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Frictionless slope

  1. Dec 11, 2013 #1
    I derived a formula for the acceleration of a ball of uniform mass rolling down a slope at some angle to the vertical, without slipping and did so in terms of kinetic energy and potential energy!

    I was then posed the question if the same ball was to slide down the same surface but this time frictionless what would acceleration then be?

    When I derived acceleration I didn't take into account friction or even fore for that matter so not sure if I am to answer in maths or using words?
     
  2. jcsd
  3. Dec 11, 2013 #2

    CAF123

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    Gold Member

    In the case of friction being present, then you would have the consider the work done by friction in your energy calculation.
     
  4. Dec 11, 2013 #3
    What's the difference between "rolling down without slipping" and "sliding down"?
     
  5. Dec 11, 2013 #4
    If it was slipping there would be no inertia so the ball would roll faster:)
     
  6. Dec 11, 2013 #5
    I do not see how slipping and inertia are connected, but, more importantly, why would the ball start rolling?
     
  7. Dec 11, 2013 #6
    Well inertia is the analogue of mass so hence it's resistance to move? So if the ball isn't rolling it's inertia doesn't play a roll, only the mass matters?
     
  8. Dec 11, 2013 #7
    Inertia is not an analogue of mass, inertia is mass. But what you seem to mean is the moment of inertia, which is indeed an analog of mass in rotary motion. Use proper terms, they matter.

    If the ball is not rolling, that is to say, not rotating, then its moment of inertia would not play any role.

    But the question, again, is: would the ball roll or not on a friction-less surface?
     
  9. Dec 11, 2013 #8
    No the ball would not roll

    It would slide.

    Could the solution be that acceleration is constant as I have just noticed it is a problem which would only be worth 1 mark

    And I apologise I meant to say moment of inertia
     
  10. Dec 11, 2013 #9
    Well friction is the reason the ball rolls so it wouldn't roll
     
  11. Dec 11, 2013 #10
    Is it perhaps

    F= mg sin θ

    a= F/m

    So

    mg sin θ/m = g sin θ
     
  12. Dec 11, 2013 #11
    You are on the right track, but is the sine function correct? You said the angle was with the vertical, so the smaller the angle, the greater the force should be, right?
     
  13. Dec 11, 2013 #12
    I mis typed in that case the angle is θ to the horizontal:)

    Would this be correct in that case?
     
  14. Dec 11, 2013 #13
    Yes, that looks good.
     
  15. Dec 11, 2013 #14
    Ah ok that is simple I suppose!!

    Critical thinking!!! Lol


    Thanks so much for you help, always appreciated:)
     
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