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In this question, I encountered a problem that if we were to conserve angular momentum about a stationary point lying on the middle axis and then conserve it from the frame of Center of mass of the rod , the equations of angular momentum conservation are coming out to be different in both cases ; Can anyone explain why ? As angular momentum seems to be conserved about both points.

As initial angular momentum of particle will remain same in both cases ; angular momentum of rod will also remains same = I꜀ₘ.ω, but the final angular momentum of particle should be different in both cases as in second case , in frame of center of mass the relative velocity of the particle would be different.

PS : This question isn't meant to be solved , I just want to clear out my confusion; that's why I also didn't included the options.

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  • $\begingroup$ You're going to need to supply more information,for example the work you've done to show conservation of momentum might be helpful. $\endgroup$
    – Triatticus
    Mar 9, 2021 at 17:15
  • $\begingroup$ Please add the steps that you followed in both in scenarios. $\endgroup$ Mar 9, 2021 at 17:53
  • $\begingroup$ You mention angular momentum is I꜀ₘ.ω²/2 but this is the expression for the rotational kinetic energy. Is there some confusion here on the terms? Please be specific and show your calculations and conventions so we can chime in. $\endgroup$
    – JAlex
    Mar 9, 2021 at 21:17
  • $\begingroup$ this is only grammar, but i believe your "an"s should be "a"s. It reads much better to my ear. $\endgroup$
    – user288901
    Mar 10, 2021 at 1:56
  • $\begingroup$ Noether’s theorem says that the angular momentum about any axis is conserved if the Lagrangian is rotationally invariant about that axis. Is the Lagrangian rotationally invariant in both cases? $\endgroup$
    – user288901
    Mar 10, 2021 at 2:35

3 Answers 3

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Yes, we can conserve angular momentum about a moving origin. I prefer not to call it a moving axis, because axis implies to me that it's the center of rotation of everything in the space, but the center of rotation of a one-dimensional free body is at infinity - movement in a straight line is identical to movement along the edge of an infinite circle.

In brief: A moving origin is an easily reversible velocity transformation of a coordinate space. Apparent discrepancies can be resolved by reversing the velocity transformation. To do this, subtract the velocity from everything and re-evaluate all quantities.

The system has two one-dimensional objects (the particle and the center of mass of the rod) one two-dimensional object (the rod), and the coordinate space (e.g. the entire rest of the universe). All three objects have a momentum, an angular momentum (dependent on the origin for the two-dimensional objects, a kinetic energy, and an angular kinetic energy.

If we lock in the coordinate space, we get a system in which a particle at some velocity strikes a rod at some other velocity. They exchange energy such that energy is conserved, momentum such that momentum is conserved, and angular momentum such that angular momentum is conserved. This gives us a system of equations that we can solve to get each quantity.

Angular velocity is conventionally absolute, not relative*. Fortunately, the angular velocity of a one-dimensional object is always zero (no matter how far it goes in a straight line, it'll never perform one circuit of its infinite-radius circle). This also means that the angular kinetic energy of the one-dimensional objects is always zero as long as they're moving in straight lines. So, we can lock the coordinate space to any free one-dimensional object we like, and everything is fine.

If we tie the coordinate space to one of the free one-dimensional objects, we impose an additional constraint that that object's energy, momentum, and angular momentum are always zero. If we stick that in our system of equations, we will get new results for the values. This isn't a paradox, it's just the relative nature of velocity. (Not to be confused with relativistic velocity - Newton's equations work just fine here.)

You could get the same results by doing your system of equations in the default coordinate space, finding the velocity of the center of mass of the rod, and then subtracting that velocity from every point in the coordinate space and re-evaluating their angular momenta.

  • re: "Angular velocity is absolute". We can't do the same thing with the two-dimensional object (locking in the coordinate system so the rod never spins and the rest of the system spins around it). Theoretically we could, but spinning the coordinate space means spinning the entirety of spacetime, which is much more messy than just shifting or speeding up the entirety of spacetime, as we do when we displace or accelerate a reference frame. And while it's a pretty good guess that there's no particular difference between spinning the entirety of spacetime and spinning yourself while spacetime stays still, we don't have the math to prove it and we certainly can't spin the whole universe in a laboratory.
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The Lagrangian is the Kinetic Energy subtract the Potential Energy. $L=K-P$. You can think of the entire bar as a set of single masses rotating side by side. So, consider one mass rotating.

The Lagrangian of one mass rotating at unit radius is

$$L=\frac{1}{2} m (\dot{x}^2 + \dot{y}^2 )$$

where

$x=\cos(\omega t)$ and $x=\sin(\omega t)$.

$$L=\frac{1}{2} m ((\frac{{d\cos(\omega t)}}{dt})^2 + \frac{{d\sin(\omega t)}}{dt})^2 )=\frac{1}{2} m ((-\omega \sin(\omega t))^2 + (\omega \cos(\omega t))^2 )$$

$$\implies$$

$$L=m\omega^2$$

Under infinitesimal rotations such that $\delta\theta$ is very small,

$$x'=x+y\delta\theta \text{ and } y'=y-x\delta\theta$$

The same rotation, from anywhere along your y axis, will be represented by

$$x'=x+y\delta\theta \text{ and } y=y'-x\delta\theta+C$$

You can show explicitly that, to first order, the Lagrangian is invariant under that transformation. That means it remains the same.

Noether's Theorem says that

Invariance under

$\delta x= h_{x}\delta\theta = y \delta\theta $

$\delta y=h_{y} \delta\theta = -x \delta\theta$

with canonical momentum

$p_x = \frac{\partial L}{\partial \dot{x}}$

$p_y = \frac{\partial L}{\partial \dot{y}}$

Implies $Q$ is a conserved quantity where

$$Q=p_x h_x + p_y h_y = m(y \dot{x} - x \dot{y})$$

And that is angular momentum.

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I got it solved , I was actually conserving Angular momentum wrt Centre of Mass before and after Collison but the point is , Centre of Mass is actually accelerated between collision so it was an inertial frame of refrence only before and after Collison and not during the collision !

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