Collision of a bullet on a rod-string system: query

[palaphys]

Homework Statement

In attachment below

Relevant Equations

Tau= I alpha, L= rxp

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.

Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string acting as a constraint. Also, it is clear from intuition that the rod will aquire an angular velocity in an anticlockwise manner just after the collision. Hence, the point P has a radial acceleration, as well as centripetal acceleration, with respect to the ground frame.

Does that mean, when observed from the frame of P, there must be a PSEUDO force on the center of mass, and hence, angular momentum is NOT conserved about that point?

But the solution to this question seems to suggest otherwise.

Please help.

 

[palaphys]

All relevant attachments are below, for those who need the solution/ question statement

 

[PeroK]

A pseudoforce is only relevant if you want to analyse the problem in an accelerating reference frame. In general, different quantities are conserved in an accelerating reference frame compared to an inertial reference frame - because of the additional pseudoforce.

 

[haruspex]

There is an instantaneous impulse from the string. This has no moment about P, so does not affect angular momentum about P.
But note that "P" is ambiguous here. Does it mean a fixed point in the rest frame or does it mean wherever that end of the rod is at any point in time? In calculating angular momentum, this can matter. It is safe to use fixed points.

 

[palaphys]

There is an instantaneous impulse from the string. This has no moment about P, so does not affect angular momentum about P.
But note that "P" is ambiguous here. Does it mean a fixed point in the rest frame or does it mean wherever that end of the rod is at any point in time? In calculating angular momentum, this can matter. It is safe to use fixed points.

A pseudoforce is only relevant if you want to analyse the problem in an accelerating reference frame. In general, different quantities are conserved in an accelerating reference frame compared to an inertial reference frame - because of the additional pseudoforce.

Well I am considering the angular momentum about point P.
Case 1: what would happen if I considered the angular momentum about the point P on the rod?
Case 2: is it possible to consider the angular momentum about some other point which is probably fixed to the GROUND, but coincides with P? Is that ''legal'' ? Because it does not feel right to me

 

[PeroK]

Well I am considering the angular momentum about point P.
Case 1: what would happen if I considered the angular momentum about the point P on the rod?
Case 2: is it possible to consider the angular momentum about some other point which is probably fixed to the GROUND, but coincides with P? Is that ''legal'' ? Because it does not feel right to me

I think @haruspex identified the source of your confusion. In any case, it all depends whether you transform to an accelerating reference frame or not. Choosing a different point (in space) for an instantaneous analysis does not do that. Of course, if you follow a point on the rod as it moves and keep this as the origin over time, then you are using an accelerating reference frame.

 

[palaphys]

I think @haruspex identified the source of your confusion. In any case, it all depends whether you transform to an accelerating reference frame or not. Choosing a different point (in space) for an instantaneous analysis does not do that. Of course, if you follow a point on the rod as it moves and keep this as the origin over time, then you are using an accelerating reference frame.

I want to keep it as simple as possible, but I want to know what the difference would be, when I consider a material point on a body vs point on ground (space)

 

[PeroK]

I want to keep it as simple as possible, but I want to know what the difference would be, when I consider a material point on a body vs point on ground (space)

A point on a body, used as the origin over time, might imply an accelerating reference frame. Newton's laws (without the addition of pseduo-forces) demand an inertial reference frame.

 

[palaphys]

A point on a body, used as the origin over time, might imply an accelerating reference frame. Newton's laws (without the addition of pseduo-forces) demand an inertial reference frame.

My main question is that how can I conserve angular momentum about P??? The solution some how includes this. But it has not specified which point P is (whether on the rod or on the ground)

 

[PeroK]

My main question is that how can I conserve angular momentum about P??? The solution some how includes this. But it has not specified which point P is (whether on the rod or on the ground)

P is taken to be a fixed point. Or, more accurately, the solution assumes we are working in the inertial reference frame where the ground remains at rest.

 

[haruspex]

Btw, I would not bother finding the CM or its velocity or moment of inertia. It is simpler to find the momentum and angular momentum of each mass separately then sum as appropriate.

 

[palaphys]

Btw, I would not bother finding the CM or its velocity or moment of inertia. It is simpler to find the momentum and angular momentum of each mass separately then sum as appropriate.

But here in the question, it seems that it is unsolvable without finding the MOI, velocity of cm etc.

