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Thread: Hope someone can help

  1. #1
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    Hope someone can help

    I have this od chestnut to solve, I have done as much as what I think is right but I need some assistance:

    a) A body of mass m kg is attached to a point by string of length
    1.25 m. If the mass is rotating in a horizontal circle 0.75 m below the
    point of attachment, calculate its angular velocity.

    So far I have:

    $\displaystyle r = \sqrt{1.25^2 - 0.75^2}$
    $\displaystyle r = 1$

    $\displaystyle \tan\theta = \frac{\omega^2r}{g}$

    $\displaystyle \tan\theta = \frac{1}{0.75}$

    $\displaystyle therefore:$
    $\displaystyle 53.13 = \frac{\omega^2}{9.81}$

    $\displaystyle \omega = 22.83$

    Have I totally done this wrong, also were do I need to go from here?

    Thanks
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  2. #2
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    From your own (correct) equations:

    $\displaystyle \tan \theta = \frac{\omega^2r}{g} $

    $\displaystyle \tan \theta = \frac{1}{0.75} $

    Equating the two is

    $\displaystyle \frac{\omega^2r}{g} = \frac{4}{3} $

    There is where you went wrong. I think you equated $\displaystyle \theta $ to $\displaystyle \tan \theta $ instead of $\displaystyle \tan \theta $ = $\displaystyle \tan \theta $
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  3. #3
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    Thanks

    so w = 3.63 ms-1 ?


    R
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  4. #4
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    Ok so part 2:

    (b) If the mass rotates on a table, calculate the force on the table when
    the speed of rotation is 25 rpm and the mass is 6 kg

    we have:

    $\displaystyle T\cos\theta = mr\omega^2$

    $\displaystyle T = \frac{6*1*25^2}{\cos\theta}$

    Which angle is $\displaystyle \theta$?
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  5. #5
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    The angle $\displaystyle \theta $ is shown in the following picture. You can find the value from your equation in the first part: $\displaystyle \tan \theta = \frac{1}{0.75}$




    Also, you need to convert 25 the rpm into radians per second.

    I also got the same answer as you did on the first part.
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  6. #6
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    thanks ive got it.
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