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Rolling motion

  1. Oct 6, 2005 #1
    An object rolls because of friction yea? So does that mean a torque is produced when the ball is given an initial push? I mean there is a force, and the force is perpendicular to the line which connects to the center of rotation, so this would be logical yea?

    But if there is a torque, then there is angular acceleration, since torque = I * alpha. And an angular acceleration in pure rolling implies accelerated motion of the center of mass. The ball can't be accelerating into infinity; its ridiculous. ARgh. Help me!

    Any thoughts at all would be appreciated. I'm sure I went wrong somewhere.
  2. jcsd
  3. Oct 6, 2005 #2

    Doc Al

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    Staff: Mentor

    In order for the ball to start rotating, a torque is needed. Friction can provide that torque. But once the ball is able to roll without slipping, the friction force needed to maintain the motion is zero (at least on a horizontal surface). Torque is not needed to maintain a constant rotational speed.

    I don't understand what you mean by "And an angular acceleration in pure rolling implies accelerated motion of the center of mass" or by "accelerating into infinity".
  4. Oct 6, 2005 #3
    But if friction is zero, how can the ball be rolling? Or is it just N's first law in action? I don't think I'm making sense.
    Yea and the latter questions are irrelevant now. Thanks.
  5. Oct 6, 2005 #4

    Doc Al

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    Staff: Mentor

    Once the ball gets rolling, friction is not needed to keep it rolling at the same speed.
  6. Oct 6, 2005 #5
    Rolling down incline


    When a ball rolls down an incline of angle theta, we say that:

    mg sin(theta) - friction = ma

    Then what we do is substitute the expression for friction with:

    moment of inertia * angular acceleration = friction * radius of ball ---> torque

    so that gives friction = (2/5) ma

    And then we get:

    a = (5/7) mg sin(theta).

    Am I right to say that friction in this case is not given by mu * N anymore? Is this a different frictional force... although I don't see how there can be another frictional force.
    OR is it an equivalent expression. In which case then, if we knew mu * N, then we can just determine acceleration from the first expression. But then, wouldn't that mean that the ball is like any other object (it could be a box, and it'd still have the same a).
    OR have I got everything wrong.

    Any advise would be good. Thanks!
  7. Oct 6, 2005 #6

    Doc Al

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    Staff: Mentor

    All good. I'm sure you realize that you are implicitly assuming the "rolling without slipping" condition, which is: [itex]a = \alpha r[/itex].

    Realize that [itex]\mu N[/itex] is the maximum available static friction force. The actual friction force will be less. (Also realize that if [itex]\theta[/itex] is too great or [itex]\mu[/itex] too low, then the static friction will not be enough to prevent slipping.)
  8. Oct 6, 2005 #7
    Thank you; really appreciate your help.
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