 

[palaphys]

P is taken to be a fixed point. Or, more accurately, the solution assumes we are working in the inertial reference frame where the ground remains at rest.

Okay, I am clear now. thanks

 

[palaphys]

I was going through the solution once more, and yet another thing struck me: in the solution, both alpha and Omega are given a +ve sign. How is this correct? Aren't the angular acceleration and angular velocity in opposite directions?? In the attatchment below, I have considered alpha and Omega to have the same signs. However aren't they in opposite (i.e alpha is clockwise and Omega is anticlockwise?)

 

[PeroK]

I was going through the solution once more, and yet another thing struck me: in the solution, both alpha and Omega are given a +ve sign. How is this correct? Aren't the angular acceleration and angular velocity in opposite directions??

If an object starts at rest, then the velocity just after an impact must be in the direction of the initial acceleration/impulse.

 

[palaphys]

If an object starts at rest, then the velocity just after an impact must be in the direction of the initial acceleration/impulse.

That sounds intuitive, but torque analysis about the CM using right hand rule tells me that angular acceleration is in the clockwise direction, and intuition tells me the rod will rotate anticlockwise, and Omega is in that direction as well ( dtheta/dt)

 

[PeroK]

That sounds intuitive, but torque analysis about the CM using right hand rule tells me that angular acceleration is in the clockwise direction, and intuition tells me the rod will rotate anticlockwise, and Omega is in that direction as well ( dtheta/dt)

If the initial angular acceleration is $\alpha(0)$, then the angular velocity after a small time interval is approximately $\omega(\Delta t) \approx \alpha(0) \Delta t$.

 

[Lnewqban]

That sounds intuitive, but torque analysis about the CM using right hand rule tells me that angular acceleration is in the clockwise direction, and intuition tells me the rod will rotate anticlockwise, and Omega is in that direction as well ( dtheta/dt)

Without the attached string, how the rotation would happen?

 

[palaphys]

Without the attached string, how the rotation would happen?

Anticlockwise. But how is that going to help?

 

[palaphys]

If the initial angular acceleration is $\alpha(0)$, then the angular velocity after a small time interval is approximately $\omega(\Delta t) \approx \alpha(0) \Delta t$.

But, I thought when solving problems we use a proper sign convention..

 

[PeroK]

But, I thought when solving problems we use a proper sign convention..

The sign convention doesn't matter. Initial acceleration and "Initial" velocity are in the same direction.

That only applies when starting from rest, of course.

 

[kuruman]

But, I thought when solving problems we use a proper sign convention..

You thought correctly. Nevertheless, ask yourself the question "If a rod, initially at rest, is given an angular acceleration $\alpha(0)$, will its angular velocity not be in the same direction as the angular acceleration but will depend on the choice of "proper" sign convention?" Nature does not follow human conventions.

 

[PeroK]

The sign convention doesn't matter. Initial acceleration and "Initial" velocity are in the same direction.

That only applies when starting from rest, of course.

Okay, I see that the initial angular velocity might be non-zero in this case and the rotation immediately slowing down.

 

[palaphys]

You thought correctly. Nevertheless, ask yourself the question "If a rod, initially at rest, is given an angular acceleration $\alpha(0)$, will its angular velocity not be in the same direction as the angular acceleration but will depend on the choice of "proper" sign convention?" Nature does not follow human conventions.

Yes, I see it now. It seems rather senseless to assign angular acceleration and angular velocity opposite directions here, especially as the rod is starting from rest.

 

[palaphys]

Alright got it, everything is clear now.

 

[haruspex]

But here in the question, it seems that it is unsolvable without finding the MOI, velocity of cm etc.

Not so. Pretty much the only time you need to find the CM of an assembly is when you are asked for something specifically related to that, such as its position, velocity, etc.

 

[Lnewqban]

Anticlockwise.

How and why does the string interfere with that rotation?

 

[WWGD]

A Rod? I thought he had retired. Ask @Charles Link , he's the main Baseball guy here.

 

[Charles Link]

A Rod? I thought he had retired. Ask @Charles Link , he's the main Baseball guy here.

