Suppose a ball is rotating due to force $F$ applied at its top (on a rough ground).There is pure rolling.

In one case we write the equation w.r.t COM i.e $F.R=I(\alpha_1)$ and $F-f=ma$ and $a=R\alpha$.

Now if we write w.r.t bottom point we write $F.2R=I'(\alpha_2)$.In such a case will $\alpha_1$ and $\alpha_2$ be equal?If yes/no please explain why.Thanks.

($I$ and $I'$ are moment of inertia) (f is friction) (R is radius of the ball) (a is the acceleration of COM of ball) ($\alpha$ is angular acceleration of ball)

  • $\begingroup$ Really confusing question. You didn't define any of your terms. $\endgroup$ – user93237 Nov 6 '15 at 18:36
  • $\begingroup$ @SamuelWeir I've edited.Please see now.What else should I add? $\endgroup$ – user74370 Nov 6 '15 at 18:39
  • $\begingroup$ Still confusing. You speak of a force F but it's not at all clear what sort of force F would cause a ball on the ground to start rotating. I can think of many different possibilities. Also, R is not defined, although someone could infer based on later information that it's probably the ball radius. What is the small 'f'? The list goes on and on. I suggest carefully drawing out a picture and presenting it with a full and clear description of the problem and all of the parameters as they are first introduced. $\endgroup$ – user93237 Nov 6 '15 at 18:51
  • $\begingroup$ @SamuelWeir ok I've edited again.Anything else? $\endgroup$ – user74370 Nov 7 '15 at 4:56

Yes, rotational velocity and acceleration is shared by all points on a rigid body. We only state that a body rotated about a point because the linear velocity is zero at that point.

See related answer here: https://physics.stackexchange.com/a/215165/392

  • 1
    $\begingroup$ This one single sentence of yours cleared a million doubts of mine :-)! Thank you ! $\endgroup$ – user74370 Nov 7 '15 at 5:02

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