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Math Help - [SOLVED] Centre of mass

  1. #1
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    [SOLVED] Centre of mass

    A uniform semicircular lamina has mass M. A is the mid-point of the diameter and B is on the circumference at the other end of the axis of symmetry. A particle of mass m is attached to the lamina at B. The centre of mass of the loaded lamina is at the mid-point of AB. Find in terms of pi, the ratio M:m


    The answer given is 3pi : (3pi-8)
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  2. #2
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    using the theorem of Pappus ...

    let R = semicircle fixed radius
    r = axis of area rotation

    \frac{\pi R^2}{2} \cdot 2\pi r = \frac{4}{3}\pi R^3

    r = \frac{4R}{3\pi}

    let point A be 0 ...

    x_{cm} = \frac{M\frac{4R}{3\pi} + mR}{M+m}

    \frac{R}{2} = \frac{M\frac{4R}{3\pi} + mR}{M+m}<br />

    \frac{1}{2} = \frac{M\frac{4}{3\pi} + m}{M+m}

    2\left(M\frac{4}{3\pi} + m\right) = M+m

    \frac{8M}{3\pi} + 2m = M+m

    m = M\left(1 - \frac{8}{3\pi}\right)

    \frac{1}{1 - \frac{8}{3\pi}} = \frac{M}{m}

    \frac{3\pi}{3\pi - 8} = \frac{M}{m}
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