thirdly, the vector sum of the torques on anybody is the rate of change of the angular momentum of that body, $\sum \vec{\tau}_i = d\vec{L}/dt$; these quantities are all still calculated with respect to the same coordinate system.

The next important result is the König theorem, that the total angular momentum of a body of mass $M = \sum m_i$ can be written as the sum of the angular momentum that a particle of mass $M$ positioned at the centre of mass would have, plus the angular momentum of the body with respect to the centre of mass frame. That's because$$\begin{align*}\vec{L} = \sum_i m_i \vec{r}_i \times \dot{\vec{r}}_i &= \sum_i m_i (\vec{R} + \vec{r}'_i) \times (\dot{\vec{R}} + \dot{\vec{r}}_i') \\ &= M\vec{R} \times \dot{\vec{R}} + \left [\vec{R} \times \left(\sum_i m_i \dot{\vec{r}}'_i \right) \right] + \left[ \left(\sum_i m_i \vec{r}'_i \right) \times \dot{\vec{R}} \right] + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' \\ &= M\vec{R} \times \dot{\vec{R}} + \vec{0} + \vec{0} + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' = \vec{L}_{\text{CM}} + \vec{L}^*

\end{align*}$$In the general case, since $\sum \vec{\tau} = d\vec{L}/dt = \frac{d}{dt} \left( \vec{L}_{\text{CM}} + \vec{L}^* \right)$, the resultant torque might cause changes in both the angular momentum about the centre of mass and the angular momentum due to the translation of the centre of mass. (Remember, again, that $\sum \vec{\tau}$ and $\vec{L}$ are coordinate variant so we're still working with respect to one chosen coordinate system).

For a coordinate system whose origin is the centre of mass, the total torque (which here is also the total real torque, because inertial forces have zero level arm) is just the rate of change of angular momentum about the centre of mass. It follows that if the lines of action of all of the forces acting on a body pass through its centre of mass, the angular momentum with respect to the centre of mass is constant. For the falling mass attached to the string, there is no resultant torque about its centre of mass, so the mass undergoes no rotation given that its initial angular velocity is zero.

The distinction is even clearer for something like e.g. a particle moving in the plane under the action of a force $\vec{F}$. The angular momentum of the particle is solely due to translation of the particle w.r.t. the coordinate system, not due to any 'rotation of the particle'. You would find that $\vec{\tau} = \vec{r} \times \vec{F} = \frac{d}{dt} \left( mr^2 \dot{\theta} \hat{z} \right) = \left( 2m r \dot{r} \dot{\theta} + mr^2 \ddot{\theta} \right) \hat{z}$. N.B. you might notice that if $\dot{r} = 0$, then the RHS is just $mr^2 \ddot{\theta} \hat{z} = I_z \alpha_z \hat{z}$; the angular acceleration $\vec{\alpha}$ in this instance is the angular acceleration of the line joining the particle and the origin.

 

[alkaspeltzar]

@alkaspeltzar are you sure you are not mixing up ideas of equilibrium and Newton's Third Law? It happens all the time for people considering linear motion. This is a similar possibility.

No I know they are in different bodies.

 

[Dale]

What i am having a hard time getting my head around is the torque on the ball.

About which axis?

there is a torque about the pivot acting on the ball.

Yes, thank you for specifying the axis.

But my question is why doesn't the ball then rotate?

About which axis?

Okay, so there is a torque acting about the pivot on the ball, so in relation to the pivot the ball loses its ang. momentum.

Yes. Remember that for any axis off the line of travel an inertial object is rotating in the sense that it has angular momentum. So due to the torque about the pivot the ball rotates less about the pivot.

hence why it travels straight and it doesn't rotate?

It does rotate, meaning that it has angular momentum. You seem to think that traveling straight is incompatible with angular momentum.

This may be a terminology issue. If by “rotate about an axis” you mean “has angular momentum about an axis” then it is correct that a torque about an axis will change your rotation about that axis. If by “rotate about an axis” you mean “travel in a circular path with the axis at the center” then it is incorrect that a torque about an axis will change your rotation about that axis.

A torque about an axis causes a change in angular momentum about that axis. That is the rule. Don’t get stuck on “rotation” if you consider rotation to be something other than angular momentum.

I find it easier to think of “rotation” as encompassing any form of angular momentum rather than to think of “rotation” as being circular motion and requiring separation of angular momentum into circular and non-circular parts.

