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Disk and blockutter frustration

  1. Nov 24, 2003 #1
    I've included a picture.

    A uniform disk of mass Mdisk = 5 kg and radius R = 0.2 m has a small block of mass mblock = 2.5 kg on its rim. It rotates about an axis a distance d = 0.17 m from its center intersecting the disk along the radius on which the block is situated.
    ==========================================

    a) What is the moment of inertia of the block about the rotation axis?
    b) What is the moment of inertia of the disk about the rotation axis?
    c) When the system is rotating about the axis with an angular velocity of 4.7 rad/s, what is its energy?

    ------------------------------------
    a) this part wasn't so bad. I=M*R^2 => (2.5kg)(.2m-.17m)^2=.0025kg*m^2

    b) This is where I get stuck. The hint says to use the parallel-axis theorem, but I don't even understand it. Can someone explain it and show me how to use it for this problem? Greatly appreciate it.

    c) Well, total energy is equal to the rotational kinetic energy of the block plus the rotational kinetic energy of the disk. So Total energy would be .5I*w^2 (disk)+.5*I*w^2 (block)
    where w =omega.
    I can figure out the KE for the blcok easily, but I don't know how to calculate the moment of inertia for the disk. Once I get that, I should have the answer down. :P
     

    Attached Files:

  2. jcsd
  3. Nov 24, 2003 #2

    NateTG

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    Science Advisor
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    Check your book for a table of moments of inertia of various shapes. You might find one.

    You could also do an integral if you're sophisticated enough.
     
  4. Nov 24, 2003 #3
    As you probably know, the parallel axis theorem states the following:

    [tex]
    I = I_{cm} + MD^2[/tex]

    Icm is the moment of inertia when the rotation axis goes through the objects center of mass... you already know that.

    M is the mass of the disk, you know that too.

    D is the distance of the axis of rotation from the center of mass (the middle). Do you know D?

    I is what you're looking for. The moment of inertia from your new rotation axis.
     
  5. Nov 24, 2003 #4
    Thanks guys. I figured it out in like 10 mins with your help. :smile:
     
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