1. Not finding help here? Sign up for a free 30min 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 hoop on an arm

  1. Jan 30, 2008 #1

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    Find the moment of Inertial of a hoop at the end of a arm. The axis is at the end of the arm and normal to the plane of the hoop.

    Let:
    M= mass of the hoop
    r= radius of the hoop
    m=mass of the arm
    d=lengh of the arm

    2. Relevant equations

    [tex] I_{hoop} = Mr^2 [/tex]
    [tex] I_{arm} = \frac {md^2} 3 [/tex]

    [tex]I_{total} = I_{hoop} + I_{arm} [/tex]



    3. The attempt at a solution
    Using the parallel axis theorm the hoop:

    [tex] I_{hoop} = Mr^2 +Md^2[/tex]

    so for the system:
    [tex]I_{total} = I_{hoop} + I_{arm} [/tex]

    Question, is this correct?
    Here's the deal, I have created a model of this system, but it is not responding as I think it should, the moment of Inertia is something I need to verify.

    No this is not homework for a class, but it is typical of a homework problem so this is where is belongs.
     
  2. jcsd
  3. Jan 30, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Does the axis of rotation pass through the centre of the hoop or is it at the opposite end of the arm?

    If the axis of rotation passes through the centre of the hoop, then your calculation is correct. One possibility for the misbehavior of the model is that both your momenta of inertia (rod and hoop) assume that their thickness is negligible. If the thicknesses of your hoop/rod is not negligible then you should use the following momenta of inertia;

    [tex]I_{hoop} = \frac{1}{2}M(r_0^2+r_1^2)[/tex]

    Where r0 and r1 are the inner and outer radii respectively. And,

    [tex]I_{cylinder} = \frac{1}{4}MR^2 + \frac{1}{3}Md^2[/tex]

    But I'm sure that you have considered this already.
     
    Last edited: Jan 30, 2008
  4. Jan 30, 2008 #3

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks Hoot,
    Actually the axis of rotation is through the opposite end. Imagine a flyswatter, edge on. (Won't get many flys this way)

    I have neglected the thickness, unfortunately that serves to increase the moment of inertia, which makes my problem worse.
    There is a torque applied near the pivot point, the model predicts angular velocity well below what my gut says.

    I think for the final Ih I missed the radius of the hoop in the application of the parallel axis theorem. It should read:

    [tex] Mr^2 + M ( r+d)^2 [/tex]

    The physical system has a pair of these paddels rotating on the same axis, I am assuming that I just double I.
    Thanks for any input.
     
  5. Jan 31, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ahh, I have a picture now :smile:
    Yes that looks correct to me. Unfortunatly, it looks as if this new addition will only serve to further compound your 'problem'.
    Again, yes, this is totally valid. I'm just sorry that I can't help you further, is this something that your building or is it purely a thought experiement?
     
  6. Jan 31, 2008 #5

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    this is an attempt to model a real system. I'll keep you up with what I learn.
     
  7. Feb 1, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks Integral, if I think of anything I'll let you know.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?