 

[alkaspeltzar]

@alkaspeltzar I think you misunderstand what a torque is. I have the feeling this could go on forever, so I thought it might help if I spell out some things (but you should really read a textbook on intro classical mechanics).

You first have to choose a coordinate system; it doesn't matter if it's inertial or non-inertial, you just need to bear in mind that if it's non-inertial then you need to take into account inertial forces. Secondly, the torque of a force with respect to this coordinate system is by definition $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is to the point of application. Thirdly, the vector sum of the torques on anybody is the rate of change of the angular momentum of that body, $\sum \vec{\tau}_i = d\vec{L}/dt$; these quantities are all still calculated with respect to the same coordinate system.

The next important result is the König theorem, that the total angular momentum of a body of mass $M = \sum m_i$ can be written as the sum of the angular momentum that a particle of mass $M$ positioned at the centre of mass would have, plus the angular momentum of the body with respect to the centre of mass frame. That's because$$\begin{align*}\vec{L} = \sum_i m_i \vec{r}_i \times \dot{\vec{r}}_i &= \sum_i m_i (\vec{R} + \vec{r}'_i) \times (\dot{\vec{R}} + \dot{\vec{r}}_i') \\ &= M\vec{R} \times \dot{\vec{R}} + \left [\vec{R} \times \left(\sum_i m_i \dot{\vec{r}}'_i \right) \right] + \left[ \left(\sum_i m_i \vec{r}'_i \right) \times \dot{\vec{R}} \right] + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' \\ &= M\vec{R} \times \dot{\vec{R}} + \vec{0} + \vec{0} + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' = \vec{L}_{\text{CM}} + \vec{L}^*

\end{align*}$$In the general case, since $\sum \vec{\tau} = d\vec{L}/dt = \frac{d}{dt} \left( \vec{L}_{\text{CM}} + \vec{L}^* \right)$, the resultant torque might cause changes in both the angular momentum about the centre of mass and the angular momentum due to the translation of the centre of mass. (Remember, again, that $\sum \vec{\tau}$ and $\vec{L}$ are coordinate variant so we're still working with respect to one chosen coordinate system).

For a coordinate system whose origin is the centre of mass, the total torque (which here is also the total real torque, because inertial forces have zero level arm) is just the rate of change of angular momentum about the centre of mass. It follows that if the lines of action of all of the forces acting on a body pass through its centre of mass, the angular momentum with respect to the centre of mass is constant. For the falling mass attached to the string, there is no resultant torque about its centre of mass, so the mass undergoes no rotation given that its initial angular velocity is zero.

The distinction is even clearer for something like e.g. a particle moving in the plane under the action of a force $\vec{F}$. The angular momentum of the particle is solely due to translation of the particle w.r.t. the coordinate system, not due to any 'rotation of the particle'. You would find that $\vec{\tau} = \vec{r} \times \vec{F} = \frac{d}{dt} \left( mr^2 \dot{\theta} \hat{z} \right) = \left( 2m r \dot{r} \dot{\theta} + mr^2 \ddot{\theta} \right) \hat{z}$. N.B. you might notice that if $\dot{r} = 0$, then the RHS is just $mr^2 \ddot{\theta} \hat{z} = I_z \alpha_z \hat{z}$; the angular acceleration $\vec{\alpha}$ in this instance is the angular acceleration of the line joining the particle and the origin.

I understand that ball had angular momentum due to its translation. What I'm trying to understand is how does it lose Ang momentum as it translates and hits door as that requires a torque. So can you respond to my questions and drawing above, that would help. I think the ball has a torque about the pivot, but not it's center of mass. Hence it looses AM AND yet travels straight

 

[etotheipi]

I think the ball has a torque about the pivot, but not it's center of mass.

Ball doesn't "have" a torque, it "has" an angular momentum (about the origin). A torque acts on the ball, which changes its angular momentum (about the origin).

Hence it looses AM AND yet travels straight

That's right. It's angular momentum about the pivot is $L_z = mlv$. If $v$ decreases, then so does $L_z$. Ball still travels in a straight line, it receives no force perpendicular to its velocity so does not change direction.

 

[alkaspeltzar]

About which axis?

Yes, thank you for specifying the axis.

About which axis?

