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Moment of inertia - is this the right integral? watch

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    "Find the moment of inertia of a thin circular disk, of total mass M and radius a, about an axis passing through its centre and lying in the plane of the disk (i.e. about the diameter)"

    I=\sigma b^2 dA and \sigma=\frac{M}{\pi a^2}, dA=\sqrt{1-x^2}dx. So I=\frac{2M}{\pi a^2}\displaystyle\int^a_{-a} x^2\sqrt{1-x^2}\ dx

    Is that right? Because it doesn't look like a nice integral...
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    (Original post by Plagioclase)
    "Find the moment of inertia of a thin circular disk, of total mass M and radius a, about an axis passing through its centre and lying in the plane of the disk (i.e. about the diameter)"

    I=\sigma b^2 dA and \sigma=\frac{M}{\pi a^2}, dA=\sqrt{1-x^2}dx. So I=\frac{2M}{\pi a^2}\displaystyle\int^a_{-a} x^2\sqrt{1-x^2}\ dx

    Is that right? Because it doesn't look like a nice integral...
    two ways

    either you use a "tailor made system" which assumes the MI of a circular hoop

    or

    use plane polars
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    (Original post by TeeEm)
    two ways

    either you use a "tailor made system" which assumes the MI of a circular hoop

    or

    use plane polars

    If I do it with plane polars, would dA=2\pi r dr so I=\frac{M}{\pi a^2}\displaystyle\int^a_{0} 2\pi r^3 dr?
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    (Original post by Plagioclase)
    If I do it with plane polars, would dA=2\pi r dr so I=\frac{M}{\pi a^2}\displaystyle\int^a_{0} 2\pi r^3 dr?
    Without writing/deriving it I think that expression is correct (assuming you did the integration wrt to theta already)
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    Since the desired axis is in the plane of the lamina, I presume you're going to use the perpendicular axis theorem.
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    (Original post by TeeEm)
    Without writing/deriving it I think that expression is correct (assuming you did the integration wrt to theta already)
    Agh no that integral wouldn't work because that's assuming the axis is perpendicular to the disk going through the centre rather than in the plane of the disk...

    Instead, could you do it with triangles so dA=\frac{1}{2}r^2d\theta and b=\frac{1}{3}a cos(\theta)?

    (Original post by ghostwalker)
    Since the desired axis is in the plane of the lamina, I presume you're going to use the perpendicular axis theorem.
    That was never mentioned in our notes but would the technique I described above work?
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    (Original post by ghostwalker)
    Since the desired axis is in the plane of the lamina, I presume you're going to use the perpendicular axis theorem.
    forgive I completely missed the axis
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    (Original post by Plagioclase)
    Agh no that integral wouldn't work because that's assuming the axis is perpendicular to the disk going through the centre rather than in the plane of the disk...

    Instead, could you do it with triangles so dA=\frac{1}{2}r^2d\theta and b=\frac{1}{3}a cos(\theta)?


    That was never mentioned in our notes but would the technique I described above work?
    forgive I completely missed the axis

    I am a bit tied up at present

    look at question 27 in this link
    http://madasmaths.com/archive/maths_...plications.pdf
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    (Original post by TeeEm)
    forgive I completely missed the axis

    I am a bit tied up at present

    look at question 27 in this link
    http://madasmaths.com/archive/maths_...plications.pdf
    Thank you!
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    (Original post by Plagioclase)
    That was never mentioned in our notes but would the technique I described above work?
    Not enough info to decide, but I have strong doubts.

    The natural theta to use is that to the desired axis, in which case I'd expect to see a sin rather than a cos.

    And I don't know where "b" has come from. I suspect a misguided centre of mass, but I'm just guessing.

    That said, I'd go with TeeEm's link.
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    (Original post by Plagioclase)
    "Find the moment of inertia of a thin circular disk, of total mass M and radius a, about an axis passing through its centre and lying in the plane of the disk (i.e. about the diameter)"

    I=\sigma b^2 dA and \sigma=\frac{M}{\pi a^2}, dA=\sqrt{1-x^2}dx. So I=\frac{2M}{\pi a^2}\displaystyle\int^a_{-a} x^2\sqrt{1-x^2}\ dx

    Is that right? Because it doesn't look like a nice integral...
    It looks fine to me, except that you want \sqrt{a^2-x^2}.

    The integral itself is straightforward: set x=a\sin\theta
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    (Original post by atsruser)
    It looks fine to me, except that you want \sqrt{a^2-x^2}.

    The integral itself is straightforward: set x=a\sin\theta
    Thanks! This is the way that makes the most sense for me.
 
 
 
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