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Math Help - uniform circular motion

  1. #1
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    uniform circular motion

    A mass m = 8.5 kg is suspended from a string of length L = 1.37 m. It revolves in a horizontal circle. The tangential speed of the mass is 2.75 m/s. What is the angle q between the string and the vertical (in degrees)?

    help?
    i know F= m v^2 /R...
    so..
    Tsinx = m v^2/Lsinx
    and
    Tcosx - mg = 0

    i'm having trouble solving for x though...
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  2. #2
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    T\cos(q) = mg

    T = \frac{mg}{\cos(q)}


    T\sin(q) = \frac{mv^2}{L\sin(q)}

    \frac{mg\sin(q)}{\cos(q)} = \frac{mv^2}{L\sin(q)}

    \frac{\sin^2(q)}{\cos(q)} = \frac{v^2}{gL}

    \sin^2(q) = \frac{v^2}{gL}\cos(q)<br />

    1 - \cos^2(q) = \frac{v^2}{gL}\cos(q)

    0 = \cos^2(q) + \frac{v^2}{gL}\cos(q) - 1

    use the quadratic formula ...

    a = 1, b = \frac{v^2}{gL} , c = -1

    \cos(q) \approx 0.75726...<br />

    q \approx 41^{\circ}

    T \approx 110 \, N
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