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Acceleration of point of contact

  1. Nov 13, 2013 #1
    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
  2. jcsd
  3. Nov 13, 2013 #2


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    Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?
  4. Nov 13, 2013 #3
    Is'n it stationary?
  5. Nov 13, 2013 #4
    The velocity of the point of contact during rolling motion is zero.
  6. Nov 13, 2013 #5
    My question is about its acceleration
  7. Nov 13, 2013 #6
    Hi quawa99...

    If the velocity of the CM is vcm,radius R and the angular speed of the sphere is ω ,then what is the relationship between them considering the sphere is undergoing pure rolling ?
  8. Nov 13, 2013 #7
    The velocity of of CM with respect to the point of contact is given by Vcm=RXw where w is angular velocity of the sphere.
  9. Nov 13, 2013 #8
    vcm is with respect to the ground frame .

    Anyways, now what is the relation between the acceleration of CM and the angular acceleration ?
  10. Nov 13, 2013 #9
    a=RXangular acceleration
  11. Nov 13, 2013 #10
    Now ,what is the value of acm in the given problem ?
  12. Nov 13, 2013 #11
    Its not a problem given in a book
    Angular acceleration and linear acceleration of cm in horizontal direction is 0
  13. Nov 13, 2013 #12
    I was referring to that question.

    Anyway, the point of contact (in fact, all the points on the surface) have acceleration toward the centre of mass of the object. Be it, uniform pure rolling or accelerated pure rolling, the acceleration will be towards the centre of mass.
    Also even if there is friction, since there is no slipping between the surfaces, the work done by it will be zero.
  14. Nov 13, 2013 #13
    Why do all the points on surface have an acceleration towards centre ?I have read that pure rolling is the same thing as pure rotation about point of contact so doesn't that mean acceleration of cm with respect to point of contact is v^2/R ?
  15. Nov 13, 2013 #14
    Yes. Pure rolling is same as rotation about point of contact.
    About the point of contact and the acceleration, mathematically you can derive for any point on the surface. And if you substitute the condition for the pioint of contact, you'll find out that the acceleration of point of contact is towards the centre.
  16. Nov 13, 2013 #15
    Since vcm = constant ,acm = 0 .

    apoint of contact = apoint of contact w.r.t CM + acm

    So,apoint of contact = ?
  17. Nov 13, 2013 #16
    You have to derive that. Also, acceleration of centre of mass is zero only in a uniform pure rolling.
  18. Nov 13, 2013 #17
    What is the acceleration of point of contact with respect to cm?
  19. Nov 13, 2013 #18
    My question is about a uniformly rolling sphere
  20. Nov 13, 2013 #19
    apoint of contact w.r.t CM = αR ,where α is the angular acceleration :smile:
  21. Nov 13, 2013 #20
    α is zero in my question
  22. Nov 13, 2013 #21
    if you consider the horizontal to be the x-axis, then, the point of contact makes an angle 3π/2. Now
    a(point of contact)= a(cm)i + a(P,cm)
    = a(cm)i + a(P, tangential) -i + a(P, radial)
    = a(cm)i - Rα(cm) + R(w^2)
    α+ angular acceleration of centre mass.
  23. Nov 13, 2013 #22
    So now what is the problem ? You have everything to answer the question .
  24. Nov 13, 2013 #23
    according to your equation the point of contact has 0 zero acceleration but then shouldn't that mean if you observe the cm in the frame of point of contact(which is stationary in present situation) it would have a centripetal acceleration of v^2/R as it is undergoing pure rotation about point of contact with angular velocity v/R
  25. Nov 13, 2013 #24
    a(point of contact) = a(point of contact w.r.t CM) + a(cm)
    = a(p, tangential) + a(P, radial) + a(cm) i
    Since the rolling is uniform, a(cm) is zero as well as α(cm) i.e; angular acceleration of centre of mass and since a(P, tangential) = Rα(cm), it becomes zero.
    Therefore, a(point of contact) = a(P, radial) = R(w^2) j
    i is positive x-axis and j is positive y-axis.
  26. Nov 13, 2013 #25
    The equation does NOT show that acceleration is zero. It is Rw^2. Since V=Rw, it is equal to v^2/R
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