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Problem in understanding model for rigid body

  1. Apr 8, 2012 #1
    A rigid body is usually modeled as a large collection of particles in a fixed arrangement. Being particles, they by definition have no size so they cannot undergo rotation about their own axes as that does not make sense. All kinetic equations for rigid bodies are derived from this model, and when a rigid body rotates about its own center of mass this rotation is accounted for by the revolving of the particles around the center of mass of the rigid body.

    Here is my problem in understanding:
    In reality the items (I use "item" to avoid confusion with the modeled particle) of which the body is composed have an angular velocity, the same as the angular velocity of the body as a whole. So these items have energy due to their own rotation

    It must be so that when you model the rigid body with items whose size approches zero, hence the number of items approches infinity, the contribution of energy due to their own angular velocity approaches zero. I don't understand why this is so, so if someone is able to show this would be greatly appreciated.
     
  2. jcsd
  3. Apr 8, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Indy1! Welcome to PF! :smile:
    Divide it (in 2D) into n2 parts, each part has a rotational energy and a kinetic energy …

    add them all up, and you get 1/2 Iω2.

    I ~ mr2, and m ~ r2, so I ~ r4, so the rotational kinetic energy of one part is ~ 1/n4, so the the rotational kinetic energy of n2 parts is ~ 1/n2, which tends to zero.
     
  4. Apr 9, 2012 #3
    Thanks tiny-tim,

    The problem for me is that you are using the formula 1/2 Iω² = Σ (1/2) m[itex]_{i}[/itex]v[itex]_{i}[/itex]², which is the sum of kinetic energies (due to curvilinear translation) of the particles of which the rigid body is thought composed of. But in reality the parts have an angular velocity the same as the rigid body, (like the moon around the earth, angular velocity of the moon is the same as that of system earth-moon, hence always the same side of the moon faces the earth). You can break the parts down in more parts forever, and end up with a larger and larger amount of smaller parts, all with the same angular velocity, but apparently together they have less and less rotational energy, and only their kinetic energy matters.

    So if we imagine a rotating object with all parts it's composed of like the gondolas of a ferris wheel (they only have curvilinear translation), and those parts are small enough and there are a lot of them, the object would have the same kinetic energy as a normal object that has the same rotation, dimensions and weight?
     
  5. Apr 9, 2012 #4

    AlephZero

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    That's not quite right. It does makes sense to think of the zero-size particles as rotating. They all rotate by the same amount as the body as a whole. But since each particle has "no size", it doesn't have any angular momentum or angular kinetic energy when it rotates.

    You can see that is true in the limit, as Tiny Tim explained. For eaxmple a square plate of mass m and side a has a moment of inertia about its center of ##ma^2/6##. If you cut it into four plates each with mass m/4 and side a/2, the MOI of each part about its own center is ##(m/4)(a/2)^2 / 6 = ma^2/96##, so the total MOI of the four plates, each rotating about its own center, is ##ma^2/24##. The difference is accounted for by the linear motion of centroid of the plates around the center of the whole plate, in a circle with radius ##a/(2\sqrt 2)##. This gives an inertia term of ##(m/4)a^2/8## for each plate and ##ma^2/8## for the four plates.

    And of course ##ma^2/8 + ma^2/24 = ma^2/6##.

    If you keep subdividing each piece into 4, the contribution from the rotation of all the pieces each about its own center (##ma^2/6##, ##ma^2/24##, ##ma^2/96##, ##ma^2/384##, etc) becomes smaller and smaller, but the total inertia about the center of the whole plate is always the same, ##ma^2/6##.
     
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