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`        Calculate the moment of inertia of a pyramid when rotated about its base.`
8 years ago

## Answers : (1)

```							Dear student,

The moment of inertia of a cone about its central axis, start with the standard Intertia equation
I = integral r^2 dm
dm = rho dV (rho is density) (dV is basically volume)
dV = r dr dtheta 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(0 to H) int(0 to 2pi) int(0 to r) r^3 dr dtheta dx
solving the integral leaves
I = rho int(0 to H) int(0-2pi) 1/4 r^4 dtheta dx
solving the second integral leaves
I = rho int(0 to H) 1/2 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 = rho int(0 to H) 1/2 pi [(R/H)x]^4 dx
time to do some housekeeping and take all the constants outside the integral
I = (rho pi R^4) / (2 H^4) int(0 to 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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by                                            answering the questions on         Discussion        Forum.    So      help         discuss     any               query   on        askiitians  forum   and      become an      Elite            Expert    League            askiitian.

Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
8 years ago
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