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Centripetal force of a roller coaster car

  1. Mar 17, 2008 #1
    I got all the other centripetal force questions but this one has me stumped...

    As a roller coaster car crosses the top of a 30m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?

    The formula I used for all of the other questions was F=mv^2/r. The problem is the only number I have is the diameter (therefore radius). I'm guessing because the apparent weight is the same as true weight the net force is zero? There's also no way to get mass. Maybe there's another formula? Any assistance with this problem would be much appreciated!
  2. jcsd
  3. Mar 17, 2008 #2
    The true weight of something is [tex]F=mg[/tex], so then you don't need to figure out the mass.
    Last edited: Mar 17, 2008
  4. Mar 17, 2008 #3


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    If the apparent weight is the same as the true weight, then the passengers at the top of the loop are experiencing the equivalent of 1 g of force from their seats.
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