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Kinetic energy of a rotating disc

  1. Mar 27, 2012 #1
    if KE=1/2mv^2

    and you have a circular object rotating, with it's mass uniformly distributed through the object (ie each part of the disc weighs the same) then obviously certain parts of the disc will be moving faster than others.

    therefore closer to the middle of the disc, you have more KE and the further from the middle you have less KE. how then do you assign kinetic energy to the whole disc?

    the obvious answer might be to take the midpoint, but is this correct?

    this just popped up in my head whilst i was strolling home, and may not have been very clearly or technically explained, but if you can understand the point i am trying to make, please help me out :)
     
  2. jcsd
  3. Mar 27, 2012 #2

    K^2

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    In rigid body physics, there is a concept of rotational kinetic energy. It depends on angular velocity and moment of inertia. Angular velocity is a vector aligned with axis of rotation and magnitude equals to 2pi * frequency. Moment of inertia is a tensor, which can be represented as a 3x3 matrix. In general, rotational kinetic energy is equal to 1/2 ωTIω, where I is moment of inertia matrix, and ω is a column matrix representing a vector, with ωT being its transpose.

    When you have a fixed axis of rotation, only the projections along that axis matter, and you can pick out a scalar moment of inertia around that axis, and a scalar angular velocity, which is just the magnitude of the vector from above. Then the rotational kinetic energy around that axis simplifies to 1/2 Iω², which starts to resemble the normal kinetic energy equation you are used to. For a solid disk, there is a simple expression for moment of inertia around rotational symmetry axis, which is typically the axis you think of disk rotating through. I = 1/2 MR², where M is mass of the disk and R is the radius. Notice that this means that in your approach, you'd have to chose a point that's R/√2 from center to get the same result, and in general, this will vary with geometry.

    If you look up Wikipedia article on Moment of Inertia, you'll be able to find values to use for other geometries, as well as the list of rules for computing the tensor and scalar versions for arbitrary geometry.
     
  4. Mar 27, 2012 #3

    AlephZero

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    That's not the correct answer, but the basic idea is useful. The distance from the axis which would give the same kinetic energy if all the mass was at that distance is called the radius of gyration. The moment of inertia of a rotating body is then ##I = Mk^2##, where ##k## is the radius of gyration. The radius of gyration is therefore ##k = \sqrt{I/M}##.
     
  5. Mar 27, 2012 #4

    OldEngr63

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    The OP seems to be confused about the distribution of velocity within the rotating disk, as evidenced by the statements: "and you have a circular object rotating, with it's mass uniformly distributed through the object (ie each part of the disc weighs the same) then obviously certain parts of the disc will be moving faster than others.

    therefore closer to the middle of the disc, you have more KE and the further from the middle you have less KE. how then do you assign kinetic energy to the whole disc?"

    If we assume that the disk is rotating about a fixed center, then for a particle in the disk at a radius r, the velocity of the particle is
    v = r ω
    so it is evident that the velocity increases with the radius. The velocity is zero at the axis of rotation, and the local kinetic energy density is zero at the axis of rotation. Continuing on for the particle at radius r, its kinetic energy, assuming a differential mass, dm, is

    dT = (1/2) (v2) dm
    = (1/2) (r ω)2 ρ r dr dθ dt
    where
    ρ = material density
    θ = angular coordinate, 0 →2π
    t = thickness of disk

    To find the total kinetic energy of the disk, we integrate,

    T = (1/2) ρ ω20t00Rr3 dr dθ dt

    = (1/2) ω2 2πtρ R4/4

    = (1/2) ω2 (πtρ R2) (R2)/2

    = (1/2) ω2 M R2/2

    = (1/2) ω2 I

    where

    M = mass of disk
    I = Mass Moment of Inertia of the disk = M R2/2
     
  6. Mar 27, 2012 #5

    sophiecentaur

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    It's the other way round. The parts of the disc nearer the outside are moving faster so the KE of an element of mass on the outside is higher. It is proportional to the radius squared (v = rω and KE is mv2/2)

    The element of mass right at the centre is not moving at all so has no KE.

    To work out the total KE, you use Iω2/2. I, the moment of inertia takes account of the 'squared' function of distance from the centre of rotation.
    I =ƩmnRn2 for all n elements of the object. Or you can do it using Integral Notation.
     
  7. Mar 27, 2012 #6

    K^2

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    sophiecentaur, OldEngr63 was just quoting OP, and that's precisely what he was correcting.
     
  8. Mar 27, 2012 #7

    sophiecentaur

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    RTFM again! I should have quoted the OP - or just gone back to sleep.
     
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