Kinetic Energy in the Center of Mass Frame

Given two bodies moving along a line, I can find the velocity of the center of mass frame. Taking the time derivative of the Galilean transformation I get

$v'=v-v_{cm}$

By definition, the total momentum of the center of mass frame is $0$. From this, I can then find the velocity of the center of mass frame.

$\begin{eqnarray*} \Sigma m_iv' = 0\\ \Sigma m_i(v_i-v_{cm}) = 0\\ \Sigma (m_iv_i-m_iv_{cm}) = 0\\ \Sigma (m_iv_i)-v_{cm}\Sigma (m_i) = 0\\ v_{cm} = \frac{\Sigma m_iv_i}{\Sigma m_i} \end{eqnarray*}$

Now, I am now trying to find the relationship between the total kinetic energy of the particles in the original frame given by $T$ and the total kinetic energy in the center of mass frame $T_{cm}$. The original kinetic energy is

$T = \Sigma\frac{1}{2}m_iv_i^2$

and I believe the kinetic energy in the center of mass frame is given by

$T_{cm} = \Sigma\frac{1}{2}m_i(v_i-v_{cm})^2\\ T_{cm} = \Sigma\frac{1}{2}m_i(v_i^2-2v_iv_{cm}+v_{cm}^2)^2\\ T_{cm} = \Sigma\frac{1}{2}m_iv_i^2 - \Sigma m_iv_iv_{cm} + \Sigma\frac{1}{2}m_iv_{cm}^2$

and substituting back in (and rearranging) gives

$T = T_{cm} + \Sigma m_iv_iv_{cm} - \Sigma\frac{1}{2}m_iv_{cm}^2$

However, this does not match the correct solution of

$T = T_{cm} + \Sigma\frac{1}{2}m_iv_{cm}^2$

I believe I am conceptually missing something when I am setting up the total kinetic energy in the center of mass frame, but I can't figure it out for the life of me. Does anybody have a hint as to what I am doing incorrectly?

1 Answer

Starting with

$T = T_{cm} + \Sigma m_iv_iv_{cm} - \Sigma\frac{1}{2}m_iv_{cm}^2$

use

$v_{cm} = \frac{\Sigma m_iv_i}{\Sigma m_i}$

which can also be written as

$\Sigma m_iv_i = v_{cm}\Sigma m_i$

and substitute back in to get

$T = T_{cm} + \Sigma\frac{1}{2}m_iv_{cm}^2$

• Thank you! I thought that had something to do with it, but for some reason it clicked when I read your response lol. – ubomb Mar 24 '16 at 3:12