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? Please explain in detail, I am new to physics.
[Physics] Work done on the centre of mass and change in the kinetic energy of the system
work
Related Solutions
$W=\Delta K_\text{system}$ and $W_\text{external}=\Delta K_\text{system}+\Delta V$ are consistent with each other iff $\Delta V=-W_\text{internal}$. The latter is the definition of the potential energy for conservative forces.
The cited equation is thus valid iff all internal forces are conservative.
The first thing you must do is define your system.
If the system is the book alone the the external forces on the book are the force that you exert on the book and the gravitational attraction on the book by the Earth.
If the book starts and finishes at rest then there is no change in the kinetic energy.
The work done by you on the book is positive as the direction of the force that you exert on the book is the same as the displacement of the book.
The work done by the gravitational force due to the Earth is negative because the gravitational force is in the opposite direction to the displacement of the book.
If the two external forces are equal in magnitude and opposite in direction then the net work done on the book is zero (equal to the change in kinetic energy).
Of course one could reason that the net external force on the book is zero so the net work done by external forces on the book is zero.
There is no mention of gravitational potential energy because it is the energy associated with the book and the Earth as a system.
So now let's consider this system of the book and the Earth.
The external force is now the force that you apply on the book.
The force that the Earth exerts on the book is an internal force and its Newton third law pair is the force that the book exerts on the Earth.
When you do positive work separating the book and the Earth that work increases the gravitational potential energy of the book-Earth system.
If you released the book the separation between the book and the Earth will decrease and the gravitational potential energy of the system will decrease.
The book (and the Earth) would then have kinetic energy.
Usually only the motion and kinetic energy of the book is considered because the mass of the Earth is so much greater than the book.
This results in the speed and kinetic energy of the Earth being very much smaller that that of the book.
Best Answer
So you have a system of particles at positions $\mathbf r_i$ experiencing some internal forces $\mathbf G_{ij} = -\mathbf G_{ji}$ and some external forces $\mathbf F_i$, Newton's laws state that $$ m_i \ddot {\mathbf r}_i = \mathbf F_i + \sum_j \mathbf G_{ij},$$where the dots over the vectors are time derivatives. (If you are a beginner, you will have to understand that I am using a lot of calculus and vectors here; when something is in boldface it is a quantity with both length and direction; a position for example is a tuple of three functions of time $\mathbf r(t) = \big[x(t), y(t), z(t)\big]$. Its first time derivative $\dot{\mathbf R}$ is the instantaneous velocity $\dot{\mathbf r} = \mathbf v = [\dot x, \dot y, \dot z]$ and the time derivative of that is the instantaneous acceleration $\ddot{\mathbf r} = \mathbf a = [\ddot x, \ddot y, \ddot z].$ If you do not understand these ideas of calculus and vectors, they are a good place to start as everything in classical mechanics makes more sense if you know them. Please do not get distracted by "vector calculus," that is also important but it is somewhat different.)
We would like to get rid of the internal force terms and this can be done by summing over $i$ using the fact that Newton's third law $\mathbf G_{ij} = -\mathbf G_{ji}$ forces $\sum_{ij} \mathbf G_{ij} = \mathbf 0.$ This lets us then find, $$\sum_i m_i \ddot {\mathbf r}_i = \sum_i \mathbf F_i .$$ This is made a little more illuminating indeed by defining the total mass $M=\sum_i m_i$ and the center of mass position $\mathbf R = \sum_i (m_i/M)~\mathbf r_i$ at which point the left-hand side says, for constant masses, $$M~\ddot{\mathbf R} = \sum_i \mathbf F_i,$$so the center of mass moves like a particle of mass $M$ and only responds to the external forces.
Now the power exerted by a force on a particle is generally that force dotted with the particle's velocity. The power exerted on any one particle would be $$P_i = m_i \ddot {\mathbf r}_i \cdot \dot{\mathbf r}_i = \mathbf F_i\cdot \dot{\mathbf r}_i + \sum_j \mathbf G_{ij}\cdot\dot{\mathbf r}_i.$$This is related to the particle's kinetic energy $K_i = \frac12 m_i~ \dot{\mathbf r}_i \cdot \dot{\mathbf r}_i$ by the relation that $P_i = \dot K_i.$ This is the work-energy theorem in its differential form.
However this becomes hard to tie to the same effective story that exists for the kinetic energy of the center of mass, $$K = \frac12 M~\dot {\mathbf R}\cdot \dot {\mathbf R},\\ P = \dot K = M~\ddot{\mathbf R} = \sum_i \mathbf F_i \cdot \dot {\mathbf R} $$ for the simple fact that we have a multiplication of two sums here and this has nontrivial structure: $$K = \frac12 \frac1M ~\sum_{ij}m_i~m_j~\dot {\mathbf r}_i\cdot \dot {\mathbf r}_j.$$Clearly the diagonal elements $i=j$ are related to these $K_i$ that we know and love, giving some formula $\sum_i (m_i/M) K_i.$ But just as clearly, the kinetic energy includes these off-diagonal terms that mean that the system is not only described by them.
This is actually a really common occurrence, even for purely rigid bodies. The Earth has some motion through space defined by its center-of-mass motion. But its total kinetic energy is not just the kinetic energy of its center-of-mass: there is also rotational kinetic energy, which this picture misses. You can give a lot of energy to the Earth by spinning it faster and faster and it will continue in the same orbit. Heck, if you spun the Earth fast enough it would cease to be rigid and break apart and fling its crust out in all directions into space -- though before you got to the point of hitting other planets you would probably form a big accretion disk that would like to, over the next many million years, reform a planet by collapsing again and converting all of that energy to heat and re-radiating a lot of the energy you gave it out into space. (If, y'know, it didn't form a toroidal planet.)
For rigid bodies the entire motion is defined by the center-of-mass motion plus an instantaneous angular velocity vector $\mathbf\Omega,$ with (I believe) the relation $$\dot{\mathbf r}_i = \mathbf\Omega\times(\mathbf r_i - \mathbf R) + \dot{\mathbf R}.$$I think with this one can find a 3x3 matrix (the "moment of inertia tensor") $\bar{\mathbf I}$ such that the total kinetic energy is $\sum_i K_i = \frac12 \mathbf\Omega \cdot \bar{\mathbf I} \mathbf \Omega + K,$ proving that for rigid bodies this is all you need to understand the kinetic energy.
For non-rigid bodies of course there can be arbitrarily more things to deal with. You build a mass out of a spring linking two balls; there will be both rotational kinetic energy as in a rigid body, plus some energy due to their vibration. Maybe the $\mathbf G_{ij}$ do not even conserve energy: consider two balls which attract each other by some force, they might bounce off each other elastically like steel or ping-pong balls, or they might stick together like if made of a soft substance like Play-Doh. But that center of mass cannot possibly know which situation is which, because its motion only depends on external forces and that's a distinction about internal forces.