Acceleration of point of contact

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In a discussion about the acceleration of the point of contact for a sphere undergoing pure rolling on a smooth surface, it is established that the velocity of the point of contact is zero at any instant. The relationship between the center of mass (CM) velocity, radius, and angular speed is defined as v_cm = Rω, where ω is the angular velocity. The acceleration of the point of contact is influenced by the centripetal acceleration due to the sphere's rotation about the CM, which is Rω². While the linear acceleration of the point of contact is zero in uniform rolling, it still experiences centripetal acceleration. The clarity of the original question regarding the definition of "point of contact" is emphasized, as it can lead to different interpretations and answers.
  • #31
quawa99 said:
For a sphere undergoing pure rolling on a smooth surface with center of mass having constant velocity,what is the acceleration of point of contact

You've got different answers because your question isn't clear. What do you mean by "point of contact"? Three possible interpretations and related answers:

1) a virtual point that moves along the surface so as to coincide always with the surface's point that touches the sphere. Its speed is constant, so it's acceleration is zero.

2) the material point of the flat surface that, at a given instant, touches the sphere. It does not move: speed zero, acceleration zero.

3) the material point of the sphere's surface that, at a given instant, touches the flat surface. This is the most interesting case. Its trajectory is an ordinary cycloid, its instantaneous speed is zero, however, its acceleration is the centripetal acceleration it had if the sphere were only rotating (no translation) around the same axis with same angular speed. Can you see why?

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TRu-TS
Buon vento e cieli sereni
 

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