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Uniform Circular Motion Confusion

  1. Mar 22, 2005 #1
    I have a question about uniform circular motion that is just confusing me. Maybe one of you can elucidate it for me.

    Say I am in a roller coaster car and I am travelling in a loop. At the top of the loop if my velocity is not great enough I should fall out of the loop but because the car is bolted to the track I will not.

    I have a seat belt on and it is keeping me from being planted into the ground.

    If the centripetal force is 576N and the force of gravity on me is 784N then wouldn't the seat belt have to support a force of 1360N?

    Thanks!
     
  2. jcsd
  3. Mar 22, 2005 #2
    I haven’t done this in while so I might be wrong (but I very much doubt it!).

    The centripetal force pushes you out words (away from gravity) so the force would be 784-576= 208N
     
  4. Mar 22, 2005 #3

    AKG

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    Have you drawn a free body diagram? Do you know how net force relates to centripetal force in uniform circular motion?
     
  5. Mar 22, 2005 #4
    No, the centripetal force pushes you inwards. What gokugreene did not consider is that the car you are riding in (and you and your seatbelt) all have an inertia when you entered the loop. You were travelling say, 50mph in the x direction, once you enter the loop, you are no longer travelling in the x direction, and all your kinetic energy in the x direction is being converted to energy in the direction you are travelling in that instant in time. What happens is all the force from hitting the loop is directed normal to the track, and the force pushing you up and along the loop in the first half of the loop is what keeps you from falling off the second half.
     
  6. Mar 22, 2005 #5

    AKG

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    No, the centripetal force would force you inwards, so the force would be 784-576= 208N. A force pushing outwards would send you flying away. The centripetal force, which is the net force, is obviously the direction of acceleration (F = ma). It is because the velocity always points tangential to the circle and acceleration always points into the circle that keeps you right on the circle.
     
  7. Mar 22, 2005 #6

    AKG

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    [tex]\vec{F} _{net} = \vec{F} _g + \vec{F} _{belt}[/tex]

    [tex]\vec{F} _c = \vec{F} _g + \vec{F} _{belt}[/tex]

    [tex]-576N\hat{y} = -784N\hat{y} + \vec{F} _{belt}[/tex]

    [tex]-576N\hat{y} - (-784N\hat{y}) = \vec{F} _{belt}[/tex]

    [tex]-576N\hat{y} + 784N\hat{y} = \vec{F} _{belt}[/tex]

    [tex]208N\hat{y} = \vec{F} _{belt}[/tex]

    I don't think this has anything to do with normal forces or inertia.
     
  8. Mar 22, 2005 #7
    The inertia is providing the tangential force. This is what I was explaining, except in terms of velocity.
     
  9. Mar 22, 2005 #8
    Ok. Say my velocity=12m/s, my mass=80kg, and the radius of the loop=20m. The minimum velocity to remain in the loop is 14m/s v=sqrt(gr).
    So wouldn't the force on the seat belt have to be equal in magnitude of Fc+mg to keep me from plummeting to my death?

    AKG, your calculations are right and that is what I got but the car isn't going fast enough to remain in circular motion if it is going 12m/s, correct?

    If I am going faster than 14m/s I shouldn't need a seat belt. The seat should provide the normal force, but because, I am going less than 14m/s the seat belt will have to restrain me.
     
    Last edited: Mar 22, 2005
  10. Mar 22, 2005 #9

    AKG

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    whozum, there is no tangential force in uniform circular motion.

    gokugreene
    As I suggested in my first post, have you drawn a free body diagram? If you've read what I've already written, your question has already been answered. Do you know that in uniform circular motion, that Fc = Fnet? Also, if I have a 20 N mass hanging from a rope that can bear a maximum of 10 N, would you say that I have to apply 20N + 10N = 30N upwards to keep it from breaking? You're really just not thinking about this question... :uhh:
     
  11. Mar 22, 2005 #10
    I see what you are saying. Thanks AKG.
     
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