Yes. Remember that for any axis off the line of travel an inertial object is rotating in the sense that it has angular momentum. So due to the torque about the pivot the ball rotates less about the pivot.

It does rotate, meaning that it has angular momentum. You seem to think that traveling straight is incompatible with angular momentum.

This may be a terminology issue. If by “rotate about an axis” you mean “has angular momentum about an axis” then it is correct that a torque about an axis will change your rotation about that axis. If by “rotate about an axis” you mean “travel in a circular path with the axis at the center” then it is incorrect that a torque about an axis will change your rotation about that axis.

A torque about an axis causes a change in angular momentum about that axis. That is the rule. Don’t get stuck on “rotation” if you consider rotation to be something other than angular momentum.

I find it easier to think of “rotation” as encompassing any form of angular momentum rather than to think of “rotation” as being circular motion and requiring separation of angular momentum into circular and non-circular parts.

Yes that is my issues. I keep hearing there is torque on the ball and think, why it doesn't go in a circular path and wonder why not? I was taught a torque was a twisting force, so I feel it should be twisting around that axis.

But what you are saying is as the ball hits, it twists the door about the pivot, but the door applies a twist about the pivot back onto the ball. The ball is not connected and still can continue straight but regardless it loses Ang. Momentum. It doesn't have to rotate to be affected.

 

[Dale]

Right. Torque is about angular momentum, not necessarily “rotation” in the sense of circular motion.

 

[kuruman]

Yes that is my issues. I keep hearing there is torque on the ball and think, why it doesn't go in a circular path and wonder why not? I was taught a torque was a twisting force, so I feel it should be twisting around that axis.

The torque on the ball lasts only during the collision. That's a very small time. In that time. the angular momentum about the pivot drops to zero. A body does not have to go in a circular path if there is a torque about a point acting on it. If you extend your right arm holding a ball horizontally and you drop the ball, there is a torque generated by gravity about your right shoulder, a different torque about your left shoulder and zero torque about your right hand that released the ball. That is why it is important to specify about what point you are considering the torque. Also note that the ball follows a straight line path, not a circular path.

 

[alkaspeltzar]

Ball doesn't "have" a torque, it "has" an angular momentum (about the origin). A torque acts on the ball, which changes its angular momentum (about the origin).

So that's probably part of my confusion. Trying to think the torque on the ball will make it rotate circularily. But it's that the torque exists about the pivot point for the ball. And the reaction force is directly acting on the ball. So it travels straight, losing AM. Correct?

 

[russ_watters]

I keep hearing there is torque on the ball...

No, I'm pretty sure the only one saying that is you.

Caveat:

The torque on the ball lasts only during the collision. That's a very small time.

It would probably be best not to bring in a non-zero duration collision, because that changes an awful lot.

 

[Lnewqban]

I understand that ball had angular momentum due to its translation. What I'm trying to understand is how does it lose Ang momentum as it translates and hits door as that requires a torque. So can you respond to my questions and drawing above, that would help. I think the ball has a torque about the pivot, but not it's center of mass. Hence it looses AM AND yet travels straight

Can we agree about a tranfer of momentum (no specific type) happening between two bodies?

The ball loses momentum and the door gains exactly the same amount of momentum.
Any change of momentum (ΔP) is considered impulse (J).

The mechanism for that transfer of momentum is the force/reaction at the point of the impact, which is applied for a time interval.

The ball is free to move along a straight trajectory: we could say that its momentum is of linear type.
It was linear before the collision and, if the ball still has velocity, is linear after the impact.

The door is slave of the hinges, it is forced to rotate around a pivot; hence its velocity is angular, just like its momentum.

It seems to me that the torque that creates your confusion is just the result of multiplying the force of impact by its perpendicular distance to that specific pivot.
The velocity of the ball is reduced by the impact; therefore its momentum (call it as you wish to refer it to a specific pivot or not) is also reduced.

Since the trajectory of the ball does not intercept the pivot of the door, we say that the ball has angular momentum respect to that pivot, whether or not a collision between ball and door happens.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-2-angular-momentum/

 

[etotheipi]

So that's probably part of my confusion. Trying to think the torque on the ball will make it rotate circularily. But it's that the torque exists about the pivot point for the ball. And the reaction force is directly acting on the ball. So it travels straight, losing AM. Correct?

I think you've nearly got it, but you're still overcomplicating it a little. In the centre of mass frame of the ball, there is zero torque on the ball, because the reaction force from the hinge acts through the centre of mass. So the ball's spin angular momentum doesn't change.

