Work done on the centre of mass and kinetic energy of system

  • #1
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If we have two objects forming an isolated system and their centre of mass is ##X_{com}##, we know by work energy theorem that work done on centre of mass will be ##\int F_{ext}.X_{com}= 0## as no external force is acting on the system. However, if there is internal forces between the objects, there will be a change in kinetic energy of both the objects and as a result it will change the kinetic energy of the system. Why is the work done on centre of mass not equal to the change in kinetic energy of the system?
 

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  • #2
Why is the work done on centre of mass not equal to the change in kinetic energy of the system?
Kinetic energy does not have to be conserved but total energy does. The total energy will remain constant for the closed system.
 
  • #3
jbriggs444
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Why is the work done on centre of mass not equal to the change in kinetic energy of the system?
The total kinetic energy of the system includes the kinetic energy of each component separately. If we ignore rotation and thermal energy for the moment and consider a simple system with two masses, m1 and m2, the figure for total kinetic energy is: $$KE=\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
The kinetic energy considering only the motion of the center of mass is given by: $$KE=\frac{1}{2}m_{tot} v_{CM}^2$$
WIth a bit of algebra, it can be shown that the former is always at least as large as the latter.

If you apply an internal force, accelerating mass 1 in one direction and mass 2 in the opposite direction, the center of mass does not change velocity. The center-of-mass kinetic energy remains constant. But the total kinetic energy changes.
 
  • #4
vanhees71
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Hint: Introduce CM and relative coordinates and rewrite the kinetic energy in terms of those:
$$\vec{X}=\frac{m_1 \vec{x}_1 + m_2 \vec{x}_2}{m_1+m_2}, \quad \vec{r}=\vec{x}_1-\vec{x}_2.$$
 
  • #5
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The total kinetic energy of the system includes the kinetic energy of each component separately. If we ignore rotation and thermal energy for the moment and consider a simple system with two masses, m1 and m2, the figure for total kinetic energy is: $$KE=\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
The kinetic energy considering only the motion of the center of mass is given by: $$KE=\frac{1}{2}m_{tot} v_{CM}^2$$
WIth a bit of algebra, it can be shown that the former is always at least as large as the latter.

If you apply an internal force, accelerating mass 1 in one direction and mass 2 in the opposite direction, the center of mass does not change velocity. The center-of-mass kinetic energy remains constant. But the total kinetic energy changes.
Apologies for the late reply. If we apply external force on the two particle system which are exerting a force on each other, in such a case, when we say the external force does work on the system, do we mean the external force does work on the center of mass of the two particle system, ## W_{ext} =\int \vec{F_{ext}}.d\vec{r_{com}} ##? Or will the work done by the external force be equal to the sum of the individual works it does on the two particles, ##W_{ext}= W_{ext, particle 1}+W_{ext, particle 2} = \int \vec{F_{ext}}.d\vec{r_{1}} + \int \vec{F_{ext}}.d\vec{r_{2}}##?
 
  • #6
jbriggs444
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Apologies for the late reply. If we apply external force on the two particle system which are exerting a force on each other, in such a case, when we say the external force does work on the system, do we mean the external force does work on the center of mass of the two particle system, ## W_{ext} =\int \vec{F_{ext}}.d\vec{r_{com}} ##? Or will the work done by the external force be equal to the sum of the individual works it does on the two particles, ##W_{ext}= W_{ext, particle 1}+W_{ext, particle 2} = \int \vec{F_{ext}}.d\vec{r_{1}} + \int \vec{F_{ext}}.d\vec{r_{2}}##?
One might mean either thing. It is ambiguous. If my [forty odd years ago] experience is any indication, classroom teachers and textbooks are not careful in describing how "work" applies to rotating or non-rigid systems. [Calling @Doc Al -- I know he has ideas on this]

Often, one is uninterested in details and is only concerned about the resulting bulk motion of the system. In which case "center of mass" work is determined by multiplying the applied force by the motion of the center of mass. This approach kind of works, but it causes problems if you expect the work energy theorem to be useful as a statement about conservation of energy.

Sometimes one is interested in the picky details. In which case the work is determined by multiplying the applied force by the motion of the exact point(s) on which the force acts.
 
  • #7
vanhees71
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Well, then you should read more theoretical-mechanics textbooks. Almost all of them get the spinning top (rigid body) and thus rotation energy right.
 

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