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Can't figure out where to go with this (Centripital Acceleration)

  1. Jan 23, 2005 #1
    An 1800 kg car passes over a hump in a road that follows the arc of a circle of radius 42 m.

    a. What force does the road exert on the car as the car passes the highest point of the hump if the car travels at 16 m/s?

    b. What is the maximum speed the car can have as it passes this highest point before losing contact with the road?

    Now for my work

    a.

    F = ma

    F = mv²/r

    F = (1800)(16²)/42

    F = 10971.4 N

    I think that is horribly wrong in my opinion... and also I have no idea how to start question b. No pictures or anything either
     
    Last edited: Jan 23, 2005
  2. jcsd
  3. Jan 23, 2005 #2
    okay I did a little more work but I don't know if the concepts are correct

    a. Force normal (from the road) would be equal to weight minus centripetal force


    F = mg - mv²/r
    F = (1800)(9.8) - (1800)(16²)/42

    F = 6668.6 N

    b.weight = centripetal force

    mg = mv²/r

    v = sqrt(mgr/m)
    v = sqrt(gr)
    v = sqrt(9.8*42)
    v = 20.3 m/s

    someone wanna confirm this work if Im going the right way?
     
  4. Jan 23, 2005 #3

    learningphysics

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    Homework Helper

    I believe you have the correct answers. Only one thing I'd point out. There isn't an independent centripetal force... The sum of the forces acting on the object results in a net force that is centripetal. Hope I don't cause any confusion. Just wanted to point out that it is the normal force acting with gravity, that results in the centripetal force. I'd have written the first equation as:

    [tex]mg-F_n = ma_y[/tex]

    This is just the sum of the forces in the y direction, taking downward as positive. The only two forces acting in the y direction here are the normal force and the weight.

    And then, since it is moving in a circle the acceleration is centripetal... so you know acceleration is v^2/r

    [tex]mg-F_n=\frac{mv^2}{R}[/tex]

    and then solve for F_n as you did.
     
    Last edited: Jan 23, 2005
  5. Jan 23, 2005 #4
    okay thanks for the reply
     
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