
#1
Dec212, 02:54 AM

P: 147

The total kinetic energy (as viewed from one inertial frame) of a free, rigid body is the sum of all the infinitesimal kinetic energies of the components that comprise the body.
How do we prove that for a rotating body [tex]E_k=\frac{1}{2}\left(M_{T} v_{c}^{2} + I_{c} ω^{2}\right)[/tex] 



#2
Dec212, 07:52 AM

P: 147

Where M_{T} stands for the total mass of all the infinitesmal components combined.




#3
Dec212, 08:25 AM

Mentor
P: 10,809

Integrate ##\int \frac{1}{2}v^2 \rho dV## (in other words, kinetic energy = 1/2m^2 for all infinitesimal m) and split v into components from translation and rotation and you will get the correct result.



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