Yesterday, with @WWGD mentioning me, was the first look that I had at this thread. I read the problem in the OP, but I don't want to look at any solutions that have been posted in the thread until after I spend at least a little time trying to solve it myself. From my first impressions of it, it looks like the string supplies both an impulse $\Delta(mv)= \int F \, dt$ along with a very short $\Delta L= \int \tau \, dt$, (for which $T$ can't be computed, but rather $\int T \, dt$), after which there is likely to be a tension $T$ supplying a centripetal force, but this looks to me to be a somewhat non-trivial problem, and it might take me a day or two or longer to come up with a solution. It looks very interesting though.

 

[Charles Link]

I now have a preliminary solution to the problem posed in the OP. I think you can make the assumption that the string can stretch a small amount before it has any effect, so that the rod can be treated as a free object for the initial impact. The rod/bullet system will thereby have an initial velocity of its center of mass, along with an angular velocity about its center of mass. It is then that the string will come into play, exerting an impulse of $\int T \, dt$ which has vector components that will cause a momentum change and thereby a change in the center of mass velocity, as well as a change in the angular momentum from the torque, and thereby a change in the rotational rate about the center of mass. The impulse $\int T \, dt$ must result in the velocity of the attachment point to be zero in the direction of the string.

Preliminary results are that the moment of inertia $I=(5/24)mL^2$, and after some lengthy calculations $\int T \, dt=4 mv_o \cos(\theta)/(27 \cos^2(\theta)+5)$.

I will try to post more detail over the next day or two. I also anticipate I may have some algebraic errors.

One additional comment is that right after this impulsive $\int T \, dt$, one could anticipate that the tension in the string is near zero because the velocity of the point of attachment along the string is zero.

Edit: It may also be worth mentioning that the torque=rate of change of angular momentum applies to either a stationary point as the origin, but also holds when referenced about the center of mass, even if the center of mass is undergoing an acceleration.

 

[haruspex]

@Charles Link
I think we are reading the question differently. It asks for the tension after the impact, not the tensional momentum delivered during the impact.
Seems to me we have to find the motions immediately after impact and deduce the tension from those.

Let the post-impact angular velocity be $\omega$ anticlockwise and the speed of the left end of the rod be $u$ down and left.
The post-impact velocities of the bullet and rod centre can each be resolved as the sum of the velocity of the left end of the rod and a velocity relative to that, the latter being purely in the positive y direction and having magnitudes $L\omega/2, L\omega$ respectively.
For convenience I'll write $s=\sin(\theta)$. For clarity, I'll write M for the rod's mass.
Conservation of linear momentum normal to the string:
$mvs=-mu+mL\omega s -Mu+ML\omega s/2$
Conservation of angular momentum about P (as a fixed point in space):
$mvL=mL^2\omega+ML^2\omega/3$
so $L\omega=\frac{3mv}{3m+M}$
Combining:
$(m+M)u=(m+M/2)sv\frac{3m}{3m+M}-mvs$
$=\frac{Mmvs}{6m+2M}$
Now setting M=m we get
$u=vs/16$
$\omega=\frac{3v}{4L}$

From here it gets messy, so less convincing…

The left end of the rod now has acceleration $u^2/D$ towards the fixed attachment point. The rod's mass centre and the bullet have acceleration $\omega^2(L/2), \omega^2L$ respectively leftwards relative to the left end of the rod.
In the direction of the string, these last have components $\omega^2(L/2)s, \omega^2Ls$ respectively.
The tension in the string is responsible for the net acceleration of the system in that direction:
$T=2mu^2/D+m\omega^2(3L/2)s$
$=mv^2s(\frac{s}{128D}+\frac{27}{32L})$

 

[Charles Link]

Very interesting, but I need to think about it a whole lot more. With the approach that I took, the system gets balanced by an impulse from the string (which changes the velocity of the center of mass as well as the rotational rate) in such a way that the point of attachment will then have zero velocity in the direction of the string and it will be moving only perpendicular to the direction of the string. The motion is a result of the motion of the center of mass and also from a rotation about the center of mass. The string thereby experiences zero force after the initial impulsive response, where for a very short instant the tension could be enormous, but only lasts for a short instant.

I calculated the impulse that would make this velocity component in the direction of the string zero, which it must, because the string is assumed to stretch only a very little.

I'm not so sure that the assumption that the point P is fixed is valid. I got my old Mechanics book by Becker from the shelf earlier today, and the reference points for the angular momentum formulas must be selected carefully. The point of attachment undergoes an acceleration, so it may not be valid to assume no torques from that point implies conservation of angular momentum from that point, as I believe you did in your calculations.

