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I want to find the moment of inertia about the axis perpendicular to the base and passing through the center of mass (z-axis?). Mass is M and height is H and base has sides $a$.

I know I can take a slab of the pyramid and take the volume of the slab dv = $b^2$ x dH where $b$ is the length of the slab and dH is the height and then get the mass dm = $\rho$ x dv = $\rho$ x $b^2$ x dh. I'm stuck at this point and would appreciate any help.

Sujjan Shaikh
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let $\rho$ be the mass per unit volume, so that $M=\frac 13\rho a^2H$

Set the origin at the vertex and let the axis through the vertex and the centre of the base be the $x$ axis.

Consider an element in the form of a thin square lamina of thickness $\delta x$ which is parallel to the base and at a distance $x$ from O. Let the side of this square be $2y$.

Then $$\frac yx=\frac {a}{2H}.$$

The volume of the element is $\delta V=4y^2\delta x$

The mass of the element is $\delta m=4\rho y^2 \delta x$

Now using the standard result for the moment of inertia of a square lamina about an axis perpendicular to the plane of the lamina and through the centre, we have the MI of the element as $$\frac 16\delta m (2y)^2=\frac 83\rho y^4\delta x$$

So the MI is given by $$\frac 83\frac {3M}{a^2H}\int_0^H\frac{a^4x^4}{16H^4}dx=\frac{1}{10}Ma^2$$

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David Quinn
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The basic integral you need is $\displaystyle I=\int_Q r^2 \, dm$. The problem with your approach so far is that the slab is not equidistant from the axis of rotation, so that $r^2$ varies over your slab. You need to take a chunk where you can approximate the $r^2$ as constant over your chunk. That means a chunk $dV=dx\,dy\,dz$. So here, $dm=\rho\,dx\,dy\,dz$, where you can get $\rho=M/V$, assuming constant density. Then you need to set up your integral as $$I=\rho\iiint_Qr^2\,dx\,dy\,dz.$$ If you draw some careful pictures, you can find out how $r^2$ varies with respect to $x,y,$ and $z$, and you can get your limits for each of the variables. Can you proceed from here?

Adrian Keister
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  • Is this the center of mass? I'm not sure how I would get the limit of x,y and z – Sujjan Shaikh Apr 04 '18 at 16:39
  • No, I am not describing the center of mass. That's a different integral entirely. You mentioned the moment of inertia about an axis, and that is what I am talking about as well. To get the limits, you need to draw a careful picture, find the relationships among $x, y,$ and $z$. That will give you both $r^2$ in terms of $z$, as well as the limits on the integrals. – Adrian Keister Apr 04 '18 at 18:30
  • Ah so if I take an arbitrary point from the z-axis then the distance between that point and the z-axis is [(x,y,z) - (0,0,z)] so r^2 = x^2 + y^2. Im unsure how to get the limits from there. – Sujjan Shaikh Apr 04 '18 at 21:31