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How does centripetal acceleration can cancel out gravity?

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I know the answer and how to get it, but it doesn't make sense because ac points down and so does g. So why do they seem to cancel each other out?

    What is the minimum constant speed a motorbiker needs in order to make it around a vertical loop with a radius of 3 meters?

    2. Relevant equations

    I know that I have to make ac = g.
    So, ac = 9.8m/s^2.
    v = √(r*ac)

    3. The attempt at a solution

    v = √(r*ac) = √(3*9.8) = 5.4m/s

    This all seems to imply that ac and g cancel out, but how can they when they are both in the same direction?
     
  2. jcsd
  3. Mar 29, 2013 #2

    Simon Bridge

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    They don't - what you have worked out is the speed in which the centripetal force is supplied entirely by gravity at the top of the loop. That is why the acceleration and gravity point in the same direction - the acceleration is supplied by the gravity.

    You experience weight by the ground pushing against you. If you were in free-fall, inside a box that is also in free-fall, the floor no longer pushes against you and you feel weightless. But gravity is not "cancelled out". It is similar for the motorcycle.

    If the motorcyclist went any faster, some additional centripetal force will be needed.
    That will come from the normal force between the bike and the track ... i.e. if he went faster, he'd feel the track pushing against him. At exactly the speed you worked out, the track does not push against him at all, giving the feeling of not having any weight.
     
  4. Mar 29, 2013 #3

    collinsmark

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    In addition to what Simon Bridge already said, it's useful here to recall Newton's second law of motion.

    [tex] m \vec a = \sum_i \vec F_i [/tex]

    Here there are two forces, the normal force exerted from the track on the motorbiker, and the force of gravity exerted on the motorbiker.

    In the critical case which you're already calculated, what is the normal force exerted from the track on the motorbiker? [Edit: when at the very top of the loop.]

    Is the motorbiker accelerating? If so, how much, and in what direction, recalling [itex] m \vec a = \sum \vec F [/itex]?
     
    Last edited: Mar 29, 2013
  5. Mar 29, 2013 #4
    F=ma, your force and acceleration should be in the same direction, as you indicated. If the net forces on a body don't equal zero, the body will accelerate as is the case for centripetal acceleration.

    The problem statement is kind of silly, because centripetal acceleration doesn't "cancel" gravity, it is simply the response of a body subjected to the force of gravity.

    [edit, I didn't see the motorcycle part. I thought it was an open ended question about orbits]
     
    Last edited: Mar 29, 2013
  6. Mar 29, 2013 #5

    SammyS

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    It's not that the centripetal force cancels the gravitational force.

    If the magnitude of the centripetal force required to make an object follow the circular path is equal to, or greater than, the gravitational force, then the track (or roadway, or whatever structure) must provide either no force, or additional downward force.

    Otherwise, if the magnitude of the centripetal force required to make an object follow the circular path is less than, the gravitational force, then the track would need to provide an upward force in order for the object to follow that circular path.
     
  7. Mar 30, 2013 #6
    I guess it would be zero.

    I understand it is 9.8m/s^2 downward.
     
  8. Mar 30, 2013 #7

    collinsmark

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    Yes! Good job. :approve:
     
  9. Mar 30, 2013 #8
    Thanks!
     
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