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Moment of inertia problem

  1. Jan 11, 2010 #1
    Hey everyone,

    Im solving a group of questions based on a system of particles, A(2,0) with a mass of 4kg, B(0,1) with a mass of 6kg and C(1,1 ) with a mass of 3kg. I have to find the centre of mass and also calculate the moment of inertia.

    To find the centre of mass im attempting the following:

    " 1/13 [(4.2) + (6.0) + (3.1)]= 3/13 "

    " 1/13 [(4.0) + (6.1) + (3.1)]= 1 1/65 "

    so centre or mass wud b? .... (3/13, 1 1/65 )

    Im not sure if this is correct?

    Also im unsure of how to calculate the moment of inertia?

    I really would appreciate any help,

    Regards,

    Richard
     
  2. jcsd
  3. Jan 11, 2010 #2

    Doc Al

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    Staff: Mentor

    I assume you mean:
    1/13 [(4*2) + (6*0) + (3*1)]= ??

    That would be correct for the x-component, but your answer is not. Redo your arithmetic.

    Similar comments for the y-component.

    The moment of inertia of a particle about some axis is mR², where R is the distance to the axis. Where's your axis?
     
  4. Jan 11, 2010 #3
    oops! few mistakes there, yeah the correct answer would be (11/13, 9/13) ?? My axis is the centre of mass. how would I apply this to the system in this equation?

    thank you so much for your help btw!

    - Richard
     
    Last edited: Jan 11, 2010
  5. Jan 11, 2010 #4

    Doc Al

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    Staff: Mentor

    Good.
    Start by finding each particle's distance (or distance squared) from the center of mass. Then calculate the moment of inertia of each particle. Add them to get the total for the system.
     
  6. Jan 11, 2010 #5
    thanks again for your quick reply! really helpful!! :D

    - for the length of each particle to the centre of mass im using:

    d = *squareroot* [ (a-x)^2 + (b - y)^2 ]

    where (x , y) = centre of mass position and (a, b) = particles position.

    Following this and using mR^2, im getting some odd numbers. for the first particle the results
    are: 28 164/169??

    thank you again for your help!!
     
  7. Jan 11, 2010 #6

    Doc Al

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    Staff: Mentor

    That should work fine.

    No, that's not right. Show how you got that result, step by step.
     
  8. Jan 11, 2010 #7
    well for the length of the first particle 'A', i have: sqroot**( (2-11/13)^2 + (0-9/13)^2)**

    which results in 3 sqroot*34*/13, is this correct so far? with mR^2, will 'm' represent the mass of each particle? if so, i get 7 41/169, different arithmetic :s thanks again!
     
  9. Jan 11, 2010 #8

    ideasrule

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    Homework Helper

    Yes, that's right. Do you really have to work in fractions? This will be much easier if you just use decimals.
     
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