1
$\begingroup$

I have a doubt about the formula of the kinetic energy of the rototranslatory motion \begin{align} K = \frac{1}{2}MV_{cm}^2 + \frac{1}{2}I_{cm} \omega^2, \end{align} does it come from Koenig's first theorem or from another proof? And if so, when should I use Koenig's theorem?

$\endgroup$
1
  • 1
    $\begingroup$ The decomposition of KE is the second theorem $\endgroup$
    – nasu
    Commented Nov 19, 2020 at 19:26

1 Answer 1

1
$\begingroup$

For a system of particles, the total kinetic energy is the sum of the kinetic energy of the center of mass (your first term) plus the kinetic energy of the motion about the center of mass. This is true regardless of the relative motions of the particles about the center of mass (system could be a gas, not necessarily a solid). Yes, this is part of Koenig's theorem. (See a good physics mechanics textbook for the derivation; such as Goldstein, Classical Mechanics.)

For a rigid body the kinetic energy of the motion about the center of mass due to rotation is your second term. (Again, see a physics mechanics book.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.