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Trying to reconcile the real world observation with first principles, here is a problem I came across.

Consider a ball initially rotating with axis parallel to the floor lowered gently onto a rough floor. Now obviously the angular velocity decreases linearly as a torque from friction is being applied along the circumference of rotation.

But from observation, the velocity also increases linearly. The force of friction seems to linearly affect the ball as well. How is this possible? Why is the force of friction affecting the center of mass?

This is a force at a distance that affects the center of mass, apparently. Thought experiment: Apply a force to the bottom of the ball in outer-space. Will the ball just rotate or move forward?

How is the same or different as the case previously discussed?

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You apply F=ma whether on the floor or at space. This will accelerate the ball linearly. That's independent from applying the torque and resultant angular acceleration. Take a ruler on a table and hit it on the edge (right angle to the ruler), it will perform some rotation but CM also moves forward.

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  • $\begingroup$ How can a force both increase angular acceleration and linear acceleration? And why is it connected to the center of mass even though the force was not applied there? $\endgroup$ – Jeffery Opoku-Mensah Jan 20 '18 at 23:02
  • $\begingroup$ @J'' Are you familiar with free body diagrams? If you vectorially add all the forces acting on a rigid object, you get the force that determines the acceleration of the center of mass, regardless of whether the net force acts through the center of mass or not. $\endgroup$ – Ben51 Jan 20 '18 at 23:07
  • $\begingroup$ I am more than familiar with free body diagrams. I am asking about how a force applied to a particle or region not on the center of mass affects the acceleration at the center of mass. I am asking why this is true, especially given it is not affecting the CoM. Is there a theory of rigidbodies where we can mathematically show that the forces applied at any set of points sum to the total force on the CoM, for example? It is not something we can resolve with a simple free body diagram, most likely I thought it would somehow involve modeling the rigidbody as a set of point masses. $\endgroup$ – Jeffery Opoku-Mensah Jan 20 '18 at 23:40
  • $\begingroup$ And just to be clear, using the assumption that a rigidbody composed of point masses has the property that every point must maintain the same relative distance to one another, if we apply a force that applies no torque, then it is trivial to show that every point must have the same acceleration, and thus the CoM effectively has the same force applied. Why is this the same for rotation? $\endgroup$ – Jeffery Opoku-Mensah Jan 20 '18 at 23:44
  • $\begingroup$ All of the forces between constituent parts of a system comes in third-law pairs: they sum to zero. Only the external force applied to the system contributes to the acceleration of the center of mass. And, by conservation of momentum, there is no way the rate of change of momentum of the center of mass can be anything other than the vector sum of the externally applied forces. $\endgroup$ – Ben51 Jan 20 '18 at 23:52

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