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I was reviewing some dynamics for rigid bodies and system of particles.

In this link the author writes that the change in velocity of the COM is independent of where a force is applied to a system of particles:

https://physics.stackexchange.com/a/118468/147291

While I can follow his derivation there is something I must be missing because for example, if you apply a pure torque on a rigid body you only induce spin and not a change in linear velocity. Therefore the linear velocity of the COM will be different depending upon where you apply the force.

Would appreciate clarification.

Qmechanic
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FourierFlux
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1 Answers1

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The acceleration of the center of mass $\vec a$ of a system is determined by $\vec F = m \vec a$ where $\vec F $ is the net external force and $m$ is the mass. If the vector sum of all the external forces is zero, the CM is not accelerated regardless of where the forces are applied.

For evaluations about the CM, $\vec N ={d \over dt} (\bf I \vec \omega)$ where $\vec N$ is the total external torque, $\bf I$ is the inertia tensor, and $\vec \omega$ is the angular velocity. For rotation in a plane this simplifies to $\vec N = I \vec \alpha$ where $I$ is the moment of inertia, and $\alpha$ the angular acceleration. Note the torque (and angular momentum) depend on the point about which they are evaluated.

By pure torque I assume you mean what is also called a couple where the total force is zero. For example, two forces equal in magnitude but opposite in direction applied to each end of a rod on a pivot at its center. Look up questions regarding "couple" on this site.

You can have no net external force, but external torque about a selected point, and vice versa.

John Darby
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  • I see, for some reason one of my books talked about the line of action directed towards the COM is the only component which would impart linear acceleration and the perpendicular component would only impart torque – FourierFlux Jan 14 '23 at 03:17
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    Comment about force seems wrong. Any net external force causes acceleration of CM. Torque depends on point about which it is evaluated. For example, ball rolling down incline. Torque from friction with respect to CM is not zero, torque due to friction with respect to point on incline is zero. See https://ethz.ch/content/dam/ethz/special-interest/mavt/mechanical-systems/mm-dam/documents/Notes/Dynamics_LectureNotes.pdf. – John Darby Jan 14 '23 at 21:28
  • Oh you're right, for some reason my mechanics book kept talking about line of action along the COM. I keep forgetting this. – FourierFlux Jan 14 '23 at 21:47