## richyw 3 years ago Compute the polar moment of inertia $$I_0$$ for a disc $$D$$ centered at the origin with radius $$a$$ and constant density $$\rho (x,y)=\rho$$. Write (\I_0\) in terms of the radius $$a$$ and the total mass $$m$$ of $$D$$

1. richyw

Compute the polar moment of inertia $$I_0$$ for a disc $$D$$ centered at the origin with radius $$a$$ and constant density $$\rho (x,y)=\rho$$. Write $$I_0$$ in terms of the radius $$a$$ and the total mass $$m$$ of $$D$$

2. richyw

I'm not sure at all how to do this. I have been working on it for like 6 hours and feel I am getting nowhere. No questions like this in my textbook. I could do all those ones fine (in about 1/4 that time haha)

3. richyw

I just don't see how I can get the total mass. I don't think it should matter really

4. eseidl

$I_{0}=\int\limits_{}^{}\int\limits_{}^{}pr^2dA$Using polar coordinates going from r=0 to r=a and theta going from 0 to 2pi (and using the fact that density, p=constant)$I_{0}=p \int\limits_{0}^{2\pi}\int\limits_{0}^{a}r^2rdr d \theta$$I_{0}=\frac{2\pi pa^3}{3}$

5. eseidl

Now, 2pi a^2 is the area of the disc. Area*density=m (total mass of the disc). We have:$I_{0}=\frac{am}{3}$

6. richyw

oh wow! thanks a lot. I got to the first step. feel real stupid now for not seeing the second. thanks again!

7. eseidl

oooops. pi a^2 is the area lol not 2pi a^2

8. richyw

yeah

9. eseidl

$I_{0}=\frac{2am}{3}$

10. eseidl

:)

11. richyw

oh wait I'm opening this again. I just noticed when I did the first part. I got a different answer. This is what I did.$I_0=\iint_R(x^2+y^2)\rho(x,y)dA$$=\rho \int^{2\pi}_0\int^a_0(r^2\cos^2\theta+r^2\sin^2\theta)\,r\,dr\,d\theta$$=\rho \int^{2\pi}_0\int^a_0r^2 (cos^2\theta+\sin^2\theta)\,r\,dr\,d\theta$$=\rho \int^{2\pi}_0\int^a_0r^3\,dr\,d\theta$$=\frac{\pi \rho a^4 }{2}$

12. richyw

and then I can now see that works out to $I_0=\frac{ma^2}{2}$

13. eseidl

you're correct. I forgot the extra factor of r when I did the integration. Not my day...

14. eseidl

I should have integrated r^3 to (1/4)4^4...I carried it through as (1/3)r^3 which is wrong.

15. richyw

alright well this formula looks familier from physics so I think i'll go with it!

16. eseidl

nice work catching another one of my patented errors :)

17. eseidl

yep, yours is correct. we ended up with the same integral...I just made a computational error.