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Rolling without slipping and acceleration

  1. Jun 20, 2013 #1
    Why don't we consider radial acceleration on the lowermost point in rolling without slipping?
     
  2. jcsd
  3. Jun 20, 2013 #2
    Your question is a little vague - can you be more specific? What is the situation? Don't consider radial acceleration while finding out what?
     
  4. Jun 20, 2013 #3
    While rolling without slipping of a solid sphere of a solid cylinder or a hollow cylinder or a hollow sphere.
     
  5. Jun 21, 2013 #4

    haruspex

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    At the moment that a point on the circumference of the wheel touches the ground, what are its acceleration and velocity?
    It might help to think about a short time before it touches the ground. Where is the instantaneous centre of rotation of the wheel, and how far from that is the point that's soon to make contact?
     
  6. Jun 21, 2013 #5
    Well, there is no force providing torque, so why should there be any angular acceleration? That is why I asked you to be specific - are you considering a situation where there is friction acting on the lowermost point or not?
     
  7. Jun 21, 2013 #6
    No matter if friction is there or not. It is rolling without slipping and the conditions needed for that have been taken care of by the question/problem/examiner who made the question.(Though those conditions haven't been specified). And I didn't ask whether there would be any angular acceleration or not. I am talking about RADIAL acceleration that a particle executing circular motion experiences which is towards the centre (centripetal: centre seeking). So that's what I am saying. In rolling without slipping, the lowermost point is instantaneously at rest, but an instant later it has kicked back up and allowed a different point to come to the lowermost position. So why does the particle kick back up? Does it have something to do with the radial acceleration of the particle? Does the radial acceleration play any role in rolling without slipping or not?
     
  8. Jun 21, 2013 #7
    Have you not answered your own question?
     
  9. Jun 21, 2013 #8
    I don't think so because I still haven't come to a conclusion as of yet.
     
  10. Jun 21, 2013 #9
    Yes you have: "I am talking about RADIAL acceleration that a particle executing circular motion experiences which is towards the centre (centripetal: centre seeking)."

    You stated quite plainly that anything in circular motion experiences radial acceleration.
     
  11. Jun 21, 2013 #10
    Yes, what effect does it have on the lowermost point in contact with the surface? By the virtue of what does it kick back up while pure rolling?
     
  12. Jun 21, 2013 #11
    My apologies, I misread your question as one of angular acceleration. Radial acceleration is present for all circular motions - it is in fact a definition. It is present because of the internal forces in the body, and hence we do not consider it explicitly as such.
     
  13. Jun 21, 2013 #12
    By the virtue of centripetal acceleration, just like you said.
     
  14. Jun 21, 2013 #13
    But the radial acceleration of the lowermost point is straight back up normal to the surface. But the lowermost point moves diagonally upwards, i.e along the circular arc. Is it because while the lowermost point tends to move up, the rolling object is also rotating so it goes along the arc? Kind of like the Coriolis Effect?
     
  15. Jun 21, 2013 #14
    The Coriolis effect is irrelevant here, but otherwise your observation is correct. In a frame co-moving with the rolling body, the lowermost point is moving horizontally backwards, and the upward acceleration tilts the velocity vector up.
     
  16. Jun 21, 2013 #15

    haruspex

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    andyrk, pls try to answer my questions in post #4. You will find that in an inertial frame the point of the wheel in contact is not accelerating.
     
  17. Jun 21, 2013 #16
    The particle does experience radial acceleration at the lowermost point, so why cant it experience acceleration at the lowermost point in an inertial frame?
     
  18. Jun 21, 2013 #17

    haruspex

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    Do the algebra. Suppose the point of interest is about to contact the ground. It still has an angle theta to go. If the radius of the wheel is r, what is its height from the ground? What does that approximate to for small theta? Writing theta = -ωt, constant ω, what does that give you for the vertical acceleration?
    Remember, it has no centripetal acceleration since it has no tangential velocity.
     
  19. Jun 21, 2013 #18
    Height of the point is r-rθ=r(1-θ)
    Ok yes I get you. Every particle executing circular motion has a centripetal acceleration but also one condition should be that at all points of time its velocity should be tangential to the circular trajectory. Then if the particle doesn't have any radial as well as tangential acceleration how does the particle go back up?
     
  20. Jun 22, 2013 #19

    haruspex

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    No, it's r(1- cosθ). What does that approximate for small θ?
    However, I now realise my earlier responses were wrong. Here's what I should have written:
    The point of contact does accelerate upwards, but whether you consider it centripetal acceleration depends on your frame of reference. If you're moving with the velocity of the wheel's centre (as someone already posted on this thread) it looks like centripetal acceleration, rω2. But viewed from the roadside, that point has no forward movement, and is itself the centre of rotation, so cannot have centripetal acceleration. But its vertical position is r(1-cosθ) ≈ rθ2/2 = rω2t2/2. Differentiating twice, it has vertical acceleration rω2, as required.
     
    Last edited: Jun 22, 2013
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