Confusions in conservation of angular momentum and torque

Question

Say someone is on a spinning object with heavy weights in his hands. He then pulls those heavy weights to himself and then the total angular speed is faster. We can explain this with conservation of angular momentum with that $I_0w_0=I_1w_1$, and $I_1 < I_0$ (because of smaller radius) , therefore $w_1 > w_0$. That must mean that there are some angular acceleration since $w_1 > w_0$, both to the heavy objects and the person. But there are no torque applied to the heavy objects. Neither do the inward force applied by the person nor the gravitational pull is tangential to its rotation. How is this (1) possible? ($\tau_z = I\alpha_z$ where $\alpha_z \neq 0$ and $\tau_z = 0$)

Also, the kinetic energy of the objects and the person will be higher than before if you put in some values. But since there is no torque, there is no work. ($W = \int\tau_z d\theta$) How is this (2) possible?

I got this problem from my textbook but I cannot answer the question myself so I am asking it here.

Answer 1

Oh, I get it. The equation $\tau_z = I\alpha_z$ is not valid for a non-rigid object (even if it is solid and doesn't change forms, if it changes in radius relative to the axis, it is non rigid, as there is no I), and only the equation $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{\tau}$ is valid for those. So it can change its acceleration without torque if it is non-rigid. Also for the energy, the equation $W = \int\tau_zd\theta$ only is valid for rigid ones as it assumes that the path taken is circular and not a spiral.

See Rotational Mechanics: Is Angular Acceleration Possible without any External Torque?

Answer 2

Nevermind my comment. But imagine that is the situation. Yes, $\vec F \cdot \vec r= 0$ so there is no torque, but $d\vec L/dt=0$, so of course there is no torque.

The springs (I refuse to use humans in any problem, and you should to) do work against centripetal force and convert that to rotational kinetic energy: $\omega$ increases because $I$ (or just $mr^2$ in the ideal limit) decreases.

Seriously, leave humans bodies out of it, and replace muscles with loaded spring, and clarity will occur.

Answer 3

Suppose that you have a body (the system) moving at constant speed $v_{\rm A}$ in a circular orbit of radius $r_{\rm A}$, centre $O$, under a (external) centripetal force $F_{\rm A}$ and by the application of a central force its position is changed such that it is now moving at constant speed $v_{\rm B}$ in a circular orbit of radius $r_{\rm B}$ under a centripetal force $F_{\rm B}$.

In moving from position $A$ to position $B$ the trajectory of the body might be something as shown in the diagram.

At position $C$ the central force $F_{\rm C}$ is not orthogonal to the velocity $v_{\rm C}$ and has a component in the direction $v_{\rm C}$ which will result in the speed and the kinetic energy of the body increasing and the direction of the new velocity such that the body moves closer to position $O$.
Whist this is happening there is a displacement of the force in the direction of the force and so the force is doing work.

Thus in this example the angular momentum about position $O$ is constant because the force is central, ie the force cannot exert a torque about position $O$, but the angular speed is increasing and the moment of inertia of the body about position $O$ is decreasing at rates which result in their product remaining constant.