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What conditions are needed so that a billiard ball slides rather than rotates?

  1. Jan 31, 2013 #1
    Or at the point at which it starts to rotate....

    My thinking is that the sliding friction is the only thing that will cause the ball to rotate F = u*N
    N = mg

    And moment of inertia I = 2/5*m*r^2

    So, when I = 2/5*m*r^2 = u*N = F it will be at the instant of the change over....

    Am i right in thinking that??
     
  2. jcsd
  3. Jan 31, 2013 #2
    As far as I know it's the static friction which causes moment and makes the ball roll.
     
  4. Jan 31, 2013 #3
    OK static friction. Am I right in equating the two equations?
     
  5. Jan 31, 2013 #4
    Well, how did you get it? It doesn't make sense to me.
     
    Last edited: Jan 31, 2013
  6. Jan 31, 2013 #5
    Well, if the table was frictionless and if you hit the ball dead center it would never rotate...

    So the frictional force will be the force that causes the rotation.....

    The sliding will occur if something opposes the frictional force.... ie the mass moment of inertia of the ball.

    If the frictional force is equivalent to the mass moment of inertia then the ball wont rotate. Right?
     
  7. Jan 31, 2013 #6
    The moment of inertia is a property of an object and it resists angular acceleration cf. mass in linear motion. It is neither a force nor moment. Right? :smile:
     
  8. Jan 31, 2013 #7
    OK so I'm wrong. Can you point me in the right direction then?
     
  9. Jan 31, 2013 #8

    haruspex

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    Take a look at the dimensions (M, L, T, ...). I = ML2, F = MLT-2. So the equation cannot make sense.
    What torque does the frictional force exert about the ball's centre?
    What kind of movement results from an unopposed torque?
    What equation connects that movement with the torque and the moment of inertia?
     
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