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    A mass of 9.50 kg is suspended from a 1.19 m long string. It revolves in a horizontal circle.
    The tangential speed of the mass is 2.28 m/s. Calculate the angle between the string and the vertical.

    My solution:

    Let the angle between the string and the vertical be \theta

    Tcos\theta=mg ------ (1)

    Tsin\theta=\frac{mv^2}{r} ------ (2)

    Dividing (1) by (2),

    tan\theta=\frac{v^2}{rg}

    r=lsin\theta

    So,

    tan\theta=\frac{v^2}{lsin\theta g}

    \frac{sin^2\theta}{cos\theta}=\f  rac{(2.28)^2}{1.19*9.8}=0.45

    1-cos^2\theta=0.45cos\theta

    cos^2\theta+0.45cos\theta-1=0

    cos\theta=\frac{-0.45\pm\sqrt{(0.45)^2+4}}{2}

    cos\theta=\frac{-0.45\pm2.05}{2}

    cos\theta=0.8 or cos\theta=-1.25

    Rejecting cos\theta=-1.25, we get

    cos\theta=0.8

    \theta=36.87^0

    Is this correct?

    Thanks
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    yep, seems fine to me.
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    Yep, it's all good. Perhaps giving a reason why you are rejecting \cos \theta = -1.25 would be necessary (even though it is quite obvious...)
    • Study Helper
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    Minor quibble:

    You've worked out 0.45 to 2 sig.fig. and then quoted your answer, theta, to 4!
 
 
 
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Updated: February 6, 2010
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