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I did find the proof of pseudo torque about the center of mass of a body in combined rotation and translation being zero, and thus I'm convinced that in this frame of reference, we can conserve angular momentum. However, although the net torque about the point of contact in rolling is zero, I don't understand how we can conserve angular momentum about the point of contact in rolling motion, even though this axis itself is not fixed and constantly translating. Any help is appreciated.

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2 Answers 2

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The point of contact moves with v. The angular momentum is calculated and constant in relation to the point of contact the object has in every moment, not the first or some fixed Point of contact.

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If the body is rolling along a horizontal surface and there is no slipping at the point of contact, the constant centre of mass translational speed $v$ is related to the constant angular speed $\omega$, $v= r\,\omega$ where $r$ is the radius of the object.

The angular momentum of the body about the centre of mass is $I_{\rm com}\omega$ and this is called the spin angular momentum.
About any other point there is an additional contribution to the angular momentum called the orbital angular momentum.
About any point on the horizontal surface the orbital angular momentum is $mv\times r$.
Thus the total angular moment of the body about any point on the horizontal surface is $mvr+I_{\rm com}\omega = mr^2\,\omega + I_{\rm com}\omega = (mr^2 + I_{\rm com})\omega$ and $mr^2 + I_{\rm com}$ is the moment of inertia about the point of contact (parallel axes theorem).
The point of contact is moving at a constant velocity and thus defines an inertial frame of reference, so the fact that the point of contact is not accelerating means that it can be used when applying Newton's second law.

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  • $\begingroup$ But that point of contact is used regardless of whether the wheel is slipping or in pure rolling motion. $\endgroup$ Commented Aug 19, 2023 at 17:06

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