Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

richyw Group Title

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

  • This Question is Closed
  1. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  2. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  3. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  4. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  5. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  6. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  7. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  8. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah

    • one year ago
  9. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  10. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    :)

    • one year ago
  11. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  12. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  13. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  14. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  15. richyw Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  16. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    nice work catching another one of my patented errors :)

    • one year ago
  17. eseidl Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.