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Force on coaster = ma(c) + ma(g) ?

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip?

    2. Relevant equations

    Force on an object traveling in a circle perpendicular to the force of gravity is centripetal acceleration plus acceleration due to gravity times the mass of the object.... i think

    3. The attempt at a solution

    16 meters per sec squared divided by 40 meters times 500 kg ...... plus 500 kg times 9.8 meters per second squared = 8.1 kN ... right?
     
  2. jcsd
  3. Nov 2, 2008 #2

    tiny-tim

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    Hi CaptainADHD! :smile:
    Right! :biggrin:
     
  4. Nov 2, 2008 #3
    Great, thanks!

    So if I wanted to calculate the force on the coaster at the top of the circle, it would be centripetal acceleration times mass MINUS the force of gravity, since gravity is in the opposite direction, right?
     
  5. Nov 2, 2008 #4

    tiny-tim

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    It depends whether the coaster is under the track or on top of it at the top of the loop …

    draw two diagrams, and do the Newton's second law equation, and you'll see what I mean. :wink:
     
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