I'm a student studying classical mechanics and ran into some trouble with the topic of rigid body rotation. The formula τ=Iα for rigid bodies, where τ is the torque exerted by external forces about the center of mass (CM) of the rigid body, I is the moment of inertia of the body about CM, and α is the angular acceleration of the body about CM, is derived under the assumption that internal forces exert no net torque about CM (that is, a rigid body cannot gain or lose angular momentum due to internal forces). We seem to apply this equation to rigid objects but is the assumption that internal forces exert no net moment on the rigid body a justified assumption for rigid bodies? For example, consider a disk that's subjected to a force F: This disk, when starting from rest, will gain an angular momentum in the direction pointing out of the screen as the entire disk starts to rotate in the counter-clockwise direction. Now consider a chunk of mass dm located away from the point of application of the force F: It is clear that this chunk dm has only internal forces acting on it. In fact, any dm that's not located at the point of application of F has only intetnal forces acting on it. But the problem that I see with this is that all chunks located away from the point of application of F, when considered as one system, gain angular momentum in the same direction (direction out of the page) when they're only acted on by internal forces. That is, the total angular momentum of these chunks increase even though the only forces acting on them are internal forces. Isn't this in direct conflict with the assumption that internal forces don't exert a net torque on the body (and hence internal forces shouldn't have an effect on the total angular momentum of the rigid body)? I am hoping someone could clarify my confusion on this. Thank you!
3 Answers
While considering "chunks" you are treating them as free bodies being acted upon by external forces. Internal and external forces are subject to frame of reference you choose.
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$\begingroup$ Right. Maybe I should've clarified but I'm asking about how when you consider all the chunks (all the chunks located away from point of application of F) together as one system, the total angular momentum of the system increases even though the only forces on the chunks in the system are the internal forces between the chunks. $\endgroup$ Commented May 24, 2021 at 11:39
Consider your object as consisting of a number of near-infinite elements $\mathrm{d}m$, each of which connected to its immediate neighbours on a mesh (grid).
Now, if we isolate the original element (upon which the force $F$ acts) with an immediate neighbour, as done with the thin green rectangle, we can calculate how the force $F$ is 'transmitted' onto that first $\mathrm{d}m$.
Each mass element has forces acting on it in the $x$ and $y$-directions (for a $2D$ object) and these forces can be found using the free body diagram of isolated elements.
(This method is used for the calculation of internal stresses of complicated shapes and is then known as Finite Element Analysis)
Using this method will show that the various elements $\mathrm{d}m$ are stationary with respect to the origin element and why.
Above, an example where stresses in the beams (the nodes are considered friction free) are determined using the 'isolation' method.
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$\begingroup$ Thank you for the reply! Are you implying that, since external force F is transmitted to all the elements, there is actually an external torque on all the particles? $\endgroup$ Commented May 24, 2021 at 13:47
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$\begingroup$ I'm not sure it would be called external. But certainly there's torque acting on each element. How do we know this for certain? ALL elements undergo angular acceleration and per Newton's 2nd Law that means torque MUST act on them. A Finite Element Analysis would reveal that. $\endgroup$– GertCommented May 24, 2021 at 15:07
You chose the frame of reference as the chunks and forces applied by any other body on the chunks would be an external force which would increase the angular momentum due to torque.