- #1

Yalanhar

- 39

- 2

- Homework Statement
- Calculate the rotational inertia of a solid hexagonal (side = R) about an axis perpendicular through its center.

- Relevant Equations
- ##I=\int r^2 dm##

What I did:

##\frac{dm}{dA} = \frac{M}{\frac{3\sqrt3 R^2}{2}}##

##dm = \frac{2M}{3\sqrt3 R^2} dA## (1)

##dA=3\sqrt3 rdr## (2)

(2) in (1)

##dm = \frac{2M}{3\sqrt3 R^2} 3\sqrt3 rdr##

Now in the integral

##I = \int \frac{r^2 2Mrdr}{R^2}##

How can I solve the integral interval? I think I need to change respect to ##\theta## and integrate over 2##\pi##. Or can I change r in the integral and integrate considering dA? Then the interval would be 0 to A (## A = \frac{3\sqrt3 R^2}{2}##.

I don't quite understand how i choose the interval. In a rod to it's center is -l/2 to +l/2, but in a ring it goes from 0 to R. Why not -R to +R?

Thanks

##\frac{dm}{dA} = \frac{M}{\frac{3\sqrt3 R^2}{2}}##

##dm = \frac{2M}{3\sqrt3 R^2} dA## (1)

##dA=3\sqrt3 rdr## (2)

(2) in (1)

##dm = \frac{2M}{3\sqrt3 R^2} 3\sqrt3 rdr##

Now in the integral

##I = \int \frac{r^2 2Mrdr}{R^2}##

How can I solve the integral interval? I think I need to change respect to ##\theta## and integrate over 2##\pi##. Or can I change r in the integral and integrate considering dA? Then the interval would be 0 to A (## A = \frac{3\sqrt3 R^2}{2}##.

I don't quite understand how i choose the interval. In a rod to it's center is -l/2 to +l/2, but in a ring it goes from 0 to R. Why not -R to +R?

Thanks