In the coordinate system centred at the hinge, and thinking in terms of the König theorem I mentioned earlier, the result is that the torque acting on the ball serves only to reduce the contribution angular momentum due to translation of the centre of mass about the origin, and doesn't affect the spin angular momentum at all (still zero).

 

[sophiecentaur]

No, I'm pretty sure the only one saying that is you.

If there is no friction and the tangential velocity of the surface that the ball strikes is equal to that of the ball at the point of contact then there will be no torque impulse. But if the two velocities are not equal and there is friction, the ball could have a change in angular momentum.

In the diagram above, there is normal contact so the velocities are equal so no transfer of angular momentum. But, if the contact time is finite and the door moves a finite amount during contact then why no transfer? The velocities can't be the same throughout the contact, can they?

Perhaps the problem hasn't been specified tightly enough and various different assumptions have been made.

 

[etotheipi]

If there is no friction and the tangential velocity of the surface that the ball strikes is equal to that of the ball at the point of contact then there will be no torque impulse. But if the two velocities are not equal and there is friction, the ball could have a change in angular momentum.

No angular impulse about the centre of mass of the ball. There will be an angular impulse about a coordinate system translated by a constant non-zero vector.

In the diagram above, there is normal contact so the velocities are equal so no transfer of angular momentum. But, if the contact time is finite and the door moves a finite amount during contact then why no transfer? The velocities can't be the same throughout the contact, can they?

Are you talking about a scenario in which the ball and rod are initially moving at the same speed, whilst in contact, before losing contact? But I think the OP was describing a collision of a ball with a rod, which is different.

 

[kuruman]

Can we agree about a tranfer of momentum (no specific type) happening between two bodies?

The ball loses momentum and the door gains exactly the same amount of momentum.
Any change of momentum (ΔP) is considered impulse (J).

The mechanism for that transfer of momentum is the force/reaction at the point of the impact, which is applied for a time interval.

The ball is free to move along a straight trajectory: we could say that its momentum is of linear type.
It was linear before the collision and, if the ball still has velocity, is linear after the impact.

The door is slave of the hinges, it is forced to rotate around a pivot; hence its velocity is angular, just like its momentum.

It seems to me that the torque that creates your confusion is just the result of multiplying the force of impact by its perpendicular distance to that specific pivot.
The velocity of the ball is reduced by the impact; therefore its momentum (call it as you wish to refer it to a specific pivot or not) is also reduced.

In this problem we cannot agree that generic momentum is conserved; we have to assert that angular momentum about the pivot is conserved. Otherwise we will not be able to answer any questions about what happens after the collision. We cannot use linear momentum conservation because, as you correctly pointed out, the rod is a "slave to the pivot." If the system is ball plus rod, the pivot exerts an external force on it. That does not conserve linear momentum. Angular momentum is also not conserved unless we specify the pivot as the point about which we calculate it because only about that point is the external torque zero. I think this point was also made in the previous thread.

 

[alkaspeltzar]

I think you've nearly got it, but you're still overcomplicating it a little. In the centre of mass frame of the ball, there is zero torque on the ball, because the reaction force from the hinge acts through the centre of mass. So the ball's spin angular momentum doesn't change.

In the coordinate system centred at the hinge, and thinking in terms of the König theorem I mentioned earlier, the result is that the torque acting on the ball serves only to reduce the contribution angular momentum due to translation of the centre of mass about the origin, and doesn't affect the spin angular momentum at all (still zero).

I got it. There is torque on the ball ABOUT the pivot, which is why it transfers AM. but looking at the ball alone, the torque doesn't affect it's c of m, so it continues to go straight or stop. I need to remember it is ABOUT THE POINT which where the torque arises. The world is good and all balances. I see the difference now

 

[kuruman]

@alkaspeltzar perhaps solving this problem will clarify things for you.

A ball of mass $m$ moving with speed $v_0$ collides with a hinged rod of length $L$ and mass $M$ at distance $d$ from the hinge. After the collision the ball is at rest relative to the hinge.
(a) Find the angular speed of the rod after the collision.
(b) If the collision lasts for time interval $\Delta t$, find the average torque exerted by the rod on the ball.

