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

[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.

 

[Charles Link]

Item of interest is that I was logged off of PF earlier the other day, so that when I viewed this thread, I got an AI summary of the discussion. The AI seemed to do a better job of summarizing the discussion and pointing out a couple of the key points than I could have. From all appearances, it seemed to actually be generating intelligent thought. It seems to have reasonably good language skills, but I have to wonder if it is capable of doing the kind of mathematics that was needed to solve the problem as presented in the OP.