3
$\begingroup$

A man holds in his hands two equal masses with outstretched arms, standing on the center of a platform that rotates with a certain angular velocity. If you drop both of the masses without moving your arms, what happens to the angular velocity of the man and the angular velocity of the masses?

I gave this a huge thought time to this theoretical question, but in short words, I initially thought that the angular velocity of the man would decrease since from the conservation of angular momentum: $$L_i=L_f$$ Initially you had two masses, which means that the moment of inertia $I=mr^2$ would make the initial angular momentum to be higher than the final one, since you are losing those masses, but it seems that it isn't the case, because it apparently remains the same? Which seems very unintuitive to me, can someone shine a light on how that's true?

$\endgroup$
7
  • $\begingroup$ How does it remain same? $\endgroup$ Commented Nov 6, 2022 at 19:03
  • $\begingroup$ That is precisely what I'm asking $\endgroup$
    – user348222
    Commented Nov 6, 2022 at 19:03
  • $\begingroup$ You said it seems $\endgroup$ Commented Nov 6, 2022 at 19:07
  • $\begingroup$ Yes by looking it on google and a little bit of research, but I'm asking why. $\endgroup$
    – user348222
    Commented Nov 6, 2022 at 19:08
  • $\begingroup$ This is a good question as it might appear counterintuitive to some. It should be re-opened. PS. The op never asked for a numeric result, but an analysis of a thought experiment. $\endgroup$ Commented Nov 7, 2022 at 14:03

2 Answers 2

3
$\begingroup$

It really depends on the way the man drops the masses, namely direction and velocity.

If it let them drop opening his hands, the component of the angular momentum aligned with the axis of rotation of the system man+masses is conserved until the masses reaches the ground.

Since there is no moment exchanged between the man and the masses around the axis of rotation, the man and the masses keep constant angular momentum independently as well.

With formulas. The dynamical equation of the angular momentum of a system reads:

$\dot{\mathbf{\Gamma}}_H = -\dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext}$,

where $\mathbf{\Gamma}_H$ is the angular momentum around the pole $H$, $\mathbf{Q}$ is the momentum, $\mathbf{x}_H$ is the position of the pole and $\mathbf{M}^{ext}_H$ is the external moment acting on the system. If the pole doesn't move $\dot{\mathbf{x}}_H = \mathbf{0}$.

We can write the equation of the independent systems, $\dot{\mathbf{\Gamma}}_H^{tot} = \dot{\mathbf{\Gamma}}_H^{man} + \dot{\mathbf{\Gamma}}_H^{mass} = \mathbf{M}_H^{ext} = \mathbf{0}$,

$\dot{\mathbf{\Gamma}}_H^{man,ext} = \mathbf{M}_H^{man,ext}$$\qquad ,\qquad$ $\dot{\mathbf{\Gamma}}_H^{mass} = \mathbf{M}_H^{mass,ext} = - \mathbf{M}_H^{man,ext}$

where in general the moment $\mathbf{M}^{man,ext}_H$ is the moment acting on the man due to the masses (external to the man system), and $\mathbf{M}^{masses,ext}_H = - \mathbf{M}^{man,ext}_H$ for the third principle of dynamics (action/reaction), is the moment acting on the mass system due to the man. If the man and the masses exchange no moment, $\mathbf{M}^{masses,ext}_H = - \mathbf{M}^{man,ext}_H = \mathbf{0}$, and thus (taking only the $z$-component of the angular momentum, around the axis of rotation),

$\dot{\Gamma}_{H,z}^{man,ext} = 0$$\qquad ,\qquad$ $\dot{\Gamma}_{H,z}^{mass} = 0$

Thus:

  • the man keeps rotating with the same angular velocity $\boldsymbol{\Omega} = \Omega_z \mathbf{\hat{z}}$, if it doesn't change its position and consequently its moment of inertia $I_z$, $\Gamma_{H,z} = I_z \Omega = \text{const.}$;
  • the masses move on a parabolic trajectory in a vertical plane, with constant tangential component $v_{\theta} = R \Omega$, and thus constant angular moment w.r.t. the axis of rotation of the man $\Gamma_{H,z} = m R v_{\theta} = m R^2 \Omega = \text{const.}$
$\endgroup$
12
  • $\begingroup$ How is there no change of angular momentum if you are losing mass? $\endgroup$
    – user348222
    Commented Nov 6, 2022 at 19:09
  • $\begingroup$ There is no momentum around the axis of rotation between masses and the man, so their angular momenta independently keep constant values. $\endgroup$
    – basics
    Commented Nov 6, 2022 at 19:34
  • $\begingroup$ About the formula $\dot{\mathbf{\Gamma}}_H = -\dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext}$. Do you have a reference for it. I do not dispute it since I derived it on my own in the past but I was wondering about its history since it is not a standard form taught in physics. $\endgroup$ Commented Nov 7, 2022 at 14:00
  • 1
    $\begingroup$ basics.altervista.org/test/Physics/Me/eom_points.html anyway, here some hand-weitten notes for a system of point masses $\endgroup$
    – basics
    Commented Nov 7, 2022 at 14:03
  • 1
    $\begingroup$ I know. I started the project few months ago, I built the infrastructure of the website and github repo. Now it's idle for quite a long time. I need to find time to work on it, and choose a style of the notes (no extra colours, and whatever makes them confusing and not easy to navigate) $\endgroup$
    – basics
    Commented Nov 7, 2022 at 14:48
3
$\begingroup$

In addition to basics complete calculations I might add an hand-wavish comparison which could give you intuition.

If you were to be on a bike going at a fixed speed, holding a 5L jug of water or some heavy object and suddenly let go of it (no throws or anything) then you would continue cruising at the same speed, feeling no acceleration. The jug would follow you at your speed (and eventually crash down on the road due to gravity).

This is essentially the same setting as your question except that it's a linear movement instead of a rotational (therefore I find it a bit easier to imagine).

You're not loosing mass per say, you're just "disconnecting" parts of your system (letting go of the object). So you're not exchanging energy with whatever you're letting go of and your own personal kinetic energy doesn't increase.

$\endgroup$
1
  • $\begingroup$ yeah, exactly the same thing, for linear momentum instead of angular momentum. +1 for the helpful analogy $\endgroup$
    – basics
    Commented Nov 6, 2022 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.