1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moment of Inertia of a Wheel

  1. Apr 14, 2014 #1
    Twelve uniform, thin rods of mass and length are welded
    together to form a “wheel” as shown in the figure. What is the
    moment of inertia of this wheel for rotation around an axis through
    its center and perpendicular to the plane of the wheel? The welds
    contribute no mass to the wheel.

    I understand the contributions from the spoke....but how would I get the contribution from the six rods around the circumference?

    I think it can be done by noting I=1/3ML^2 for a rod with the pivot center at the end and I=1/12ML^2 when the pivot center at the middle.


    If i try and integrate the from the center along the circumference I have

    I=∫M/L (x^2)dx

    but how would I get my limits of integration? I would have to integrate through an angle zero to a fixed distance from the center? SO ((3)^1/2)/2Lcos(Θ)dΘ integrate from 0 to (360)/6=60 degrees, correct?
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Are you familiar with the parallel axis theorem? If so, can you use it to get I for one of the perimeter rods?
     
  4. Apr 14, 2014 #3
    That's what my first attempt was......but where would a convenient point be to compute the moment of inertia? If I move along one of the spokes to the outside I feel like I am in the same situation as I was before.
     
  5. Apr 14, 2014 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The parallel axis theorem states ##I = I_{cm} + Md^2##. What does ##d## represent here?

    Can you find ##d## for one of the perimeter rods?
     
  6. Apr 14, 2014 #5
    would d be (L)root(3)/2.....(the vertical distance from the center?)
     
  7. Apr 14, 2014 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes!
     
  8. Apr 14, 2014 #7
    I figured it out! I =(1/12l^2 +3/4L^2 )6 + (1/3Ml^2(6)=7mL^2

    Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Moment of Inertia of a Wheel
Loading...