I did assume an instantaneous capture of the bullet by the system before the string responded to that result. Perhaps this isn't completely justified, but it served to simplify the problem conceptually.

 

[haruspex]

With the approach that I took, the system gets balanced by an impulse from the string (which changes the velocity of the center of mass as well as the rotational rate) in such a way that the point of attachment will then have zero velocity in the direction of the string and it will be moving only perpendicular to the direction of the string.

Quite so.

The motion is a result of the motion of the center of mass and also from a rotation about the center of mass. The string thereby experiences zero force after the initial impulsive response,

Not so. Since the left end of the rod will have a speed, it is then describing an arc around the fixed point of the string, so is accelerating. This implies a continuing tension.
What you calculated was not, in fact a tension: it was a momentum, so different dimensions.

I'm not so sure that the assumption that the point P is fixed is valid.

I'm not assuming point P is fixed. I am defining my reference point for conservation of momentum as a fixed point located where P is in the diagram.

I got my old Mechanics book by Becker from the shelf earlier today, and the reference points for the angular momentum formulas must be selected carefully.

Quite so, again. It is safe to use a fixed point in space (as I did) or the mass centre of the rigid body in question. More generally, any point for which the line of acceleration passes through the mass centre. If not, you can make it right by including the usual fictitious force.

I did assume an instantaneous capture of the bullet by the system before the string responded to that result.

We do have to assume the capture is, in the limit, instantaneous.

 

[Charles Link]

Not so. Since the left end of the rod will have a speed, it is then describing an arc around the fixed point of the string, so is accelerating. This implies a continuing tension.
What you calculated was not, in fact a tension: it was a momentum, so different dimensions.

I computed $\int T \, dt$, and with what I computed, the velocity of the point of attachment is moving perpendicular to the string immediately after the impulse. I have yet to compute what it does after that. Perhaps I should write out my expressions in detail, but (my computations are) for that instant just after the impulse response of the string, which is the response to the rod which absorbed the bullet and starting spinning in such a way that an impulse from the string was then needed to change these initial conditions to new initial conditions. With the new initial conditions just after the impulse of the string, the string is no longer being stretched at all, so it will not supply any force. I think I have an alternative interpretation. I'm only computing things for a brief instant, but even if the point of attachment is then moving in the start of a circle, it is not a point mass of 2m that is attached. Instead I have the object rotating and translating such that the string is no longer being tugged on. I have a steady angular momentum/rotation about the center of mass, along with a translation.

 

[Charles Link]

and a follow-on: again I'm not so sure about your point of reference. The center of mass is stated in Becker to be one of the reference points where the formulas apply without any correction terms. If the point of attachment were fixed, what you computed for angular velocities would certainly work. I'll need to look at Becker some more to see what the correction term is. With my computation, I believe I have the system rotating at a slightly different rate than what your calculations gave.

 

[wrobel]

Homework Statement: In attachment below
Relevant Equations: Tau= I alpha, L= rxp

My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?

The point P is a point of the lab system that coincides with the end of the rod before the collision. If the collision happens at the moment $t$ then the angular momenta of the system about P at the time $t-0$ and time $t+0$ are the same.
A complete system of equations for the velocities of the system right after the collision follows from theory of impact. See for example Pars: A Treatise on ANALYTICAL DYNAMICS

 

[haruspex]

With the new initial conditions just after the impulse of the string, the string is no longer being stretched at all, so it will not supply any force.

After the impulse, the string ensures that the left end of the rod stays at constant distance from the string's fixed point. This means the left end of the rod is, immediately, accelerating towards that fixed point. For that to happen, there must be tension in the string.

and a follow-on: again I'm not so sure about your point of reference. The center of mass is stated in Becker to be one of the reference points where the formulas apply without any correction terms. If the point of attachment were fixed, what you computed for angular velocities would certainly work. I'll need to look at Becker some more to see what the correction term is. With my computation, I believe I have the system rotating at a slightly different rate than what your calculations gave.

My point of reference is a fixed point in space. That is always a valid choice.
The impulsive tension acts through that point, so does not alter the angular momentum of the rod+bullet system about that point.
To express the post-impact angular momentum, we only need the instantaneous velocities after the impact.

