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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 \(D\)
 one year ago
 one year 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 \(D\)
 one year ago
 one year ago

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richywBest ResponseYou've already chosen the best response.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 \(D\)
 one year ago

richywBest ResponseYou've already chosen the best response.1
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)
 one year ago

richywBest ResponseYou've already chosen the best response.1
I just don't see how I can get the total mass. I don't think it should matter really
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
\[I_{0}=\int\limits_{}^{}\int\limits_{}^{}pr^2dA\]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)\[I_{0}=p \int\limits_{0}^{2\pi}\int\limits_{0}^{a}r^2rdr d \theta \]\[I_{0}=\frac{2\pi pa^3}{3}\]
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
Now, 2pi a^2 is the area of the disc. Area*density=m (total mass of the disc). We have:\[I_{0}=\frac{am}{3}\]
 one year ago

richywBest ResponseYou've already chosen the best response.1
oh wow! thanks a lot. I got to the first step. feel real stupid now for not seeing the second. thanks again!
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
oooops. pi a^2 is the area lol not 2pi a^2
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
\[I_{0}=\frac{2am}{3}\]
 one year ago

richywBest ResponseYou've already chosen the best response.1
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.\[I_0=\iint_R(x^2+y^2)\rho(x,y)dA\]\[=\rho \int^{2\pi}_0\int^a_0(r^2\cos^2\theta+r^2\sin^2\theta)\,r\,dr\,d\theta\]\[=\rho \int^{2\pi}_0\int^a_0r^2 (cos^2\theta+\sin^2\theta)\,r\,dr\,d\theta\]\[=\rho \int^{2\pi}_0\int^a_0r^3\,dr\,d\theta\]\[=\frac{\pi \rho a^4 }{2}\]
 one year ago

richywBest ResponseYou've already chosen the best response.1
and then I can now see that works out to \[I_0=\frac{ma^2}{2}\]
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
you're correct. I forgot the extra factor of r when I did the integration. Not my day...
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
I should have integrated r^3 to (1/4)4^4...I carried it through as (1/3)r^3 which is wrong.
 one year ago

richywBest ResponseYou've already chosen the best response.1
alright well this formula looks familier from physics so I think i'll go with it!
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
nice work catching another one of my patented errors :)
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
yep, yours is correct. we ended up with the same integral...I just made a computational error.
 one year ago
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