Question

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 \(D\)

Answer 1

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 \(D\)

Answer 2

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)

Answer 3

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

Answer 4

\[I_{0}=\int\limits_{}^{}\int\limits_{}^{}pr^2dA\]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)\[I_{0}=p \int\limits_{0}^{2\pi}\int\limits_{0}^{a}r^2rdr d \theta \]\[I_{0}=\frac{2\pi pa^3}{3}\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yeah

Answer 9

\[I_{0}=\frac{2am}{3}\]

Answer 10

:)

Answer 11

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.\[I_0=\iint_R(x^2+y^2)\rho(x,y)dA\]\[=\rho \int^{2\pi}_0\int^a_0(r^2\cos^2\theta+r^2\sin^2\theta)\,r\,dr\,d\theta\]\[=\rho \int^{2\pi}_0\int^a_0r^2 (cos^2\theta+\sin^2\theta)\,r\,dr\,d\theta\]\[=\rho \int^{2\pi}_0\int^a_0r^3\,dr\,d\theta\]\[=\frac{\pi \rho a^4 }{2}\]

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

nice work catching another one of my patented errors :)

Answer 17

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