question 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.


This is where the parallel axis theorem comes in. People I have discussed this with, as well as the answer key, 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?


Your Answer

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

Browse other questions tagged or ask your own question.