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Circular acceleration

  1. Oct 24, 2008 #1
    Okay, so i understand that circular acceleration and force point to the middle of the circle. Now lets say a roller coaster is doing a loop, at the top of the circle, why wouldn't the roller coaster just fall down? force of gravity is pointing down, the Fc is pointing down. What force is keeping the roller coaster from falling down? and why don't we ever lable this force on free body diagrams?

    THANKS!
     
  2. jcsd
  3. Oct 24, 2008 #2

    rcgldr

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    There are wheels on the side and bottom of a roller coaster than prevent it from falling away from the track. This and harnesses on the passengers allow inverted maneuvers that are "negative" g.

    Near the top part of a positive g loop, a roller coaster is accelerating downwards faster the rate of acceleration due to gravity, so the passengers experience a downwards force from the seat, which relative to the passenger orientation, feels like a normal positive g situation where the seat and floor are pushing "up" with respect to the passengers orientation. If a passenger were to hold a ball and releases it under these conditions, the ball would "fall" towards the floor of the roller coaster whenever it's under a "positive" g situation.
     
  4. Oct 24, 2008 #3
    So in reality you are falling, and the belt is what's keeping you in place (the force that the belt applies to you = mg+fc?)?

    There wouldn't be any normal force sine both forces are pulling the passenger away from the chair, right (at the top)?

    So is it possible to feel weightless at the top of the loop?

    or at the buttom of the loop?

    What my understanding is right now: At the buttom the mg is point "down" the "fc" is point "up", and to be weightless there won't be any normal foce since fc = mg.

    But i have a feeling that is wrong, and that you can feel weightless only at the top?
     
  5. Oct 25, 2008 #4
    I'm not very good at English, so if there's sth wrong with my words, plz tell me!
    In my opinion, the Fc KEEPS the roller-coaster LOOPING. The centripetal force keeps objects orbit, just so. Because this force continuously change its direction. Don't misunderstand that it makes the roller coaster falls down. If The roller coaster is the system of reference, there's a centrifugal force f ( we admit it) and we have f=-Fc( vector) f+P=0(vector) when the roller coaster at the top!
     
  6. Oct 25, 2008 #5

    Doc Al

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    One key thing you are missing is that the coaster is moving at the top of the loop. Were it not for the force of the track pushing down on the coaster (the "normal" force), it would be shooting off into the air like any other projectile.

    At the top of the loop, the forces on the coaster are: (1) gravity, which acts down, and (2) the normal force of the track on the coaster, which also acts down. These two forces add up to the net centripetal force on the coaster.

    Fc is not a separate force--it should never appear on a free body diagram!
     
  7. Oct 25, 2008 #6

    Doc Al

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    Assuming the coaster is designed to have sufficient speed at the top, the belt is not holding you in--you are being "pushed" into the seat (that's the normal force).

    Sure there's a normal force on the coaster. It acts down.

    If the coaster is designed to have just enough speed to barely hold you in the seat, then your acceleration would be g downward--just like you were in freefall and you'd feel "weightless" at the top of the loop. And the normal force on you would be zero.

    At the bottom of the loop an upward normal force is required to produce an upward centripetal acceleration. So you won't have a chance to feel weightless at the bottom, only at the top.

    Again, "centripetal force" is not a separate force. It is due to the normal force and weight.
     
  8. Oct 25, 2008 #7
    Thank you so much! that makes it so much more clear, can't believe the prof forgot to mention that... he would always have Fc on his diagrams.

    So at the buttom the normal force is > then gravitational force? If it is, why is that?
     
  9. Oct 25, 2008 #8

    rcgldr

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    The normal force (Fc) is path and speed dependent and independent of gravity. Most spirals on coasters involve "negative g", Fc is less than gravity while inverted, so the coaster hangs from the track, and the restraints keep the passengers from falling out. Most, but not all loops are "positive g" so that Fc is greater than gravity at the top.
     
  10. Oct 25, 2008 #9

    Doc Al

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    A standard free body diagram should only show actual forces. (Normal and gravity, in this case.) :grumpy:

    Of course, free body diagrams are not the only kind of diagrams one can draw. :wink:
    At the bottom of a loop the centripetal acceleration is upward and thus the required centripetal force is upward. Since Fc = Normal - weight, the only way to make that positive is if the normal force is greater than the weight. So you'll feel heavier than usual when traversing the bottom of a loop.
     
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