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

[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. :)

 

[Charles Link]

and a follow-on: I'm taking a very close look at Becker's discussion and derivation of the angular momentum law, and this one is looking to be a very subtle one. I think I may have located the source of the difficulty, but I need to look it over carefully. We have angular momentum $J=\sum r_i \times m_i \dot{r}_i$. You @haruspex discovered above (post 45) that this angular momentum is more than just $I \omega$, because there are the $u$ terms as well in the velocity $\dot{r}_i$, i.e. $\dot{r}_i=r_i \dot{\theta}+u$, where $\omega=\dot{\theta}$. I just recognized that also. I think we might find ourselves to now be in agreement with each other, other than perhaps a couple of algebraic errors that we need to correct. Very good. :)

($u$ is the velocity of the body at the point of attachment. This $\theta$ of $\omega= \dot{\theta}$ is not to be confused with the other $\theta$ of the angle of the string. The two $\theta$'s are different.)

and note, with this added $u$ being the same for each particle, we have $\sum r_i m_i =0$ at the center of mass, so these missing terms don't appear in the calculation when done from the center of mass. Hope you agree @haruspex with all of this. Cheers. :)

and note you did have it correct, that your frame of reference is indeed an inertial one. My mistake here. But we do not have $J=I \omega$ for this case with the reference being the point of attachment, and that seems to be the thing that was our stumbling block.

and note, in the case of a door on a hinge, we do have $J=I \omega$, because $u=0$.

additional note: The moment of inertia $I$ of course needs to be computed from whatever reference point that is used.

and note also that $\omega$ is the same for a rigid body, whether it is referenced from the center of mass or from some other point.

 

[Charles Link]

@haruspex Please see my post 51. Together I think we successfully solved it=at least the first part. What it does after the initial impulsive response, I still have yet to figure out, but perhaps your solution in post 31 is how the second part works.

 

[haruspex]

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.

Angular momentum of a body relative to a reference point has, in principle, two components: the spin angular momentum (its angular momentum about its own mass centre) and orbital angular momentum (arising from its linear motion on a line displaced from the reference point).
In the scenario you describe, pushing up harder above the (fixed) offset reference point increases the magnitude of each, but they have opposite signs and the increases cancel.

 

[Charles Link]

Angular momentum of a body relative to a reference point has, in principle, two components: the spin angular momentum (its angular momentum about its own mass centre) and orbital angular momentum (arising from its linear motion on a line displaced from the reference point).
In the scenario you describe, pushing up harder above the (fixed) offset reference point increases the magnitude of each, but they have opposite signs and the increases cancel.

The rotation rate will increase, but in any case, I think you successfully located what was causing the big dilemma, and I found it also somewhat independently afterward when I studied Becker's derivation and looked carefully at his $r_i \times m_i \dot{r}_i$ terms. See my post 51.(I really was still at the drawing board in post 50). and thank you so much for your patience in the matter. I was incorrect when I thought the problem was that the point was not an inertial frame, but we did have a major inconsistency, and you discovered it first when you spotted (post 45) the missing terms in the angular momentum consideration. Cheers. :)

Edit: It should be noted before you spotted the correction terms of post 45, you basically were writing (post 31) $J=I \omega$ at the point of attachment, which we both see now does not work if the velocity at the point of attachment $u$ is non-zero. It needed the two additional corrections terms with the $u$ in them to have the correct expression for the angular momentum.

 

[kuruman]

I have (had) a different approach to this problem, which I have not gotten to work, and I think I know why. So please hear me out.

Step 1. Initial considerations
Referring to the original diagram, we are looking for the tension $T(\theta)$. This tension has a vertical and a horizontal component. Since the impulse delivered to the rod is vertical, only the vertical component $T(0)$ comes into play at the moment immediately after the collision. Then $T(\theta)=T(0)/\cos\theta.$

Step 2. Find $T(0).$
The approach to that is to solve the "no string attached" problem first, then find the tension needed to keep the left end of the table at rest relative to the table.

a. The first part is a standard problem of linear and angular momentum conservation. We note that
the CM has velocity $V_{\text{cm}}=\frac{1}{2}v_0$ throughout the collision
the angular momentum relative to a point on the table coinciding with the CM can be written as $L_{\text{cm}}=mv_0\frac{1}{4}L$ before the collision and as $L_{\text{cm}}=I\omega$ where $I=\frac{5}{24}mL^2.$ Thus, conservation of angular momentum yields $$\omega=\frac{6}{5}\frac{v_0}{L}.$$ We conclude that immediately after the collision, the CM will be moving up with speed $\frac{1}{2}v_0$ and simultaneously rotate about the CM with angular speed $\omega.$ Specifically, the speed of the left end of the rod will instantaneously rise from zero to $~v_{\text{left}}=\frac{3}{4}L\omega=\frac{9}{10}v_0$ directed down.

