Kinetic energy from c.o.m frame

[pardesi]

we know
kinetic energy of a system from a certain frame of refrence is the sum of kinetic energy in the c.m framee of refrence that moves translationally to the previous frame and the kinetic energy of c.m from that frame.

but what if the i choose the frame attached to the c.m such that it rotaes about a point in the previous frame is there a closed formulae for that

 

[jostpuur]

Do you know how to prove the first statement you made? (and which you btw. didn't say very clearly)

I'm not sure what kind of answer you are seeking, but at least precisely the same result doesn't work with rotating frames . You can verify this by trying to prove it, and you'll see that the calculations give something more complicated.

 

[pardesi]

yes soory for the non clarity(was too bored to write in latex...)
well i tried doing the same way as first but ended up in a much complex formulae...is there any known result

 

[jostpuur]

I'm not sure about well known results, but it is not difficult to calculate something about this. If we, for simplicity, assume that the center of mass is not moving, then velocities of the particles in rotating frame are

$$\dot{x}_k-\omega\times x_k$$

still written with the inertial frame vectors. Omega is the angular velocity of the rotating frame. If correct kinetic energies are

$$E_k = \frac{1}{2}m_k |\dot{x}_k|^2$$

and if we define some kind of effective kinetic energies

$$E_{k,eff} = \frac{1}{2}m_k |\dot{x}_k- \omega\times x_k|^2$$

then, assuming I calculated this correctly, we have

$$E_{k,eff} = E_k + \frac{1}{2}m_k |\omega\times x_k|^2 - \omega\cdot L_k$$

where L is the angular momentum of the particle. Then

$$E = \sum_{k=1}^N E_k = E_{eff} - \frac{1}{2}\sum_{k=1}^N m_k |\omega\times x_k|^2 + \omega\cdot L$$

I'm not sure. I was bored, and looking for something small to do. Check the result yourself if you want to know if it's correct.

What ever you were looking for, I believe that the result will involve equations that look something like that.