Find the moment of inertia of a solid cone (mass M, base radius R) around its symmetry axis.

Question

anonymous · Answer

$$I=(3/10)MR^2$$

anonymous · Answer

can you explain to me how you get there?

anonymous · Answer

yeah let me type it out, one sec...

anonymous · Answer

The moment of inertia of a cone about its central axis, start with the standard Inertia equation:

$$I = \int\limits r^2 dm$$

$$dm = \rho dV$$ (rho is density) (dV is basically volume)

$$dV = r dr d \theta dx$$

not going to prove that here but you will see in the integral that this does indeed form the volume. integral will be refered to as int from here on.

This now forms the triple integral

$$I = \rho \int\limits_{0}^{H}\int\limits_{0}^{2\pi}\int\limits_{0}^{r} r^3 dr d \theta dx$$

solving the integral leaves

$$I = (1/4)\rho \int\limits\limits_{0}^{H}\int\limits\limits_{0}^{2\pi} r^4 d \theta dx$$

solving the second integral leaves

$$I = (1/2)\rho \int\limits_{0}^{H} \pi r^4 dx$$

ok so now you have to sub in the equation for r (the radius) of the cone

$$r = (R/H)x$$

this is the radius at the base divided by the height of the cone multiplied by the distance along the x axis. this equation gives you r at any point this gives you

$$I = (1/2)\rho \int\limits_{0}^{H} \pi [(R/H)x]^4 dx$$

this can now be solved and simplified to

$$I = (\rho \pi R^4 H) / 10$$

At this stage your solution is complete, however you can tidy up the equation by taking out the mass term.

$$m = (\rho \pi H R^2) / 3$$

split the Inertia term up to serperate out the mass term

$$I = [(\rho \pi H R^2) / 3]*[ (3R^2)/10 ]$$

this is now the complete solution in terms of mass

$$I = (3mR^2)/10=(3/10)mR^2$$

anonymous · Answer

hope that explained it

anonymous · Answer

thanks so much!