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Math Help - Circular Motion Problem

  1. #1
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    Circular Motion Problem

    Ok,

    A car of mass 1.5 tonnes pass over a bridge formed the arc of a circle of radius 20m.

    a) Find the force exerted by the car on the road at the top of the bridge if the car is travelling at 10m/s.

    b) What speed would cause the car to be on the point of lifting up off the bridge at its highest point

    What I've done:

    F = mv^2 / r
    F = 1500 * 10^2 / 20
    F = 7500 N
    ---- Don't know if this is anything (the answer in the book is 7200N)

    tan(t) = v^2/rg
    tan(t) = 10^2/20*9.8
    tan(t) = 0.510
    (t) = tan^-1(0.510)
    (t) = 27.02 deg

    Force of car on road = mg sin (t)
    = 1500*9.8sin (27.02)
    = 6678.23 N

    However as I said previously the answer is 7200 N.

    b) No idea really... not sure of the formula to use

    The answer for b) is 14m/s.
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  2. #2
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    Quote Originally Posted by jeta View Post
    Ok,

    A car of mass 1.5 tonnes pass over a bridge formed the arc of a circle of radius 20m.

    a) Find the force exerted by the car on the road at the top of the bridge if the car is travelling at 10m/s.

    b) What speed would cause the car to be on the point of lifting up off the bridge at its highest point

    What I've done:

    F = mv^2 / r
    F = 1500 * 10^2 / 20
    F = 7500 N
    ---- Don't know if this is anything (the answer in the book is 7200N)

    tan(t) = v^2/rg
    tan(t) = 10^2/20*9.8
    tan(t) = 0.510
    (t) = tan^-1(0.510)
    (t) = 27.02 deg

    Force of car on road = mg sin (t)
    = 1500*9.8sin (27.02)
    = 6678.23 N

    However as I said previously the answer is 7200 N.

    b) No idea really... not sure of the formula to use

    The answer for b) is 14m/s.
    a) Fnet = mg - N = 14700 - N

    where N is the normal reaction force of the road on the car and the downwards direction is positive.

    But Fnet = mv^2/r = 7500.

    Therefore 7500 = 14700 - N => N = 7200 Newton.

    From Newton III, size of force exerted by car on road = 7200 Newton.


    b) When the car is on the point of "lifting up off the bridge" N = 0.

    Therefore mv^2/r = mg

    => 1500 v^2/20 = 14700 => v = ....
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  3. #3
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    Thanks so much... How do you just know everything about maths? You really do...

    So yeah, thanks again.
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