2
$\begingroup$

For some context, consider an idealized situation with a person, rigidly attached to the shaft of a bicycle wheel that can spin. sitting on a chair that can rotate. All bearings are frictionless. Initially, the person and chair are at rest but the wheel is spinning, say counterclockwise at rate $\omega_s$, around a vertical axis.

initial condition

Next, the wheel is flipped over, so the angular momentum of the wheel is now negative. Obviously, the person must start rotating counterclockwise to conserve angular momentum. Since the person is rigidly attached to the shaft, the center of mass of the wheel would start translating in a circle around the AOR.

final

This is where the parallel axis theorem comes in. People I have discussed this with believe that $\overrightarrow L_f = \omega_f(I_{s,p}+I_w+md^2)-\omega_sI_w$ where m is the mass of the wheel, d is the distance from the COM of the wheel to the AOR, $I_{s,p}$ is the moment of inertia of the person and chair, $\omega_f$ is the final speed of the person-chair-wheel system, $I_w$ is the moment of inertia of the wheel about its axis of rotation.

Here is my rationale: assuming the mass of the wheel is entirely concentrated in its rim, and the bearing it is running on is frictionless, there is no way a vertical torque could be applied to the spinning wheel and thus there would not be a component of the angular momentum related to the wheel rotating around its own center of mass due to the translation of the shaft/wheel CoM (as would be the case for a rigid body; the only components of angular momentum related to the rotation of the wheel are $\omega_sI_w$ and $\omega_fm_wd^2$, and not $\omega_fI_w$. This gives me $\overrightarrow L_f = \omega_f(I_{s,p}+md^2)-\omega_sI_w$.

My understanding is that the parallel axis theorem is only valid for rigid bodies in which the body is rotating around an axis not at its CoM, which this system clearly is not.

What is the correct interpretation?

$\endgroup$
2
$\begingroup$

My understanding is that the parallel axis theorem is only valid for rigid bodies

That theorem belongs to geometry, not to dynamics. There's no reason why it should not hold for non-rigid bodies. On the contrary: take positions of all masses at a given instant of time and compute relevant moments of inertia using those positions. You'll see the theorem - obviously - holds.

The only point where rigidity comes into play is about dependence of moments of inertia on time. But this may also happen with rigid bodies.

$\endgroup$
  • $\begingroup$ This is—of course—correct, but it might not answer the question intended. The parallel axis theorem is correct for any mass distribution, but the equilbrium mass distribution of a non-rigid body may be different in the states the OP wishes to consider meaning that he can't compute the one state of the basis of the other. $\endgroup$ – dmckee Apr 28 at 18:06
  • $\begingroup$ @dmckee I replied to a precise OP's question. As to his contraption I've said nothing because I wasn't able to understand his description nor his drawing. $\endgroup$ – Elio Fabri Apr 29 at 6:27
  • $\begingroup$ @ElioFabri Thanks for your answer, my main problem was the specific application of the concept; I have started a new question that specifically addresses it and is more descriptive. $\endgroup$ – jgorton Apr 29 at 13:56
1
$\begingroup$

. . . the bearing it is running on is frictionless, there is no way a torque could be applied to the spinning wheel

You would be correct if the wheel was mounted in a gimbal.

A frictionless bearing of the type mentioned in your question is an object whose outer casing cannot apply a torque to its inner parts in a direction along the central axis of the bearing but the outer casing can apply torques to the inner parts of the bearing if the direction of the torque is at right angles to the central axis of the bearing.

Thus, an external torque can be applied on the wheel provided that the direction of the torque on the outer casing of the bearing is in the plane of the bearing.

The wheel & person & chair system connected to the ground via a horizontal bearing cannot have vertical external torques applied to it from the ground or the vertical forces as a result of gravitational attraction.

This means that the vertical component of the angular momentum wheel & person & chair system is conserved.
The horizontal components of the angular momentum of the wheel & person & chair system can change because the bearing can exert horizontal external torques on the wheel & person & chair system.

To flip the wheel over an external horizontal torque acting on wheel & person & chair system is applied by the bearing as a result of the person using their arms.
During this process when the wheel bearing is inclined to the horizontal there are internal vertical torques acting: vertical torque on wheel due to person and chair, and equal magnitude and opposite direction vertical torque on person and chair due to wheel.

When the conservation of vertical angular momentum is used in such an example the initial and final states of the wheel & person & chair system are "rigid".

$\endgroup$
  • $\begingroup$ Edited the third to last paragraph; this is my main problem. $\endgroup$ – jgorton Apr 28 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.