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Circular Motion/Gravity Problems

  1. Mar 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A car of mass 1000kg has just reached the top of a circular bridge of radius 6.0m (See the figure provided). Calculate where the car will reach the ground level if the car is traveling at
    a) 5.0 m s-1
    b) 7.67 m s-1
    c) 20.0 m s-1
    at the top of the curve
    Image for first question:
    [​IMG]


    At what minimum speed should the Earth spin to enable you to float away from the Earth's surface at the equator?

    2. Relevant equations
    [tex]
    \begin{array}{l}
    W = mg \\
    a = \frac{{v^2 }}{r} = \frac{{4\pi ^2 r}}{{T^2 }} = \frac{{2\pi v}}{T} \\
    g = \frac{F}{m} = \frac{{GM}}{{r^2 }} \\
    F = ma \\
    F = \frac{{GMm}}{{r^2 }} \\
    \end{array}
    [/tex]


    3. The attempt at a solution
    I have no idea on where to start with these problems, so all help is greatly appreciated, many thanks in advance

    unique_pavadrin
     
  2. jcsd
  3. Mar 10, 2007 #2
    Ad the second question - use the equation FG = Fcentripetal. Then rearrange it to get the formula for angular speed.
     
  4. Mar 10, 2007 #3

    mjsd

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    mmm... sounds like a projectile motion problem...?? do we assume that the car stick to the ground or what?
     
  5. Mar 10, 2007 #4
    mjsd, that is what ai am trying to find out, if the car will stick to the ground, past the peak of the bridge traveling at the different velocities
     
  6. Mar 11, 2007 #5

    mjsd

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    ok, so in order for it to stick it mustn't move too fast... so that it will remain in uniform circular motion following the arc. obviously mv^2/r comes in here. if v is too high it will fly off
     
  7. Mar 11, 2007 #6
    so i have to find when the reaction force along the top of the bridge is zero?
     
  8. Mar 11, 2007 #7

    mjsd

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    yes, that will give you the cut-off point
     
  9. Mar 11, 2007 #8
    okay thanks for that, ill try it and see how i go
     
  10. Mar 11, 2007 #9

    mjsd

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    although in the question you said the circular bridge is of radius 6m, the diagram seems to indicate the radius is 10m instead??
     
  11. Mar 11, 2007 #10
    for the velocity of 5 meters per second:
    [tex]
    \begin{array}{l}
    a = \frac{{v^2 }}{r} = \frac{{5^2 }}{{10}} = 2.5\,m\,s^{ - 2} \\
    F = ma = 2.5\left( {1000} \right) = 2500 \\
    W = mg = 9.8\left( {1000} \right) = 9800 \\
    W > F \\
    \end{array}
    [/tex]

    therefore the car will not leave the road

    when the car is traveling at 7.67 meters per second:
    [tex]\[
    \begin{array}{l}
    a = \frac{{v^2 }}{r} = \frac{{7.67^2 }}{{10}} = 5.88289\,m\,s^{ - 2} \\
    F = ma = 5.88289\left( {1000} \right) = 5882.89 \\
    W = mg = 9.8\left( {1000} \right) = 9800 \\
    W > F \\
    \end{array}
    [/tex]

    therefore the car will not leave the road

    for the car traveling at 20 meters per second:
    [tex]
    \begin{array}{l}
    a = \frac{{v^2 }}{r} = \frac{{20^2 }}{{10}} = 40\,m\,s^{ - 2} \\
    F = ma = 40\left( {1000} \right) = 40000 \\
    W = mg = 9.8\left( {1000} \right) = 9800 \\
    W < F \\
    \end{array}
    [/tex]

    therefore the car will leave the road.
    where the car will land:
    [tex]
    \begin{array}{l}
    s = ut + \frac{1}{2}at^2 \\
    s_{vertical} = 6 \\
    u_{vertical} = 0 \\
    a_{vertical} = 9.8 \\
    6 = \frac{1}{2}\left( {9.8} \right)t^2 \\
    t = \sqrt {\frac{{2\left( 6 \right)}}{{9.8}}} = 1.1065 \\
    s_{horizontal} = ut = 20\left( {1.1065} \right) = 22.1313\,m \\
    \end{array}
    [/tex]

    therefore the car will land approximately 22 (horizontally) meters from the top of the bridge.

    does that look correct? thanks.
     
  12. Mar 11, 2007 #11

    mjsd

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    haven't check your numerical values, but method looks good (which is the more important bit)
     
  13. Mar 11, 2007 #12
    you legend! thanks
     
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