b. Given the above, the question is, what instantaneous tension $T(0)$ is required to keep the left end of the rod instantaneously at rest? This translates to keeping the rod from rotating about the CM and can be easily answered, if the instantaneous force $F$ at the right end is known. Both forces are up and we can balance torques about the CM, $$T(0)\frac{3}{4}L-F\frac{1}{4}L=0\implies T(0)=\frac{1}{3}F.$$ Step 3. Conclusion
This doesn't solve the problem, because on the right end of the rod we have to convert an impulse $J$ to a force $F$ without knowing the time interval over which the force is acting. The problem states that the impact is instantaneous but that doesn't help. Also note that regardless of the approach, one still has to convert an impulse to a force without knowing $\Delta t$. This reminds me of @haruspex's pet observation that one cannot find the average force given the change in velocity and the displacement but not $\Delta t.$ Here we are asked to find the force given zero displacement and zero $\Delta t$ and an impulse $J.$ I don't see how it can be done and that is why I decided, finally, to post my thoughts.

 

[Charles Link]

@kuruman Somewhat good, but you forgot to include $v_{cm}$ to the velocity of the left end of the rod for the first part. Once you make this correction, I think your numbers will agree with what @haruspex and I previously computed for the zero attachment case. See posts 43 and 45. We get $u=(2/5) v$.

and looking further in your post, (I see you are somewhat puzzled), both @haruspex and I have agreed on what the impulse $\int T \, dt$ is from the string just after impact. I have not attempted to solve what the system does after that, but suggestion is to read through our latest inputs carefully through the thread. I think we have come up with solutions that agree using both the center of mass as the reference and the point of attachment as a reference. Suggest you read my post 51 before post 50, because on post 50, I was still trying to resolve an inconsistency.

I do agree with your calculations that $I_{cm}=(5/24)mL^2$, and also that $\omega=(6/5)v/L$ for the free case.

 

[kuruman]

@kuruman Somewhat good, but you forgot to include $v_{cm}$ to the velocity of the left end of the rod for the first part. Once you make this correction, I think your numbers will agree with what @haruspex and I previously computed for the zero attachment case. See posts 43 and 45. We get $u=2/5 v$.

Yes, I foolishly thought that the CM velocity need not be added.

and looking further in your post, (I see you are somewhat puzzled), both @haruspex and I have agreed on what the impulse $\int T \, dt$ is from the string just after impact.

That agreement vis-à-vis what the problem was asking is another point I missed. Thanks for the clarifications. I think I understand now.

 

[Charles Link]

In post 31 @haruspex seems to offer a solution to what occurs after the impulse response.

The discussion though focused mostly on the impulse response and @haruspex discovered a very important oversight in his calculations which showed how the impulse response of the string affected his $\omega$ result. (Without the $u$ correction terms in the $J$ formula, he had $J=I \omega$, independent of what the force was at the point of attachment. Thereby $\omega$ that he computed was simply was $J_o/I$ in every case, regardless of what the string did, and that had to be incorrect.) Before his correction, his calculations were showing that the torque at the point of attachment didn't matter at all in the result he got for $\omega$.

In his post 53, he seems to suggest that that is still the case, but I think he needs to look over his corrected formula (post 45), and he will see he has now included the torque, in an indirect way, into his formula, and he now has a correct solution as well. I don't think he has had the time to look carefully over his new corrected result. See also my post 51. (Note there is no torque from the string when using the point of attachment as a reference, but with the center of mass as a reference there is a torque from the string, so that the string must somehow enter into his angular momentum calculation, which it now does (post 45), but it hadn't yet in his post 31 solution).

[Edit: (8-30-25) Hoping @haruspex has the time to take a second look at his post 53. If we look at the resulting $\omega$, see post 59 below, we see it depends on $\theta$ and thereby on the force from the string.]

Note: The calculations done with the center of mass as the reference were showing very clearly that the force from the string would affect the resulting $\omega$, so that $\omega$ would certainly not be independent of $\theta$. Thereby, there had to be an inconsistency somewhere, and @haruspex found it in post 45.

 

[Charles Link]

an update to my post 46 that I just included an "Edit": I recomputed $u$, and I now agree with @haruspex post 45 result for $u=2v/(32-27 \sin^2{\theta} )$. Looks like we are now in complete agreement on our results, using both the point of attachment as the reference and the center of mass as a reference.

One minor additional item is the result for $\int T \, dt=4mv \cos{\theta} /(5+27 \cos^2{\theta})$ See post 41. I think we might now be in agreement on that as well, (see post 45 and post 47 by @haruspex where he needs one more $s$ in the numerator to concur completely). It should also read (post 45) $L \omega=(3/4)v(1+ \frac{3s^2}{32-27 s^2})$ with a $3s^2$ in the numerator, and then we are in complete agreement. (Note: $s=\sin{\theta}$).

 

[kuruman]

This problem is from the Indian National Physics Olympiad – 2020. A "tentative" solution appears here on page 15.