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M2 problem (again)

Discussion in 'Mathematics' started by Maths_Mike, Aug 26, 2011.

  1. Maths_Mike

    Maths_Mike New commenter

    OK I have had no problem getting the correct answers but my understanding is a bit unclear.
    Part a) Mass on a turntable at point of slipping. Given rpm and radius from centre find coeffcient of friction.
    No problem just assume frictional force is equaly to the force causing the acceleration towards the center ie mrw^2 _ I got this OK
    But then the turntable is slowed down. Find the resultant force of the object.
    The answer is mrw^2 (and I undersatnd that this force must exist and must be directly towards the centre because it causes the accelration and prevents motion in a straight line)
    My dilema is what has happened to the friction?
    It must still exist and my first idea was that it needed to be subtracted from mrw^2 to find the resultnat force.
    I think this is wrong but am still not sure why?
     
  2. Maths_Mike

    Maths_Mike New commenter

    So to try and expalin more clearly
    The frictional force can be found at the point of slipping by it being equaly to the force causing the circular motion (mrw^2)
    BUT
    unlike linear motion the resultnat force is not 0 - the friction does not cancel out so were does it go?
     
  3. David Getling

    David Getling Senior commenter

    Of course the friction cancels out, it's just that the direction in which it acts is continuously changing. And, of course, below limiting friction the amount of frictional force generated is no more than needed to exactly balance the resultant of any opposing force. Or, in this case, to supply exactly the calculated acceleration towards the centre.
     
  4. briceanus

    briceanus New commenter

    The friction is still there, just less of it. Your 'w' though is now less than it was because the turntable is going more slowly.
    This will keep going as the turntable gets slower, until it stops, when w = 0 so friction = 0.

    BUT, if you speed it up, friction reaches its maximum (as per part a), and any faster, there isn't enough friction force to hold the particle in place, and it skids off.
    Bric
     
  5. Maths_Mike

    Maths_Mike New commenter

    If it cancels with a force that constantly changes direction - what happens to the centrpetal force - which is directly to the centre - it must be equal and opposite to something. If not friction what?
     
  6. briceanus

    briceanus New commenter

    The friction doesn't cancel out. The friction is what provides the force toward the centre, which in turn enables circular motion. If there was no friction, there would be no circular motion, the particle would skid right off.
    Bric
     
  7. Maths_Mike

    Maths_Mike New commenter

    I undersatnd this but why is the resultant force mrw^2 - regardless of the friction?
    In the first part of the question my assumption that the friction opposed the force towards the centre seemed to work.
    But in the second the friction has no effect?
     
  8. Maths_Mike

    Maths_Mike New commenter

    But the force towards the centre can be much greater than the friction ?
    Sry I am confused!
     
  9. Maths_Mike

    Maths_Mike New commenter

    OK if the surface is smooth we still have F=mrw^2.
    ANd this is the RESULTANT force always even if there is friction (i think) as it causes the acceleration.
    The frictional force when it exists must however be equal and opposite to something - but what?
     
  10. briceanus

    briceanus New commenter

    In the first part, F = uR as it is at the limit. i.e. the most amount of friction possible is being used to hold the particle 'there', which in the case of circular motion means it keeps going round on the turntable. ny faster and it would skid off. The 'resultant force' is the total amount of force required to make the particle go round in circles, which is your answer.

    With the turntable slower, less friction is required to hold it in place. BUT the 'resultant force' (which is now less than before) is still mrw^2, but as the turntable is slower, w is less, so the 'resultant force' holding it in place (in circular motion) in less.

    Is that clearer?
    Bric
     
  11. mmmmmaths

    mmmmmaths New commenter

    Try this one Mike



    http://physics-forevery-one.blogspot.com/2010/05/newtons-second-law-for-uniform-circular.html
     
  12. briceanus

    briceanus New commenter

    Friction is the ONLY force acting on the particle toward the centre, unless there is a string or somesuch other attached.
     
  13. Maths_Mike

    Maths_Mike New commenter

    I get it thank you so much. The force towards the centre is the friction! (up to its max value) I ha d
    the friction going in the opposite direction.
     
  14. briceanus

    briceanus New commenter

    No probs. Feel free to ask about vertical circular motion when you get onto that. Even more fun.
    Bric
     

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