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Translational and Rotational energy in rigid bodies

  1. Dec 3, 2014 #1
    Suppose I have some sort of rigid body, a solid sphere lets say. For simplicity's sake lets assume that the sphere can only rotate about a single arbitrary axis through the center of mass. If the center of mass of the sphere is travelling with a constant velocity (with respect to some arbitrary reference frame) and the sphere also has a non-zero constant angular velocity about a single axis, will the total kinetic energy of the sphere be equal to the sum of the Rotational and Translational kinetic energy? Intuitively, I feel like the answer should be yes, but when I try to verify my intuition, the math seems to suggest otherwise.

    Etotal =1/2 ∑miv2i

    In the reference frame of the center of mass, the velocity of any particle is defined by its angular velocity and the distance from center of mass.

    v'i = rw(cosΘ)ey - rw(sinΘ)ex

    If the center of mass if moving with velocity v with respect to the reference frame we are observing the sphere at then we can apply a simple Galilean transformation to find an expression for vi

    v = aey + bex

    vi = (rw(cosΘ) + a)ey + (b - rw(sinΘ))ex
    v2i = (v'i)2 + v2 +2v'i ⋅v

    If I plug this back into the expression for the total kinetic energy, I get something odd.

    Etotal = 1/2∑mi(v'i)2 + 1/2∑miv2 + ∑miv'i ⋅v

    Etotal = Erotational + Etranslational + ∑miv'i ⋅v

    I'm not sure if I didn't properly define translational and rotational energy in this case, but the above equation seems to imply that the total kinetic energy is more than the sum of the translational and rotational kinetic energy. I'm a little confused about mechanics concerned rotation right now, so any guidance would be greatly appreciated.

    EDIT: Change "sphere" to flat disk, I wanted to work in 2 dimensions for simplicity's sake
     
    Last edited: Dec 3, 2014
  2. jcsd
  3. Dec 3, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    At first, you should not that(using standard spherical coordinates and taking the centre of mass to be at the centre of the sphere) for a particle at [itex] (r,\theta,\varphi) [/itex], [itex] v_i'=r\omega \sin\theta(\cos\varphi \hat x+\sin\varphi \hat y) [/itex].
    The next point is, those sums should be replaced by integrals for actual calculations because we're dealing with a continuum here. But as you can see, you have, in [itex] v_i' [/itex], sine and cosine of [itex] \varphi [/itex] which gives zero when integrated from [itex] 0 \ to \ 2\pi [/itex]. So that extra term is actually zero.
     
  4. Dec 3, 2014 #3
    Ah, my bad, I should have specified a flat disk instead of a sphere, but I get your point. The reason I didn't start with integral in the first place was to show that there was that I didn't intend to calculate the energies, but now that you mention it if I consider the extra term as an integral it does work out.

    ∑miv'i⋅v
    = ∑r(w)(acosΘ - bsinΘ)mi

    dm = krdΘdr
    ∑miv'i⋅v

    Integrate from 0 to R and 0 to 2π
    =wk∫∫r2(acosθ - bsinΘ)dΘdr
    =wk∫r2(b - b)dr
    =0

    I'm honestly not too familiar with using integrals in these situations, but if thats right then everything makes sense, thanks for the reply.
     
    Last edited: Dec 3, 2014
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