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

[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.

 

[haruspex]

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

The answer looks rather different. Strange that theta does not appear in the final equation for T.
This is troubling:
"1. Angular momentum is conserved only about P."
but later:
"We now calculate the acceleration of P on the rod in the direction of the string"
So is P accelerating or fixed? But this could just be sloppy wording.

 

[Charles Link]

Strange that theta does not appear in the final equation for T.
This is troubling:
"1. Angular momentum is conserved only about P."

They solved for $u$ using $\cos{\theta}=1/3$. (Their result does agree with what we both got). I anticipate that's why there is no $\theta$ dependence. They apparently plugged in that value whenever $\theta$ occurred.

Meanwhile there are no torques about P. (The angular momentum does depend on the choice of reference point). I believe they are using a fixed P like you did, where I thought you might have it incorrect, but it turns out you were right after all. :)

 

[haruspex]

They solved for $u$ using $\cos{\theta}=1/3$. (Their result does agree with what we both got). I anticipate that's why there is no $\theta$ dependence. They apparently plugged in that value whenever $\theta$ occurred.

Meanwhile there are no torques about P. (The angular momentum does depend on the choice of reference point). I believe they are using a fixed P like you did, where I thought you might have it incorrect, but it turns out you were right after all. :)

ok, thanks for checking it all out.

 

[Charles Link]

It might be worth taking an extra look at what we have observed in this problem from working it from a couple of different reference points. The angular momentum appears to be kind of a strange bird. It is different when observed from different reference points in the same inertial frame of reference. In addition we have the angular momentum $J= I \omega$ with reference point as the center of mass, even when it is moving, but when the reference point is an inertial one that isn't the center of mass, additional terms appear in computing
$J$, because $\dot{r}_i=\omega \times r_i +u$ where $u$ is the velocity of body at the reference point. (With the center of mass as reference, the $u$ terms sum to zero when computing $J=\sum r_i \times m_i \dot{r}_i$).

Meanwhile, the moment of inertia $I$ needs to be computed from the reference point that is used, and it differs from point to point, while the angular velocity $\omega$ is independent of the reference point.

We also see that torques are different for the different reference points, and the torque is even zero if it is located at the reference point.

For this last item we see something that occurred in the Olympiad solution of post 60: "We only have angular momentum conservation with point P as the reference", (because the only external force is at point P where the torque vanishes if computed from the reference point P. Otherwise the torque is non-zero, and angular momentum will no longer be conserved).

I found this problem to be very educational, and it really illustrates some of the very detailed concepts that can arise in working with angular momentum.

 

[kuruman]

found this problem to be very educational, and it really illustrates some of the very detailed concepts that can arise in working with angular momentum.

Indeed. This problem was educational for me in that it jarred me out my complacency of "having seen it all."

 

[Charles Link]

To add to post 64, it may be worth mentioning the center of percussion. If you have a door swing on a hinge or are swinging a baseball bat, there will be a reference point on the door or bat called the center of percussion about which the angular momentum is zero. Thereby if an object strikes the door or bat at this point, it will change its angular velocity, perhaps even bringing it to a stop. However, the angular momentum will still be zero after the impact, so that there is no force or torques acting on the hinge or the batter's hands as a result of the impact.

Note that the projectile also has zero angular momentum when the reference point is the point of impact, both just before and just after the impact. The projectile can also be viewed as an external force that has zero torque when the reference point is the point where the force is applied.

This is perhaps a little bit of detail, but it might even be found to be fairly straightforward after working through the details of the exercise presented in the OP, with the complete solution in @kuruman 's post 60.

 

[Charles Link]

It may be worth mentioning that as odd as angular momentum can be there is the case where we get the same results from all of the various reference points in a stationary frame when the body is rotating freely about the center of mass with no linear translational velocity. In that case angular momentum $J=\sum r_i \times m_i \dot{r}_i =r_{ref \, to \, cm} \times \omega \sum m_i r_{io}+\omega \sum m_i r^2_{io}= \omega \sum m_i r^2_{io}=I_{cm} \omega$ is independent of the point of reference, where $r_{io}$ is the vector from the center of mass to the ith point on the body.

(Note: I should really be writing these as vector formulas, but it would take a lot of extra work to include the various vector signs).

 

[kuruman]

It might also be worth mentioning that the spin angular momentum $J=I_{cm}\omega$ may be calculated about a point on the body that is moving relative to the lab frame. I will illustrate with an example specific to the rod mentioned in this thread. We are looking at the case after the bullet is embedded and no string is attached.

The figure on the right is drawn to scale using the problem parameters. The position vectors of the two ends relative to the CM,are

$\mathbf x_B=\frac{1}{4}L~\mathbf{\hat x}~;~~ \mathbf x_P=-\frac{3}{4}L~\mathbf{\hat x}.$
Note that $\mathbf x_B-\mathbf x_P=L~\mathbf{\hat x}$ as expected.

