# Compute the moment of inertia

• Mar 4th 2009, 11:24 PM
jackiemoon
Compute the moment of inertia
Hi,

Can anybody help with this question please:

Compute the moment of inertia of the object in the attached figure (a thin disk missing a portion, where the angle is θ is 45) with respect to the centre point, P.

Note: 1/(cosx)^4 .dx = [sinx / 3(cosx)^3] + [2tanx/3]. The integral can be split into two integrals, one over a sector of disk one over a triangle.
• Mar 14th 2009, 01:10 AM
chisigma
The general formula for the moment of inertia is…

$\displaystyle I_{z}= \int_{V} \rho \cdot r^{2} \cdot dv$ (1)

The object you have proposed is a sort of cilinder that i suppose homogeneous and of height $\displaystyle H$. In this case the (1), developped in cilindrical coordinates, becomes…

$\displaystyle I_{z}= H \cdot \rho \cdot (\int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \int r^{3} \cdot d \phi \cdot dr + 2 \cdot \int_{0}^{\frac{\pi}{4}} \int_{0}^{R \cdot \cos \phi} r^{3} \cdot d \phi \cdot dr)$ (2)

The first integral in (2) gives…

$\displaystyle H \cdot \rho \cdot \int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} |\frac {r^{4}}{4}|_{0}^{R} \cdot d\phi= \frac{3\cdot \pi}{8} \cdot R^{4} H \cdot \rho$

For the second integral in (2) you have to keep in mind that…

$\displaystyle \int \cos^{4} \phi \cdot d \phi = \frac{1}{32} \cdot (12 \cdot \phi + 8 \cdot \sin 2 \phi + \sin 4 \phi) + c$

… so that…

$\displaystyle H \cdot \rho \cdot \int_{0}^{\frac{\pi}{4}}|\frac {r^{4}}{4}|_{0}^{R \cdot \cos \phi} \cdot d \phi= \frac{1}{2} \cdot H \cdot \rho \cdot R^{4} \cdot \int_{0}^{\frac{\pi}{4}} \cos^{4} \phi \cdot d\phi= \frac{1}{2} \cdot H \cdot \rho \cdot R^{4} \cdot (\frac{3 \cdot \pi}{32} + \frac{1}{4})$

The requested moment of inertia so is…

$\displaystyle I_{z}= H \cdot \rho \cdot R^{4} \cdot (\frac{27 \cdot \pi}{64} + \frac{1}{8})= H \cdot \rho \cdot R^{4} \cdot 1,4503594 \dots$

In case of a ‘perfect cilinder’ the moment of inertia would be…

$\displaystyle I_{z}= H \cdot \rho \cdot R^{4} \cdot \frac{\pi}{2}= H \cdot \rho \cdot R^{4} \cdot 1,5707963\dots$

… and that is not a surprise…

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 14th 2009, 09:41 PM
chisigma
My previous message did contain errors (Headbang)... i apologize for that and reply the message with the new solution... i hope without errors (Itwasntme)...

The general formula for the moment of inertia is…

$\displaystyle I_{z}= \int_{V} \rho \cdot r^{2} \cdot dv$ (1)

The object you have proposed is a sort of cilinder that i suppose homogeneous and of height $\displaystyle H$. In this case the (1), developped in cilindrical coordinates, becomes…

$\displaystyle I_{z}= H \cdot \rho \cdot (\int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \int r^{3} \cdot d \phi \cdot dr + 2 \cdot \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac {R}{\sqrt {2} \cdot \cos \phi}} r^{3} \cdot d \phi \cdot dr)$ (2)

The first integral in (2) gives…

$\displaystyle H \cdot \rho \cdot \int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} |\frac {r^{4}}{4}|_{0}^{R} \cdot d\phi= \frac{3\cdot \pi}{8} \cdot R^{4} H \cdot \rho$

For the second integral in (2) you have to keep in mind that…

$\displaystyle \int \cos^{-4} \phi \cdot d \phi = \frac{1}{3} \cdot (sec^{2} \phi +2) \cdot tan \phi + c$

… so that…

$\displaystyle H \cdot \rho \cdot \int_{0}^{\frac{\pi}{4}}|\frac {r^{4}}{4}|_{0}^{\frac {R}{\sqrt {2} \cdot \cos \phi}} \cdot d \phi= \frac{1}{4} \cdot H \cdot \rho \cdot R^{4} \cdot \int_{0}^{\frac{\pi}{4}} \cos^{-4} \phi \cdot d\phi= \frac{1}{3} \cdot H \cdot \rho \cdot R^{4}$

The requested moment of inertia so is…

$\displaystyle I_{z}= H \cdot \rho \cdot R^{4} \cdot (\frac{3\cdot \pi}{8} + \frac{1}{3})= H \cdot \rho \cdot R^{4} \cdot 1,5114305 \dots$

In case of a ‘perfect cilinder’ the moment of inertia would be…

$\displaystyle I_{z}= H \cdot \rho \cdot R^{4} \cdot \frac{\pi}{2}= H \cdot \rho \cdot R^{4} \cdot 1,5707963\dots$

… and that is not a surprise…

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$