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Circular Motion

  1. Aug 31, 2015 #1
    1. The problem statement, all variables and given/known data
    a motor travelling at a constant speed is just able to completely go round a vertical circular track of radius 8m The total mass of motor is 210kg

    i)Compute the angular velocity at the top of the track
    ii)compute the maximum force on the track

    2. Relevant equations


    3. The attempt at a solution
    i know that when the motor is at the top of the track the formula should be Fn+m(v^2/r) = mg
    but i just cant found v.
     
  2. jcsd
  3. Aug 31, 2015 #2

    CWatters

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    The problem statement says "just able". What does that mean for Fn ?
     
  4. Aug 31, 2015 #3
    Fn=0 ? But why?
     
  5. Aug 31, 2015 #4

    CWatters

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    Fn is the normal force. Have a think about three different values for Fn and what it means for the car...

    a) Fn > 0
    b) Fn = 0
    c) Fn < 0 (Can Fn be <0 in this situation?)
     
  6. Aug 31, 2015 #5
    I really had no idea about this :frown:
     
  7. Aug 31, 2015 #6

    CWatters

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    Imagine an empty party balloon sitting on a desk. The balloon exerts a force Fballoon on the desk, and the desk exerts a normal force FN on the balloon. The magnitude of these forces is the same...

    |FBalloon| = |mg| = |FN|

    where m is the mass of the balloon.

    If you pump some hydrogen into the balloon FBalloon and FN will start to reduce.

    What happens when FN reaches zero?

    Can FN go negative? What would that mean?
     
    Last edited: Aug 31, 2015
  8. Aug 31, 2015 #7
    When Fn = 0 means the ball is floating ?
    If fn<0 mean the ball is falling down?

    So in the question that "just able" means that the velocity of the motor is just enough to make the motor in the state of " floating" on the top ofthe track?
     
  9. Sep 1, 2015 #8

    CWatters

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    On top of the track ??

    I think you mean on the inside of the track. I assumed it was a loop like this..



    It has to go fast enough that it "just" doesn't fall down at the top.
     
  10. Sep 1, 2015 #9

    CWatters

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    FN > 0 means the car is pressing against the track.

    FN approaching 0 means the car is only just barely pressing against the track.

    FN < 0 can't really happen in this situation because it implies the car is pulling downwards on the track. It could happen on a roller coaster as most modern coasters have retaining wheels to stop the carriage falling off the track if it stops upside down at the top of a loop. Normal road cars don't have these extra wheels.
     
  11. Sep 1, 2015 #10
    You are aware that this formula is supposed to be the result of applying Newton's second law to the motor as it is passing through the locations at the top of the track and at the bottom of the track, correct? Have you drawn a free body diagram on the motor at these locations and shown the forces acting on it? In your equation, if Fn is has a positive value at the top of the track, it implies that the normal force exerted by the track on the motor is acting upward. Are you aware of that? For the bottom of the track, your equation is incorrect, since it implies that the direction of the acceleration of the motor (upward) is in the same direction as g (downward). Were you aware of that? These difficulties are what happens when one does not draw a proper free body diagram for the problem.

    Chet
     
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