Question

The reference frame, in which the centre of inertia of a given system of particles is at rest, translates with a velocity $\overrightarrow{V}$ relative to an inertial reference frame $K$. The mass of the system of particles equals $m$, and the total energy of the system in the frame of the centre of inertia is equal to $\tilde{E}$. The total energy $E$ of this system of particles in the reference frame $K$ is given as $E=\tilde{E}+\frac{1}{x}m{V}^2$. Find $x$

Answer 1

To find the relationship between the values of the mechanical energy of a system in the $K$ and $C$ reference frames, let us begin with the kinetic energy $T$ of the system. The velocity of the $i$-th particle in the $K$ frame may be represented as $\overrightarrow{v_i}=\widetilde{\overrightarrow{v_i}}+\overrightarrow{v_C}$. Now we can write
$\sum \frac{1}{2}{m}_i{v}_i^2=\sum \frac{1}{2}{m}_i(\widetilde{\overrightarrow{v_1}}+\overrightarrow{v_C}).(\widetilde{\overrightarrow{v_i}}+\overrightarrow{v_C})$
$=\sum \frac{1}{2}{m}_i\widetilde{v_i^2}+\overrightarrow{v_C}\sum m_i\widetilde{\overrightarrow{v_1}}+\sum \frac{1}{2}{m}_i{v}_C^2$
Since in the $C$ frame $\sum m_i\widetilde{\overrightarrow{v_i}}=0$, the previous expression takes the form
$T=\tilde{T}+\frac{1}{2}m{v}_C^2=\tilde{T}+\frac{1}{2}m{V}^2$ (since according to the problem $v_C=V$) (1)
Since the internal potential energy $U$ of a system depends only on its configuration, the magnitude $U$ is the same in all reference frames. Adding $U$ to the left and right hand sides of equation (1), we obtain the sought relationship
$E=\tilde{E}+\frac{1}{2}m{V}^2$

Comparing the above equation with the equation given in question, we get $x=2$.