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Moments of Inertia Trouble

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Attached in a picture!

    2. Relevant equations
    I = mr^2
    m = (x*y)
    cm = (x/2, y/2)

    3. The attempt at a solution
    I was able to solve Part 1 fairly fast, but Part II has drawn a blank for me overall. I was thinking of that rotational thing from Calculus where you spin a 2D figure around the axis to make a 3D shape, but I'm really just not sure. To be honest, I'm having a very difficult time even understanding the concept of moment of inertia >.< So even if I somehow guessed how to do part II right, I really don't know what I'm doing in that section.

    I) mass 1 = (20 cm)(10 cm) = 200 cm^2
    mass 2 = (10 cm)(20 cm) = 200 cm^2
    center of mass 1 = (20 cm i, 5 cm j)
    center of mass 2 = (5 cm i, 10 cm j)

    Center of Mass = (summation of (mass)(coordinates) / summation of masses) = [density of object*(200*(20 i + 5 j))+(200*(5 i + 10 j)] / density of object*(200 + 200)
    = (4000 i + 1000 j + 1000 i + 2000 j) / 400
    = 12.5 cm i + 7.5 cm j


    II) total mass of object = 4 kg
    I = summation of mass*r^2
    = (1 kg*100 cm^2)+(3 kg*300 cm^2)
    =(100 kgcm^2)+(900 kgcm^2) = 1,000 kgcm^2

    Thank you for any help and have a nice day! :-)
     

    Attached Files:

  2. jcsd
  3. Nov 22, 2015 #2

    Doc Al

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    Staff: Mentor

    This is the moment of inertia of a point mass, not an extended object.

    Hint: Can you find the moment of inertia of a rod rotating about one end?

    Read this: Moment of Inertia
     
  4. Nov 22, 2015 #3
    Oh okay, so I should be using 1/3*mr^2?
     
  5. Nov 22, 2015 #4

    Doc Al

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    Sure, if you use it correctly. Do you understand that the MOI of a plate about one edge is the same as that of a rod about one end?
     
  6. Nov 22, 2015 #5
    I didn't know that, but I would suppose so since the area-of-force on the object is going to be farther from the center of mass for both a plate and a rod. It'd be higher than if the point of force was closer to the center of mass.
     
  7. Nov 22, 2015 #6

    Doc Al

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    Staff: Mentor

    Moving mass parallel to the axis of rotation will not affect the moment of inertia about that axis. So you can squeeze the mass of the plate into a rod and get the same MOI.
     
  8. Nov 22, 2015 #7
    Oh okay. I think I confused center of mass and axis of rotation there.

    So for this problem, would I be splitting the equation into two rods spinning about the y axis, and then add it all up, since the top square would be parallel to the bottom rectangle?
     
  9. Nov 22, 2015 #8

    Doc Al

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    Yes. That's just what I would do.
     
  10. Nov 22, 2015 #9
    ah so would this be right?

    II) total mass of object = 4 kg
    I = summation of (1/3) mass*r^2
    = (1/3)(1 kg*100 cm^2) + (1/3) (3 kg*900 cm^2)
    =(100/3 kgcm^2)+(900 kgcm^2) = 933.333333 kgcm^2 = 933 kgcm^2
     
  11. Nov 22, 2015 #10

    Doc Al

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    Looks good. But I would first convert the distances to standard units (meters instead of cm).
     
  12. Nov 22, 2015 #11
    ah okay, 0.093 kgm^2

    and rotational inertia is different for different objects, right? Should I just memorize those formulas?

    thank you so much!
     
  13. Nov 22, 2015 #12

    SteamKing

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    Staff Emeritus
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    Homework Helper

    There are formulas for the MOI for different objects. These formulas can be derived using calculus. Most people keep a table of these formulas handy:

    https://en.wikipedia.org/wiki/List_of_moments_of_inertia
     
  14. Nov 22, 2015 #13

    Doc Al

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    Staff: Mentor

    Sure.

    It's a good idea to have a few basic ones memorized... just in case. If you have a bit of calculus under your belt, be sure to understand how they are derived.
     
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