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Thread: Moment of Inertia of a Solid Cone

  1. #1
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    Moment of Inertia of a Solid Cone

    How to find the moment of inertia about the z axis of the top half of a solid cone using Spherical Coordinates?

    This what I did!

    I = ∫ R^2 dm

    where I is the moment of inertia and R is the perpendicular distance from the axis of rotation to the slant height of the cone

    changing dm with density, ρ, we get

    I = ρ ∫∫∫ R^2 r^2 sin Φ dr dΦ dθ = ρ ∫∫∫ r^4 (sin Φ)^3 dr dΦ dθ

    for dθ, I know that θ will go from 0 to 2pi

    If my triple integral setting is correct, how to set the limits of integration for dr and dΦ?
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    Re: Moment of Inertia of a Solid Cone

    It's hard to describe without drawing pictures, but here goes:

    Let $\displaystyle h$ be the height of the cone and $\displaystyle r$ the radius of the base. Position the cone so that the apex is at the center of the coordinate system and the center of the base is at $\displaystyle \phi=0$.

    As you said, it is clear that $\displaystyle \theta$ goes from $\displaystyle 0$ to $\displaystyle 2\pi$.

    The variable $\displaystyle \phi$ goes from the central axis $\displaystyle \phi=0$ to the lateral surface of the cone $\displaystyle \phi=\phi_\text{max}$. Setting up a triangle gives $\displaystyle \tan(\phi_\text{max})=\frac{r}{h}$, so the limits are $\displaystyle 0$ and $\displaystyle \tan^{-1}\frac{r}{h}$.

    The variable $\displaystyle \rho$ goes from the apex $\displaystyle \rho=0$ to $\displaystyle \rho=\rho_\text{max}$ at the base. Setting up a different triangle gives $\displaystyle \cos{\phi}=\frac{h}{\rho_\text{max}}$. So the limits are $\displaystyle 0$ and $\displaystyle \frac{h}{\cos{\phi}}$.

    To get the moment of inertia, we integrate the square of the distance from the axis to the volume element. Yet another triangle gives this distance as $\displaystyle \rho\sin{\phi}$.

    So the integral is $\displaystyle \int_0^{2\pi} \int_0^{\tan^{-1}\frac{r}{h}} \int_0^{\frac{h}{\cos{\phi}}} (\rho\sin{\phi})^2 \rho^2\sin{\phi}\,d\rho\,d\phi\,d\theta$.

    Integrating this gives the correct value $\displaystyle \frac{1}{10}\pi r^4h$.

    Your integral is correct. In your analysis, R should be the perpendicular distance from the mass element to the axis of rotation (instead of from the axis of rotation to the slant height).

    - Hollywood
    Thanks from joshuaa, HallsofIvy and topsquark
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    Re: Moment of Inertia of a Solid Cone

    thanks a lot Hollywood. you have written a very clear explanation. i Like it.
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    Re: Moment of Inertia of a Solid Cone

    Quote Originally Posted by hollywood View Post
    It's hard to describe without drawing pictures, but here goes:

    Let $\displaystyle h$ be the height of the cone and $\displaystyle r$ the radius of the base. Position the cone so that the apex is at the center of the coordinate system and the center of the base is at $\displaystyle \phi=0$.

    As you said, it is clear that $\displaystyle \theta$ goes from $\displaystyle 0$ to $\displaystyle 2\pi$.

    The variable $\displaystyle \phi$ goes from the central axis $\displaystyle \phi=0$ to the lateral surface of the cone $\displaystyle \phi=\phi_\text{max}$. Setting up a triangle gives $\displaystyle \tan(\phi_\text{max})=\frac{r}{h}$, so the limits are $\displaystyle 0$ and $\displaystyle \tan^{-1}\frac{r}{h}$.

    The variable $\displaystyle \rho$ goes from the apex $\displaystyle \rho=0$ to $\displaystyle \rho=\rho_\text{max}$ at the base. Setting up a different triangle gives $\displaystyle \cos{\phi}=\frac{h}{\rho_\text{max}}$. So the limits are $\displaystyle 0$ and $\displaystyle \frac{h}{\cos{\phi}}$.

    To get the moment of inertia, we integrate the square of the distance from the axis to the volume element. Yet another triangle gives this distance as $\displaystyle \rho\sin{\phi}$.

    So the integral is $\displaystyle \int_0^{2\pi} \int_0^{\tan^{-1}\frac{r}{h}} \int_0^{\frac{h}{\cos{\phi}}} (\rho\sin{\phi})^2 \rho^2\sin{\phi}\,d\rho\,d\phi\,d\theta$.

    Integrating this gives the correct value $\displaystyle \frac{1}{10}\pi r^4h$.

    Your integral is correct. In your analysis, R should be the perpendicular distance from the mass element to the axis of rotation (instead of from the axis of rotation to the slant height).

    - Hollywood
    Hello,
    But z- moment of inertia of a solid cone is $\frac {3}{10}*m*r^2$ where m is mass and r is a radius.Would you explain what is $\rho,\phi and \theta ?$
    Last edited by Vinod; Jul 29th 2018 at 02:51 AM.
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    Re: Moment of Inertia of a Solid Cone

    Quote Originally Posted by Vinod View Post
    Hello,
    But z- moment of inertia of a solid cone is $\frac {3}{10}*m*r^2$ where m is mass and r is a radius.Would you explain what is $\rho,\phi and \theta ?$
    With density set to 1, the mass is the same number as the volume $\displaystyle \frac{1}{3}\pi r^2h$. So $\displaystyle \frac {3}{10}mr^2=\frac {3}{10}\left(\frac{1}{3}\pi r^2h\right)r^2=\frac{1}{10}\pi r^4h$.

    The variables $\displaystyle \rho$,$\displaystyle \phi$, and $\displaystyle \theta$ are spherical coordinates: $\displaystyle \rho$ the distance from the center, $\displaystyle \phi$ the angle from the z-axis, and $\displaystyle \theta$ the angle from the x-axis when projected onto the xy-plane.

    - Hollywood
    Thanks from Vinod
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