Solution
(a) Conserve angular momentum about the hinge. The angular momentum of the rod before the collision is zero because it does not rotate about the pivot and the angular momentum of the ball after the collision is zero because it does not rotate about the pivot.$$L_{ball}=m v_0 d=\frac{1}{3}ML^2\omega~\Rightarrow~\omega=\frac{3mv_0d}{ML^2}.$$
(b) Let $\bar F$ be the average force exerted by the rod on the ball. The impulse on the ball is $$J=\frac{\Delta p}{\Delta t}=\frac{0-mv_0}{\Delta t}.$$ The average torque is $$\bar{\tau}=\bar F d=-\frac{mv_0 d}{\Delta t}= \frac{\Delta L_{ball}}{\Delta t}.$$See how it works?

 

[sophiecentaur]

Are you talking about a scenario in which the ball and rod are initially moving at the same speed, whilst in contact, before losing contact? But I think the OP was describing a collision of a ball with a rod, which is different.

My comment was about a general case but, if the impact has finite duration. the door will change its angle. The ball will then be rolling along a plane at an angle and also be deflected. The ball will have angular momentum due to contact for a finite time and a tangential component of force on the ball.
I don't think there would be any argument if the impact was not normal to the door; there will be a torque due to friction along the surface of the ball.
This won't apply in an ideal case of an instantaneous contact - which actually involves an infinite force for any Impulse if the time is zero.

 

[alkaspeltzar]

@alkaspeltzar perhaps solving this problem will clarify things for you.

A ball of mass $m$ moving with speed $v_0$ collides with a hinged rod of length $L$ and mass $M$ at distance $d$ from the hinge. After the collision the ball is at rest relative to the hinge.
(a) Find the angular speed of the rod after the collision.
(b) If the collision lasts for time interval $\Delta t$, find the average torque exerted by the rod on the ball.

Solution
(a) Conserve angular momentum about the hinge. The angular momentum of the rod before the collision is zero because it does not rotate about the pivot and the angular momentum of the ball after the collision is zero because it does not rotate about the pivot.$$L_{ball}=m v_0 d=\frac{1}{3}ML^2\omega~\Rightarrow~\omega=\frac{3mv_0d}{ML^2}.$$
(b) Let $\bar F$ be the average force exerted by the rod on the ball. The impulse on the ball is $$J=\frac{\Delta p}{\Delta t}=\frac{0-mv_0}{\Delta t}.$$ The average torque is $$\bar{\tau}=\bar F d=-\frac{mv_0 d}{\Delta t}= \frac{\Delta L_{ball}}{\Delta t}.$$See how it works?

I think so. What I'm seeing is that there is a torque about the door pivot to the ball, and that's why AM is conserved. But the balls center of mass sees no torque which is why it doesn't rotate itself(which was my initial confusion). I'm also seeing how everything relates to the door and ball around that point, hence that always is the center of attention in the problem. Thank you

 

[etotheipi]

@sophiecentaur if we are both talking about the collision with the rod, then there are a few things that don't make sense in your last post

If there is no friction and the tangential velocity of the surface that the ball strikes is equal to that of the ball at the point of contact then there will be no torque impulse.

Two things cannot collide if they have the same initial velocity.

But if the two velocities are not equal and there is friction, the ball could have a change in angular momentum.

If the two velocities are not equal then there can be a collision, but the ball can undergo a change in angular momentum even without friction, in all coordinate systems where the normal force between the rod and the ball has non-zero lever arm.

In the diagram above, there is normal contact so the velocities are equal so no transfer of angular momentum.

During a collision, the velocities of each body are not the same whilst the normal force is present (apart from one instant sometime during the collision), but instead each is accelerating from their respective initial to final velocities. That is to say, the normal force must be present even when the relative velocities are not zero (the bodies are deforming). The force between them is a delta function as the collision duration approaches zero, which results in both a linear and an angular impulse during the collision.

But, if the contact time is finite and the door moves a finite amount during contact then why no transfer? The velocities can't be the same throughout the contact, can they?

Angular momentum will also be transferred between the two objects. The calculation will be a little different, since there will now be a component of impulse perpendicular to the initial velocity, so the assumption that the velocity of the ball doesn't change direction needs to be updated.

 

[alkaspeltzar]

Does the reaction torque about the pivot from the rod onto the ball change the work done on the ball? Work is fxd, and the ball only has kinetic energy. So if looking at a torque on the ball, does that change things?