[MUSIC] Hi, this is another supplementary video about internal, the work done by internal forces. When we are calculating the work done to the this rigid bar, instead of having the net force applied to the rigid body, let's split it out as a force by each particle, okay? So if we separate out the rigid body, split it out the rigid body as a particle. The first particle is this hard which is fixed at the hinge joint, will be under the external force like RX and RY and then interaction force from the second particle, which is noted as F2 here. And the second particle is the force is applied to the second particle like F2, which is going to be action reaction force between the particle 1 and 2. And another action reaction force from the next particle as F3 this way and the gravity or I missed the gravity here. Okay, those internal forces are for the particles are directions are aligned to the lines between the two particles because the particle doesn't apply the twerks. So in the similar way, I can actually draw all the internal forces between the particles up to the particle number n. And the displacement of each particle is denoted as a TR1, TR2, etcetera, up to TRn. And then depending on the position of the particle TR1, which is located here at the very end of the Hindu join has a zero displacement. Whereas, the particle at the end of the rigid body the bar is the greatest, right? So there are different TR for each particle. Now, if I establish the F equals ma the Newton's law, then I have reaction force, internal force and the gravity will generate the ma. And then those internal forces are same magnitude and did the opposite direction. So if I sum up the all the forces for each particles will be only leave the reaction force and the gravity force which will generate the ma, same as the total mass and the center of mass acceleration. Now the work done however is the calculated by the work done per each particle and it's integral. So I should do the F to the inner product with the displacement or a single particle and do the integral. So it seems like you have a force I and I plus 1 to the particle number I and then it's integral over the dri and the next particle. You have a force FI of I plus 1 and F of I plus 2 and integral over the particle I plus 1 displacement. Now, similar to the force equation Newton's equation, F equals ma are these components. Same work done by a force f plus 1 components here and they're saying value the opposite sign will be cancelled out or not and why is that so? If those are not cancelled, then the total sum of all the particles then should be also non zero two. Why or why not, think about it. In the following slides, I am going to explain the answer. So take a pause of this video and think about it by yourself. And once you get the answer and continue listening to the video. Okay, have you thought about that? To explain it, let me just split it out an X and Y direction, with these are the forces applied to the particle number 1 in X direction and it's X direction or displacement. So the force, the work done on the left hand side is going to be f equals ma and this integrated over the displacement of x 1 so that will be the kinetic energy of that particle at one half MV square. Now as I said, the first particle displacement is 0 but next 1 x 2 and xn plus 1 and X1 etcetera are not 0 and generally different values. So if even though you have an app same f 2 here and there's a magnitude in the opposite signs showing in the consecutive particles equations of motion the work energy relationship. Those multiples displacements are different. So in general, these are not cancelled out. So if you sum them up, okay, the left hand side will be the total internal force work by the X direction. And which is not generally zero and that's going to be a single particles kinetic energy summation. And it will be turns out to the translational kinetic energy for the center of mass in X Direction and Omega Square I term in X Direction. Okay, similarly, I can do the same thing for the y direction. In this case, I should include the gravity term, okay? And I'm going to have a similarly have of work done on the left hand side in y direction of work and kinetic energy of the particle in y direction. And again the first, even though the first displacement will be 0, the next displacement series are not generally 0 and have a different magnitudes. So even though you have a same and opposite sign of the internal force here and there, those integrals are not generally cancelled out because d y 2D y 3s are different. So if you sum them up or the internal force in y direction, which is not generally zero and the gravitational potential energy. I mean the gravitational work turns out to be y directional kinetic energy of the particle and its sum. And then that's going to be expressed by the y direction or translational kinetic energy and rotational kinetic energy. If you bring up the work done in X Direction, what you can have is all the work done by the internal forces, X directions and Y directions. And the gravity worked on will turns out the total kinetic energy of the rigid pendulum. Now, we know that the typical work done kinetic energy switch is going to be gravitational energy into the kinetic energy. So this green box to term the sum of the internal forces are supposed to be 0, right? Is that really so? So what you keep what we can have finally the work done by the gravity is going to result in the change for the kinetic energy of the rigid body. When one half MV square plus I Omega Square. Okay, to examine how those internal work done by the internal force will be either cancelled out or not. Let's look at zoom in the two particles and I petition. So what I'm really interested in is this F of I plus 1 dri and if I plus 1 D RI plus 1, how this term will be some expanded. Okay, for the two particles, internal force are aligned, okay? Along the line that is connecting the two particles. So the direction for the f i plus 1 is going to be parallel to the displacement difference between ri1 minus r of i. And then each particle has its own displacement dri and tri plus 1. So again, those direction, the position for this particle and its difference is that parallel to this force direction and then there are independent displacement. Now if we look examine this I plus 1 displacement part again, so if you have those vector bringing it up and then those difference will be dri plus 1 minus ti, okay? So your tri plus 1 is going to be expressed by the try + tr of i, tri plus 1 minus tr of i. So if you plug that in here, so that you have a ri plus 1 as ri and t, its differences, those term cancelled out. So what you can finally have is force with the inner product of tri plus 1 and dr i, okay? Now note that with the rigid body, the displacement between the two particle, square his invest, should be constant and never changes. So the displacement difference between ri plus 1 and ri should be constant value. And if you take the derivative, if you differentiate them, you have a two ri plus 1 minus r of i and then t difference of this, okay? So if you have a two vector with the inner product turns out to be 0 means, those two vectors are in perpendicular, right? So displacement difference is perpendicular to the finite different infinitesimal change for those displacement vector. And again previously, we know that the force direction is going to be parallel to the displacement differences. Therefore, with these two, what we can have is f i + 1 and dri plus 1 minus l of i is going to be perpendicular, right? So the inner product that product will be 0. So this how it ends up having internal force work done is going to be 0. Again, even though the internal work is 0, the vector sum of f vector and its displacement difference vector, those component term in X directions and Y direction may not be 0, right? The sum is 0 but each component is not necessarily zero. So this is how we ended up. Work done by the X direction or internal force is not zero and work done by the Y direction or internal force is not 0. But the total internal force work done is 0. So that you don't have to consider the work done by the internal forces when you are working on work energy relationship of the rigid body and the particle is rigid body, okay? Okay, thank you for listening.