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Math Help - find moments of inertia and its period

  1. #1
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    find moments of inertia and its period

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  2. #2
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    For part (i), are you familiar with the parallel axis theorem? As to part (ii), a circular hole in a circular disk can be treated as the disk without the hole plus a disk corresponding to the hole with negative mass equal to the subtracted mass of the hole. You should also be familiar with the physical pendulum.

    --Kevin C.
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  3. #3
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    then?

    Quote Originally Posted by TwistedOne151 View Post
    For part (i), are you familiar with the parallel axis theorem? As to part (ii), a circular hole in a circular disk can be treated as the disk without the hole plus a disk corresponding to the hole with negative mass equal to the subtracted mass of the hole. You should also be familiar with the physical pendulum.

    --Kevin C.

    i got :
    I= (m_a * a^2 + 2 m_c * c^2 - m_b * b^2) /2

    then how can we get the period?
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  4. #4
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    Quote Originally Posted by szpengchao View Post
    i got :
    I= (m_a * a^2 + 2 m_c * c^2 - m_b * b^2) /2

    then how can we get the period?
    Where are you getting an m_c from? The moment of inertial of the main disc is
    I_a = \frac{1}{2}Ma^2 + Mc^2 = \frac{1}{2}\rho (\pi a^2 - \pi b^2)(a^2 + c^2)
    and the "moment of inertia" of the hole is
    I_b = \frac{1}{2}mb^2 = \frac{1}{2} \rho \pi b^2 \cdot b^2
    where \rho is the density of the material.

    So
    I_{tot} = \frac{1}{2}\rho (\pi a^2 - \pi b^2)(a^2 + c^2) - \frac{1}{2} \rho \pi b^2 \cdot b^2

    To find the period for small oscillations use
    T = 2 \pi \sqrt{\frac{I_{tot}}{MgL_{cm}}} = 2 \pi \sqrt{\frac{I_{tot}}{(\pi a^2 - \pi b^2)gL_{cm}}}
    where L_{cm} is the distance from the center of mass to the axis of rotation. I leave it to you to find L_{cm}.

    -Dan
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