- #1

bowma166

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## Homework Statement

A cone of mass

*M*has a height

*h*and a base diameter

*R*. Find its moment of inertia about its axis of symmetry

## Homework Equations

[tex]I=\int R^{2}dm[/tex]

## The Attempt at a Solution

I tried starting by finding

*dm*. It's equal to the mass per unit volume, [tex]\frac{M}{\frac{1}{3}\pi R^{2}h}[/tex], times a small

*dV*. I broke up the cone into small disks of area [itex]\frac{1}{2}\pi x^{2}[/itex] (x being the radius of the disk) and height

*dy*. I then found x in terms of y through similar triangles: x/y = R/h, so x = (Ry/h). The equation for dV then becomes [tex]dV=\pi\left(\frac{yR}{h}\right)^{2}dy[/tex].

*dm*then becomes (after simplifying) [tex]dm=\frac{3My^{2}}{h^{3}}dy[/tex], so the whole integral for I is [tex]\int_{0}^{h}R^{2}\frac{3My^{2}}{h^{3}}dy=\frac{3R^{2}M}{h^{3}}\int_{0}^{h}y^{2}dy[/tex], which evaluates to just MR

^{2}. And I definitely don't think that's right, because it seems like a cone would have a smaller moment of inertia than a hoop.

So because that method didn't work for some reason, I tried polar coordinates, breaking up the cone into a series of right triangles rotated around the center. The area of the triangle is just .5*R*h, so [tex]dm=\left(\frac{3M}{\pi R^{2}h}\right)\left(\frac{1}{2}Rhd\theta\right)[/tex] and the integral becomes [tex]\int_{0}^{2\pi}R^{2}\frac{3M}{2\pi R}d\theta=\frac{3RM}{2\pi}\int_{0}^{2\pi}d\theta=3RM[/tex] which doesn't seem right at all to me.

I'm sure I screwed up the polar coordinates somehow (I've never actually learned them... I've just kind of seen them here and there and heard a little about how they work) but it seems like my first attempt should have worked... Can someone tell me what I did wrong? Thanks!

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