For completeness, I have now calculated the impulse from the string. I get $mv\cos(\theta)/8$.

 

[haruspex]

The point P is a point of the lab system that coincides with the end of the rod before the collision. If the collision happens at the moment $t$ then the angular momenta of the system about P at the time $t-0$ and time $t+0$ are the same.
A complete system of equations for the velocities of the system right after the collision follows from theory of impact. See for example Pars: A Treatise on ANALYTICAL DYNAMICS

See post #25.

 

[Charles Link]

For completeness, I have now calculated the impulse from the string. I get mvcos⁡(θ)/8.

This is super. We now have something to compare to, and I think it shows some very interesting features. I'm not going to try to claim that I have a completely correct result, but I obtained (post 30) that $\int T \, dt=4mv_o \cos{\theta}/(27 \cos^2{\theta}+5)$. Our solutions agree precisely for $\theta=0$, and I think I know why. I'm assuming zero velocity along the string, just after the collision. Meanwhile, by you specifying the rotation to be $\omega$ about the point of attachment, you fix that point. They agree for $\theta=0$, when there will be zero velocity after the impulse perpendicular to the string. For other angles I do get a finite perpendicular velocity, but you have made the point of attachment the center of the rotation, so you seem to have fixed it. You do get a finite $u$, but this seems to come from other calculations, like linear momentum considerations. The result is that following the two impulses, (from the bullet and the string), your initial conditions are slightly different from mine.

I really made no attempt to compute any motion after the two impulses. I saw that my system was balanced, and at least for a brief moment, it would not be tugging on the string.

Edit: To carry it one step further, there is an additional term involving a cross product of $\ddot{\vec{r_o}}$ that surfaces, (comes from Becker Mechanics text). When we have $\dot{r_o}=0$ for both of us, we then have $\ddot{r_o}=0$ as well. I think this might in fact be the underlying detail that results in what is seemingly an inconsistency, but I'm not certain.

 

[haruspex]

by you specifying the rotation to be ω about the point of attachment, you fix that point

No, I don’t fix that point. I wrote:

Let the post-impact angular velocity be $\omega$ anticlockwise and the speed of the left end of the rod be $u$ down and left.

I then expressed the motions of the rod and bullet in terms of these:

The post-impact velocities of the bullet and rod centre can each be resolved as the sum of the velocity of the left end of the rod and a velocity relative to that, the latter being purely in the positive y direction and having magnitudes $L\omega/2, L\omega$ respectively.

That is, in the lab frame, the mass centre of the rod has x, y velocity components $(-u\sin(\theta), -u\cos(\theta)+\omega L/2)$.

What I did fix is the reference point for conservation of angular momentum.
Perhaps it would have been clearer if I had used the fixed end of the string for that. The algebra is the same.

Btw, I notice there is a problem viewing post #31. If you click the blurred text to make it visible then the latex is not rendered. To view it, click reply, then click the magnifying glass icon to render the latex, then click the text to unblur.

 

[Charles Link]

@haruspex Your solution in post 31 looks to be a very good one. Thank you. :)

I've looked over my solution carefully=I will need to continue to look at it, but I don't seem to be able to make the $27 \cos^2{\theta}+5$ into a $32$ in the denominator of the impulsive response $\int T \, dt$ for arbitrary $\theta$ to agree with your result.
( I have $\int T \, dt=4 mv \cos{\theta}/(27 \cos^2{\theta}+5)$ and you have $\int T \, dt = mv \cos{\theta}/8$).

It is possible I missed on some algebra, but I have to wonder if the answer to this is something very fundamental that is yet to be uncovered. Otherwise, I would think our two methods, if both are fundamentally correct, should give us the same results for all angles $\theta$. Cheers. :)

 

[haruspex]

@haruspex Your solution in post 31 looks to be a very good one. Thank you. :)

I've looked over my solution carefully=I will need to continue to look at it, but I don't seem to be able to make the $27 \cos^2{\theta}+5$ into a $32$ in the denominator of the impulsive response $\int T \, dt$ for arbitrary $\theta$ to agree with your result.
( I have $\int T \, dt=4 mv \cos{\theta}/(27 \cos^2{\theta}+5)$ and you have $\int T \, dt = mv \cos{\theta}/8$).

