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For a two-body system in absence of external forces, the kinetic energy as measured from ground frame is $K_0$ and from center of mass frame is $K_{cm}$. Find the correct statements:

$1.$ The kinetic energy as measured from center of mass frame is least.

$2.$ Only the portion of energy $K_{cm}$ can be transformed from one form to another due to internal changes in the system.

$3.$ The system always retains at least $K_0-K_{cm}$ amount of kinetic energy as measured from ground frame irrespective of any kind of internal changes in the system.

$4.$ The system always retains at least $K_{cm}$ amount of kinetic energy as measured from ground frame irrespective of any kind of internal changes in the system.

I'm confused that internal forces changes kinetic energy with respect to center of mass frame or ground frame? I found a similar question here. But it didn't address my concern.


My attempt:

Say we have a system of particles with masses $m_i$ and velocities $\vec v_i$. The total kinetic energy is then given by,

$$E=\frac{1}{2}\sum m_iv_i^2$$

Let’s define a couple useful quantities: total mass, $M\equiv\sum m_i$, velocity of the center of mass, $\vec V\equiv\frac{1}{M}\sum m_i\vec v_i$. Notation for the particle velocities in the center of mass frame: $\vec u_i=\vec v_i-\vec V$

Using these, we can rewrite the kinetic energy as $\begin{align} E&=\frac{1}{2}\sum m_i(\vec V+\vec u_i)^2\\ &=\frac{1}{2}\sum m_i(V^2+u_1^2+2\vec V.\vec u_i)\\ &=\frac{1}{2}MV^2+\frac{1}{2}\sum m_iu_i^2+\vec V.\sum m_i.\vec u_i \end{align}$

I know that the first term is the kinetic energy from the motion of the center of mass, the second term is the kinetic energy from the motion relative to the center of mass and the third term is linear momentum of system w.r.t. the center of mass.

My question : How will the third term affect the energy of the system if it is acted by an internal force?

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Internal forces change kinetic energy with respect to both the centre of mass frame and the ground frame.

Suppose a skater with mass $M$ and moment of inertia $I$ is travelling at speed $v$ and rotating with arms outstretched at angular velocity $\omega$. Then with respect to the COM frame their kinetic energy is

$K_{cm}=\frac 1 2 I \omega^2$

and with respect to the ground frame their kinetic energy is

$K_o = K_{cm} + \frac 1 2 Mv^2 = \frac 1 2 I \omega^2 + \frac 1 2 Mv^2$

If they now draw their arms in to their body so as to, say, reduce their moment of inertia to $I'= \frac 1 2 I$ then their angular velocity increases to $\omega'$ and conservation of angular momentum tells us that

$I \omega = I' \omega' \\ \Rightarrow \omega' = \frac {I}{I'} \omega = 2\omega$

Now their kinetic energy in each frame is

$K'_{cm}=\frac 1 2 I' \omega'^2 = \frac 1 2 (\frac 1 2 I)(4 \omega^2) = I \omega^2$

$K'_o = K'_{cm} + \frac 1 2 Mv^2 = I \omega^2 + \frac 1 2 Mv^2$

Note that $v$ does not change because the internal forces that cause the skater's arms to draw in always net to zero for the skater as a whole (because of Newton's Third Law) and so do not change the motion of their COM. However, both $K_{cm}$ and $K_o$ have increased by $\frac 1 2 I \omega^2$, which is equal to the work done by these internal forces.

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For an ensemble of moving particles, the total kinetic energy measured in the lab frame is: $$K_o=\frac{1}{2}M_tV_{cm}^2 + K_{cm},$$ where:

  • $M_t$ is the total mass of all the particles,
  • $V_{cm}$ is the velocity of the center of mass of the system in the lab frame,
  • $K_{cm}$ is the total kinetic energy of the system measured in the center-of-mass frame.

Internal forces can not change the velocity of the center of mass, thus they do not affect the first term of the right-hand side of the equation above. Only $K_{cm}$ is changed and this change will be the same for $K_o$.

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