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    The diagram I drew for this is just an ice cream cone shape with vertex on O and z as the axis of symmetry.

    Initially I thought of combining the two in such a way that the inertia is given by:

    \displaystyle I = \frac{M_1+M_2}{\frac{1}{3}\pi a^2 h + \frac{2}{3}\pi a^3} \left( \int_{D_1} x^2 + y^2. dxdydz + \int_{D_2} x^2 + y^2. dxdydz \right)

    where D_1, D_2 are spaces occupied by the cone and hemisphere respectively, but this seems like a nightmare to deal with.

    Having looked at the reference answer of I = \dfrac{3}{10}M_1 a^2 + \dfrac{1}{5}M_2a^2 I thought that you'd just add the inertia's from the two separate objects since inertia of the cone about the same axis is \frac{3}{10}M_1a^2, however the inertia of the hemisphere worked out to be \dfrac{2}{5}M_2a^2 instead.

    Is there a shortcut to this problem? Is addition of inertia's in these types of problems allowed, so the ref-answer answer is incorrect?

    For inertia of the hemisphere I just evaluate \displaystyle I = \frac{3M_2}{2\pi a^3} \int_{D_2} x^2 + y^2 .dxdydz which yields it, but I can't see where I'd be going wrong here.
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    (Original post by RDKGames)
    For intertia of the hemisphere I just evaluate \displaystyle I = \frac{3M_2}{2\pi a^3} \int_{D_2} x^2 + y^2 .dxdydz which yields it, but I can't see where I'd be going wrong here.
    MofI of the hemisphere should be \dfrac{1}{5}M_2a^2

    And yes they are additive.

    Edit: Looks as if you slipped a factor of 2 in that intergral soimewhere.
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    (Original post by ghostwalker)
    MofI of the hemisphere should be \dfrac{1}{5}M_2a^2

    And yes they are additive.
    Hm.. not sure where I'm tripping up then

    \begin{aligned} I & = \frac{3M_2}{2\pi a^3} \int_{D_2} x^2 + y^2 .dxdydz \\ & = \frac{3M_2}{2\pi a^3} \int_{0}^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a r^2 \sin^2 \theta \cdot r^2 \sin \theta .dr d\theta d\phi \\ & = \frac{3M_2}{2\pi a^3} \cdot 2 \pi \cdot \frac{1}{5}a^5 \int_0^{\frac{\pi}{2}} (1-\cos^2 \theta)\sin \theta .d\theta, \quad u=\cos \theta \Rightarrow du = -\sin \theta d\theta \\ & = \frac{3}{5} M_2a^2 \int_1^{0} (u^2-1) .du \\ & = \frac{3}{5} M_2a^2 [\frac{1}{3}(-1)-(-1)] \\ & = \frac{2}{5} M_2a^2\end{aligned}
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    (Original post by RDKGames)
    Hm.. not sure where I'm tripping up then

    \begin{aligned} I & = \frac{3M_2}{2\pi a^3} \int_{D_2} x^2 + y^2 .dxdydz \\ & = \frac{3M_2}{2\pi a^3} \int_{0}^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a r^2 \sin^2 \theta \cdot r^2 \sin \theta .dr d\theta d\phi \\ & = \frac{3M_2}{2\pi a^3} \cdot 2 \pi \cdot \frac{1}{5}a^5 \int_0^{\frac{\pi}{2}} (1-\cos^2 \theta)\sin \theta .d\theta, \quad u=\cos \theta \Rightarrow du = -\sin \theta d\theta \\ & = \frac{3}{5} M_2a^2 \int_1^{0} (u^2-1) .du \\ & = \frac{3}{5} M_2a^2 [\frac{1}{3}(-1)-(-1)] \\ & = \frac{2}{5} M_2a^2\end{aligned}
    Not well up on 3D coordinate change of variables, but shouldn't the second line be:

    =\displaystyle\frac{3M_2}{2\pi a^3  } \int_{0}^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a  r^2 \cos^2 \theta \cdot r^2 \sin \theta .dr d\theta d\phi \\ &
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    (Original post by ghostwalker)
    Not well up on 3D coordinate change of variables, but shouldn't the second line be:

    =\displaystyle\frac{3M_2}{2\pi a^3  } \int_{0}^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a  r^2 \cos^2 \theta \cdot r^2 \sin \theta .dr d\theta d\phi \\ &
    Don't think so :dontknow: (although using that does indeed yield 1/5 as required)

    The relation I'm given is (x,y,z) = (r \sin \theta \cos \phi, r \sin \theta \sin \phi , r \cos \theta) for r\in [0,\infty), \theta \in [0, \pi] and \phi \in [0,2\pi] , as well as being told that dx dy dz = r^2 \sin \theta .dr d\theta d\phi

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    (Original post by RDKGames)
    Don't think so :dontknow: (although using that does indeed yield 1/5 as required)
    Fair enough. I wasn't sure which version of spherical you were using.

    Nope, can't see the problem. :sad:
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    (Original post by RDKGames)
    ...
    Nope, I'm being stupid.

    The MofI of the hemisphere is 2/5Ma^2, not 1/5.

    I was working from the MofI of a sphere, and halving it, but forgot the mass is halved too, so the formula is the same. Doh!

    :facepalm2:

    :getmecoat:
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    (Original post by ghostwalker)
    Nope, I'm being stupid.

    The MofI of the hemisphere is 2/5Ma^2, not 1/5.

    I was working from the MofI of a sphere, and halfing it, but forgot the mass is halved too, so the formula is the same. Doh!

    :facepalm2:

    :getmecoat:
    :rip:

    So then my lecturer has indeed made a typo there with the reference answer?
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    (Original post by RDKGames)
    :rip:

    So then my lecturer has indeed made a typo there with the reference answer?
    I reckon so - they probably did the same thing I did.

    Apologies for wasting your time.
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    (Original post by ghostwalker)
    I reckon so - they probably did the same thing I did.

    Apologies for wasting your time.
    I'll drop 'em an email then. Thanks anyhow!
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    (Original post by RDKGames)
    I'll drop 'em an email then. Thanks anyhow!
    For confirmation:

    See here.
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    (Original post by RDKGames)
    I'll drop 'em an email then. Thanks anyhow!
    In your email, I should drop the spare t from your work. It's moment of inertia, not moment of intertia.
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    (Original post by tiny hobbit)
    In your email, I should drop the spare t from your work. It's moment of inertia, not moment of intertia.
    Haha I didn't even notice including that!

    Thankfully it was not written like that in the email
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    (Original post by RDKGames)


    The diagram I drew for this is just an ice cream cone shape with vertex on O and z as the axis of symmetry.

    Initially I thought of combining the two in such a way that the inertia is given by:

    \displaystyle I = \frac{M_1+M_2}{\frac{1}{3}\pi a^2 h + \frac{2}{3}\pi a^3} \left( \int_{D_1} x^2 + y^2. dxdydz + \int_{D_2} x^2 + y^2. dxdydz \right)

    where D_1, D_2 are spaces occupied by the cone and hemisphere respectively, but this seems like a nightmare to deal with.

    Having looked at the reference answer of I = \dfrac{3}{10}M_1 a^2 + \dfrac{1}{5}M_2a^2 I thought that you'd just add the inertia's from the two separate objects since inertia of the cone about the same axis is \frac{3}{10}M_1a^2, however the inertia of the hemisphere worked out to be \dfrac{2}{5}M_2a^2 instead.

    Is there a shortcut to this problem? Is addition of inertia's in these types of problems allowed, so the ref-answer answer is incorrect?

    For inertia of the hemisphere I just evaluate \displaystyle I = \frac{3M_2}{2\pi a^3} \int_{D_2} x^2 + y^2 .dxdydz which yields it, but I can't see where I'd be going wrong here.
    misleading post - thought it was about ice cream yet you only mention ice cream once!! poor keyword density pal
 
 
 
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