It is possible I missed on some algebra, but I have to wonder if the answer to this is something very fundamental that is yet to be uncovered. Otherwise, I would think our two methods, if both are fundamentally correct, should give us the same results for all angles $\theta$. Cheers. :)

Would you be happy to post your working, in a PM maybe?

 

[Charles Link]

and a follow-on: We both agree that $\int T \, dt =0$ for $\theta=90$ degrees. You have a result (post 31) that $u=v \sin{\theta}/16$.
I wind up with a very complex expression for $u$, but at 90 degrees, it simplifies to $u=(2/5) v$.

For this case, (90 degrees), with $\int T \, dt=0$, the string is out of the picture. I am working using the center of mass as my reference point, and I think you should readily be able to verify that my $2/5$ is correct, and that something is incorrect with the $1/16$ result. I urge you to try this simpler case with the center of mass as the reference.

and to respond to the above, PM would work ok, but I may post my results in the thread. It will take a little time, but it might be worth it, so that you and others can see how I computed it.

Note: For the moment of inertia about the center of mass for the rod with bullet attached, I get $I=(5/24)mL^2$.

 

[Charles Link]

and a follow-on: It just occurred to me with your angular momentum calculation, that when using the reference point the way you did that any forces that are applied at that point are totally excluded from your result. I can see how that would be the case when the force is applied at the center of mass. In that case no rotation results. With your method though, you get the same angular velocity for the system if you apply an enormous force on the reference point or no force at all. I do think you might be leaving off a correction term, e.g. of the kind that Becker shows in his text if the point is accelerating.

In the case of a door swinging on a hinge, you can apply whatever force you want on the hinge and it doesn't affect the result. In the case we are treating above, the point of attachment is not fixed like a hinge.

 

[haruspex]

Ha! Found my error. Missed two terms…

Let the post-impact angular velocity be $\omega$ anticlockwise and the speed of the left end of the rod be $u$ down and left.
The post-impact velocities of the bullet and rod centre can each be resolved as the sum of the velocity of the left end of the rod and a velocity relative to that, the latter being purely in the positive y direction and having magnitudes $L\omega/2, L\omega$ respectively.
For convenience I'll write $s=\sin(\theta)$. For clarity, I'll write M for the rod's mass.
Conservation of linear momentum normal to the string:
$mvs=-mu+mL\omega s -Mu+ML\omega s/2$
Conservation of angular momentum about P (as a fixed point in space):
$mvL=mL^2\omega+ML^2\omega/3$
$-muLs-Mu(L/2)s$ (the missing terms)

Plugging in M=m:
$vs=-2u+\frac 32L\omega s$
$v=\frac 43L\omega-\frac 32us$
Hence $u=\frac{2v}{32-27s^2}$, $L\omega=\frac 34v(1+\frac{3s}{32-27s^2})$.

Check: s=1 gives $u=\frac 25v, L\omega=\frac 65v$.

The left end of the rod now has acceleration $u^2/D$ towards the fixed attachment point. The rod's mass centre and the bullet have acceleration $\omega^2(L/2), \omega^2L$ respectively leftwards relative to the left end of the rod.
In the direction of the string, these last have components $\omega^2(L/2)s, \omega^2Ls$ respectively.
The tension in the string is responsible for the net acceleration of the system in that direction:
$T=2mu^2/D+m\omega^2(3L/2)s$
For s=1 that gives $\frac 1{25}mv^2(\frac 8D+\frac{54}L)$.

For the momentum the string imparts during impact, I now get $\frac 18mvc(1+\frac{27s}{32-27s^2})$, where $c=\cos(\theta)$.

 

[Charles Link]

Looks very good. We now agree at 90 degrees, but in a way, that is expected. We agreed for $\int T \, dt =0$ for 90 degrees, and no forces are being applied at the reference point. Please read my post 44 carefully.

Looks like you now have a corrected form for $u$, but my expression for $u$ is still much more complicated than what you have for your corrected version. [Edit: I recomputed my $u$ and yes, I now agree with your result. 8-29-25 8:00 PM]

Edit: and I looked at your post 45 carefully now. Those missing terms were easy to miss. We now are in agreement for the case where no forces are acting on the reference point, but we still need to get our results to agree when we have forces acting on the reference point. We do agree for $\theta=0$ degrees as well, but for that case, the point of attachment is basically fixed. Please see post 44.

