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Rotational Motion and the Law of Gravity

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A roller coaster vehicle has a mass of 500 kg when fully loaded with passengers. a) if the vehicle has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point (point a sits at the bottom of a drop of the roller coaster where the radius is 10 meters) B) what is the maximum speed the vehicle can have at point b for gravity to hold it on the track ( Point b sits at the top of the rollercoaster hill where the radius is 15 meters)


    2. Relevant equations



    3. The attempt at a solution
    Fc = MAc = M(v^2/r)
    n=m(v^2/r)-mg
    n=500(20^2/10)-(500x9.80) = Force of track on car at a = 15,100 N is this right?

    Vtop= "square roote of" gr
    ="Square roote of" (9.80 x 15) = max velocity at top at point b is 12 m/s is this right?
     
  2. jcsd
  3. Mar 24, 2010 #2
    For a), if the coaster is at the bottom of the drop, then the centripetal acceleration should be upwards. In that case, the normal force from the rails due to weight (upwards) needs to be added to the centripetal force applied by the track (upwards), not removed.

    b looks fine.
     
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