We have already found the angular speed about the CM $\omega=\frac{6}{5}(v_0/L).$ We attach a Cartesian frame to the rod as shown in the figure and write the angular velocity vector $\boldsymbol{\omega}=\omega~\mathbf {\hat z}.$

Now the velocity of point B relative to point P is $$\begin{align} \mathbf v_{BP}= & \mathbf v_B-\mathbf v_P=\boldsymbol{\omega}\times \mathbf x_B-\boldsymbol{\omega}\times \mathbf x_P=\boldsymbol{\omega}\times( \mathbf x_B-\mathbf x_P)=\boldsymbol{\omega}\times \mathbf L=\omega L~(\mathbf{\hat z}\times \mathbf{\hat x}) \nonumber \\ & \therefore~~ \mathbf v_{BP} = \omega L \mathbf{\hat y}. \end{align}$$ It follows from equation (1) that all points on the rod rotate with angular speed $\omega$ relative to point P. The angular momentum about point P is simply $$\mathbf J_P=I_P~\boldsymbol {\omega}=\left(\frac{1}{3}+1\right)mL^2\left(\frac{6v_0}{5L}\right)\mathbf {\hat z}=\frac{8}{5}mv_0L~\mathbf{\hat z}.$$ Contrast this to the angular momentum about point B, $$\mathbf J_B=I_B~\boldsymbol {\omega}=\left(\frac{1}{3}\right)mL^2\left(\frac{6v_0}{5L}\right)\mathbf {\hat z}=\frac{2}{5}mv_0L~\mathbf{\hat z}.$$ More generally, the angular momentum about a point at position $x$ from the midpoint of the rod $~(-\frac{L}{2} \leq x \leq \frac{L}{2}~)$ can be written as $$\mathbf J(x)=I(x)~\boldsymbol {\omega}~;~~~I(x)=\frac{1}{12}mL^2+mx^2+m\left(\frac{L}{2}-x\right)^2.$$

 

[Charles Link]

One comment to the above is by attaching the observation point to the rod, it is now in an accelerated frame. If you keep the reference point in a stationary frame, you then have correction terms to $J_B$ that make it the same as $J_{cm}=I_{cm} \omega$. In any case, yes, you do get $J_B =I_B \omega$ from this accelerated frame, but its value is somewhat limited, and it does differ from $J_{cm}=I_{cm} \omega$. (cm=center of mass)

 

[kuruman]

$\dots$ but its value is somewhat limited, and it does differ from $J_{cm}=I_{cm} \omega$. (cm=center of mass)

In what way is it limited? I would argue that $~ J_{cm}=I_{cm}~\omega ~$ is a special case of the more general equation in post #68, $$\mathbf J(x)=I(x)~\boldsymbol {\omega}~;~~~I(x)=\frac{1}{12}mL^2+mx^2+m\left(\frac{L}{2}-x\right)^2$$ where $x$ is the distance between the midpoint of the rod and the chosen axis of rotation. If one chooses $x=\frac{1}{4}L$, where the center of mass is, $$I(L/4)=\frac{1}{12}mL^2+m\left(\frac{L}{4}\right)^2+m\left(\frac{L}{2}-\frac{L}{4}\right)^2=\left( \frac{1}{12}+\frac{1}{16}+\frac{1}{16} \right)mL^2=\frac{5}{24}mL^2=I_{cm}.$$Writing the angular momentum as $I_{cm}~\omega$ about the CM is particularly useful because it allows the separation of the body's total angular momentum into (a) "angular momentum about the center of mass", which some call spin angular momentum and (b) "angular momentum of the center of mass", which some people call orbital angular momentum. Likewise, the total kinetic energy $KE$ of the body can be thought of as rotational KE about the CM and translational KE of the CM.

The parallel axis theorem guarantees that the moment of inertia of a rigid body has its smallest value about the CM. If you subtract $I_{cm}~\omega$ out of the total angular momentum, what's left cannot be anything other than linear momentum of the CM. Likewise, if you subtract $\frac{1}{2}I_{cm}~\omega^2$ out of the total $KE$, what's left cannot be anything other than translational KE of the CM.

 

[Charles Link]

It works for the case where there are no torques, but the expression for the torques gets much more complicated with any accelerations at the reference point. That's where @haruspex kept his expression simple with conservation of angular momentum, (by keeping his reference point fixed and free from the body), where the force from the string at the reference point didn't introduce any torques during the impact. The one thing he needed in post 31 that he corrected in post 45 was to add the correction terms to the angular momentum formula, (it is no longer just $J_P=I_P \omega$), when the body is moving relative to the reference point at the reference point. With the addition of the correction terms he actually has a fairly straightforward solution.