 

[haruspex]

and a follow-on: It just occurred to me with your angular momentum calculation, that when using the reference point the way you did that any forces that are applied at that point are totally excluded from your result. I can see how that would be the case when the force is applied at the center of mass. In that case no rotation results. With your method though, you get the same angular velocity for the system if you apply an enormous force on the reference point or no force at all. I do think you might be leaving off a correction term, e.g. of the kind that Becker shows in his text if the point is accelerating.

In the case of a door swinging on a hinge, you can apply whatever force you want on the hinge and it doesn't affect the result. In the case we are treating above, the point of attachment is not fixed like a hinge.

I'll say it again: my reference point for the angular momentum calculation is fixed. It is the point in space where the attachment between the rod and string happened to be when the impact occurred. It does not move, let alone accelerate.
Although I understood perfectly well what I needed therefore to include when I wrote out the equation, I embarrassingly omitted precisely the two terms implied by that.

Btw, note that, as foreseen in my posts #11 and #26, I avoided finding the CoM and the MoI of the merged system.

I see that if I were to change the $27s$ in the numerator in my expression for the impulse to $27s^2$ I would get your result. That would mean the sign of $\theta$ does not affect it. That feels right. I'll see if I can find a dropped factor s.

 

[Charles Link]

I'll say it again: my reference point for the angular momentum calculation is fixed. It is the point in space where the attachment between the rod and string happened to be when the impact occurred. It does not move, let alone accelerate.

But the rod moves at the point of attachment. For the case of $\theta=0$ degrees, we are also in agreement, because the string keeps the point of attachment from moving. When $\theta$ is some arbitrary angle, our results differ, and I believe the problem is the torque at the reference point, which gets incorrectly excluded from your computation, but the force on your "fixed point" does get considered if you use the center of mass for a reference.

Meanwhile forces on the center of mass don't count as torques when the center of mass is the reference because they don't cause any rotation, so a calculation with the center of mass as reference will not consider them. You do need to consider forces on your reference point though. Becker does display the type of correction term that is needed, but it is somewhat complex.

I do think in your computation, you are failing to take fully into account the effects of forces/torques applied at your reference point, i.e. from the string.

 

[haruspex]

But the rod moves at the point of attachment.

Yes, the point of attachment moves, but my reference point doesn't. Let’s say there is a mark on the table at P. That is my reference point.

Becker does display the type of correction term that is needed, but it is somewhat complex.

I am aware that if the reference point accelerates then one has to include the usual fictitious force associated with accelerating reference frames. But if the mass centre lies on the line of acceleration of the reference point then the fictitious force exerts no momentum about the mass centre, so can be ignored.
Since my reference point is fixed, the issue does not arise.
Note that the two terms I accidentally omitted are the difference between taking the reference point as fixed at the initial attachment point and taking it as dynamically where the attachment point is (and failing to plug in a fictitious force).

 

[Charles Link]

Yes, the point of attachment moves, but my reference point doesn't. Let’s say there is a mark on the table at P. That is my reference point.

This one seems to be open to interpretation if the point on the body that sits there accelerates, and it does in this case. I would have worked it how you did, had I worked it using the reference point you did.

Consider balancing a yardstick on your finger, and accelerating it upward. It does not rotate regardless of how fast you move it, and the force you apply only winds up in the linear momentum calculation. If you are off center though, the force you apply does affect the rotation rate.

The problem we have with the string attached I think is one of those special ones that we don't encounter very often. If you do the calculation from the center of mass, the applied force from the string does wind up in the angular momentum formula. It does affect the rotation rate.

When the applied force is through the center of mass, that is really a special case, in that it does wind up in the angular momentum calculation, e.g. a door on a hinge, if you use the hinge for the reference point, but it doesn't wind up in the formula if you use the center of mass as reference. That seems to be a special case though, and we can not ignore the torque from the string if we have an acceleration going on in the body at that point. Even the textbook Becker is difficult to interpret on this case, but I do think we have a case where the "fictitious" force, and it really isn't fictitious in this case, needs to be included in the angular momentum consideration.

I thank you for your patience on this one. This seems to be one of those that only come up on rare occasion, but I think it is one that really puts us at the drawing board to interpret it correctly. Cheers. :)