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Landau Lifshitz claim in their Mechanics book (39.11) that for a uniform rotation we have

$ E = \frac{mv^2}{2} - \frac{m}{2} (\omega \times r)^2 + U,$ where the rotation is given by $v' = v + \omega \times r.$

This does not make sense to me, since this centrifugal term is negative( I mean, compared to the energy of the rotating system the energy in the inertial system should be HIGHER, since there is an additional kinetic energy of the rotation) as I had exected something like $ E = \frac{m(v + \omega \times r)^2}{2} + U$ in the inertial frame.

Then they claim that in the rotating system(in that case they use the notation $v_0$ instead of $v'$ we have $ E = \frac{mv'^2}{2} - m v' \omega \times r+ U,$ ( This looks reasonable to me, as kinetic energy is less than the inertial one), but still,in that case I had expected something like

$ E = \frac{m(v' - \omega \times r)^2}{2} + U,$

so where am I wrong(especially about my reasoning in the first place) and why is it wrong, just to plug in the velocity of the transformation.

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1 Answer 1

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The energy formula

$$\tag{39.11} E~=~\frac{1}{2}mv^2 -\frac{1}{2} m({\bf \Omega} \times {\bf r})^2 + U $$

in Ref. 1 (of a point particle, as seen in a rotating reference frame $K$) consists of three terms:

  1. Kinetic energy: $\frac{1}{2}mv^2$.

  2. Centrifugal potential energy: $-\frac{1}{2} m({\bf \Omega} \times {\bf r})^2$.

  3. Other potential energies $U$.

In particular, the minus sign in front of the second term is the correct one. It is a centrifugal potential, so it encourages the system to increase its radial coordinate $r$. Phrased equivalently, it costs work (against the centrifugal potential) to reduce the radial coordinate $r$.

References:

  1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976.
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ah, so this one refers to the rotating reference frame, is it then true that my first attempt to correct this equation refers to the inertial system? –  Xin Wang Sep 29 '13 at 15:04

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