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Moment of inertia about a ceiling fan?

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data

    A ceiling fan consists of a small cylindrical disk with 5 thin rods coming from the center. The disk has mass md = 3.3 kg and radius R = 0.26 m. The rods each have mass mr = 1.2 kg and length L = 0.72 m.

    What is the moment of inertia of each rod about the axis of rotation?
    2. Relevant equations

    I = mr^2
    3. The attempt at a solution
    I tried the above equation, using m = 1.2 kg and L = 0.72, but for some reason I keep getting it wrong. Please help!
     
  2. jcsd
  3. Oct 8, 2014 #2

    462chevelle

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    are you not supposed to be using radius? as in .26? or the length of the rod+length of the disc? <-havent studied this yet so take it for what its worth.
     
  4. Oct 8, 2014 #3

    berkeman

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    Can you show more of your work? There are 5 rods. What is the moment of inertia of a rod with respect to its end?
     
  5. Oct 8, 2014 #4
    Also, what is the moment of inertia of the entire fan? is it just the individual moment of inertias added together? thanks
     
  6. Oct 8, 2014 #5
    I finally got this problem by doing 1.2(0.72^2)/3 using yahoo answers. I just don't know why i had to divide by 3! this whole concept of moment of inertia kind of confuses me =\. Thanks.
     
  7. Oct 8, 2014 #6

    berkeman

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    If you answer my post #3, I think you will see how the equations give you the answer.

    Do you always do your homework by asking Yahoo answers? Do you think that will make you smarter, a better student, and a great technical employee?
     
  8. Oct 8, 2014 #7
    Well, so I know that I = sum m x r^2. I have that equation, I just don't know why I can't use it to get the correct answers.

    Berkeman, I really don't appreciate the tone of your response. I would love to understand physics through the concepts and the math, but I need the answer for my online homework. I don't have tutoring resources available right now so I'm trying my best. Instead of questioning my intentions and my work ethic (which you are making very grand judgements about after reading only two sentences i've written), can you please just help me with this problem? thanks.
     
  9. Oct 8, 2014 #8

    berkeman

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    I did not mean anything bad by my response. I apologize if you were offended. I was just trying to focus you on the need to look up the equation for the moment of inertia of a rod with respect to its end. That seems to me to be a key in understanding how to solve this (and many other) problems. The moment of inertia for a rod with respect to its end is definitely not mr^2. Please do a wikipedia search or check your textbook, and see if that helps you to solve the problem on your own without resorting to Internet answers.

    In my own technical learning (college, real-life engineering), it is so important to learn how to learn. Because when it's time to take the tough final exam, or pull an all-nighter to figure out a technical problem that is holding up shipments of millions of units for revenue, you are often on your own. And you need to have already learned how to learn, and how to think your way through the problem. Helping students to learn how to learn is one of the key foundations of the PF, IMO.
     
  10. Oct 8, 2014 #9

    SteamKing

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    You have to be careful just blindly using formulas. I = mr2 for a rod, but does this formula calculate the moment of inertia of the rod about its centroid or some other location? This is where the parallel axis theorem comes into play.
     
  11. Oct 8, 2014 #10
    What is the parallel axis theorem?
     
  12. Oct 8, 2014 #11

    SteamKing

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    It's used to calculate the MOI of a body about a different reference location from the one at which the MOI was originally determined.

    Did you look up in your textbook to see the reference location about which I = mr2 is calculated?
     
  13. Oct 8, 2014 #12
    upload_2014-10-8_23-20-3.png
    So i read up a bit about the parallel axis theorem and I seem to have found this figure that seems to explain why the moment of inertia for the cylinder alone was 1/2mr^2. Is it just that this moment of inertia concept is an integral and so to calculate discrete sums it would be very difficult without calculus so we have handy equations for certain shapes? From what equation would I be integrating to calculate the moment of inertia? I don't get the interplay of calculus and math, I guess. How can I find the moment of inertia from the rotation of axis? why does that matter?
     
  14. Oct 8, 2014 #13
    Also the equation I = Sum mr^2 can only be used if "a rigid body consists of a few particles", according to my book. What does that mean? Would the body be on 2D? Would it be a rectangle? Thanks!
     
  15. Oct 8, 2014 #14

    SteamKing

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    I don't know how to break this news to you, but ... calculus is math. If you are having trouble understanding calculus now, your troubles are only going to get worse until you remedy this gap in your knowledge, especially if you take more physics courses.

    You should check out the formulas in this table of MOI, especially those for the thin rods:

    http://en.wikipedia.org/wiki/List_of_moments_of_inertia
     
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