Lecture 02

Added: 2026-05-12

[00:00:07][music] and welcome to the next lecture of this classical mechanic course. So uh in the last uh class uh lecture we have uh gone through this uh review of the Newtonian dynamics and uh uh where we study the the motion of this uh mass mass particle and uh we talk about this uh position vector and the velocity and then we talk about this uh linear momentum and then uh we move to uh the Newton's second law of motion from where we get this differential equation by solving this differential equation you get your uh what is uh this r and uh based on this r uh we can define the dynamics of the system.

[00:01:18]So another thing which we go through is uh like inertial or galian system and which basically preserve uh the nature of the Newton's law that is what the inertial system means and then if we have uh the conservation theorem of the linear momentum that if the external force acting on the total system the total external force acting on the system then this uh linear momentum of the particle is going to be a conserved quantity. So this much we have we had covered in our first lecture with along with the the introduction part and then we'll start this uh with the angular momentum concept. So when you have uh the same particle uh when which is uh rotating under certain torque. So then we define the angular momentum quantity that is that is this L. So angular momentum is defined as this uh the cross productduct of the two vector that is one one is my position vector and the second is my linear momentum that is r cross p.

[00:02:24]So uh why why some some object is going to have an angular momentum unless and until there is a torque is acting on that certain object. So when the torque is acting on that object it is going to rotate and when this is going to have rotational motion then there is going to have the angular momentum. So what is the direction of this L this angular momentum. So what you do is like when you have your finger in the R direction and curl this finger into uh from R to P then the direction of this thumb will tell you what is the direction of the angular momentum. So that's why the order of the factor is important. So if you're writing R cross P or writing P cross R. So when you're writing P cross R or R cross P then this is going to be different. So what you do is you put your finger along your radius vector and round it along this where the momentum is there. So this is going to give you uh the angular momentum direction. For example, in this case, you see if the if the force is acting on this particle of mass m and this is going to have a moment velocity in this direction and the corresponding momentum in this direction or if if the moment if the velocity is in this direction then it's going to have the corresponding momentum in this direction. So if we put uh uh uh our finger in the direction of R and curl in the direction of P then this uh this uh this thumb is going to be in the direction of the L. So that is going to be always perpendicular to the plane in which R and P is there told you that from where this angular momentum is coming because this particle is having a rotational motion. So that that is because of the torque and that is defined by this R cross F. So when you are applying an F at a certain distance from uh the center of the motion or the axis of motion this is the distance this is the distance from the center of the motion and this if this f has to be perpendicular. So F has to be perpendicular. If if for example if you if you see this is some of uh some attached to this is some seesaw attached to this stand and if I apply a force in this direction and what is this? This is my R. So this force has to be always perpendicular. Okay. So then I cross R cross S is going to give me this uh torque the motion. So N is R cross F in R R we know this is position vector from the axis of rotation through which it is going to rotate or if I apply a force along this direction with this much of R this is going to rotate along this. So f is uh from Newton's second law if ddt of this mv and then I can write this uh if if if you take ddt of r cross r cross mv. So if if from the differentiation part so if I take d r by dt this is d r by dt. So this is going to be v. This is the first part of keeping this mv constant and d r by dt is going to the v uh velocity and then keeping r constant and then I'm differentiating ddt of mv.

[00:05:37]So this part v cross v is going to be zero because you know that cross productduct of two vector is always going to be zero. So either I write ddt of r cross mv is exactly same of writing r cross ddt of mv. This is exactly same.

[00:05:52]So I can write this n which is from here. If you if you refer to this this r cross r cross ddt of mv is I can write as ddt of uh r cross mv. So this r cross mv is nothing but this is my p. So this is r cross p. This is r cross p and this r cross p is what is is my uh angular momentum. So and this dydt is there. So rate of change of angular momentum I told you that I'm going to use this L dot concept in lagian and Hamiltonian mechanics to write the time derivative of certain physical quantity. So this n is going to be ddt of l. So if you if you try to see from uh the translational motion to the rotational motion you see that f is uh replaced by this torque this n and this p is replaced by this angular momentum. So as as as you have this f is going to be db by dt or this going to be p dot. So similarly are going to have n = dl by dt. This is going to be l dot. So once we have time derivative of the linear momentum when f effective is zero. So then I have p is a constant quantity.

[00:07:18]This is going to be conserved quantity for the system. In the same way if uh if this torque the effective torque if the effective torque is going to be zero. So if the effective tors is zero then I have L dot is zero. So the angular momentum this L r cross P is going to be a conserved quantity. So you see that in linear motion the linear momentum is going to conserve and in the angular momentum concept uh the angular momentum is going to conserve quantity as it's uh when the effective force which is producing the angular momentum uh as a torque in this case is going to be zero. this uh uh angular momentum quantity is a conserved quantity.

[00:08:02]So with these two conservations of linear momentum and angular momentum, let's move to a very important concept which is which comes out of uh uh the definition of the work done by an external force uh on a particle. [snorts] When when you apply a force on a particle going from position A to position B, how you define the work done is is the integration. So you define the work done. W12 is integration from going from one position to the two position. And that is basically the dotproduct of the force which you are applying and the distance it is going to move. This is because I'm integrating.

[00:08:39]So if if if you apply a force from position one to position two. So this is this is going to travel a distance from position one to position two. So this ds is basically the smallest part of uh this path. So over which I'm integrating from position one to position two. So this is how you calculate the work done.

[00:08:58]And because we're considering mass as a constant quantity and so in this integral uh this f I can write f is equal to you what you write is f is equal to uh m a that is going to be m dv by dt that is acceleration is my dv by dt. So that is place of uh uh force I have written dv by dt and this d s is basically v dt and this d s is how much uh distance covered with this velocity for this small interval of time this this ds. So this is uh now uh when I move forward. So you have seen that this is I have written this uh dv by dt dot v dt as uh ddt of this is half v² I've written this here. So because when when you differentiate this when you differentiate this this this is going to be twice v dot dv by dt and then you have this half factor is there. So this is going to be cancel out. So this is going to be uh v dot dv by dt. So this is uh exactly same uh when you multiply by dt. So I can write this uh dv by dt and dot product with the v dt as uh 1x2 factor is coming from here mass is constant I get outside from the integration. So I have ddt of v². So if I am integrating over this factor will get out. So what I'm going to have is like uh from going 1 to 2 this is dv² and this is nothing but mx2 and this is 1x m I'm going to have this uh uh v² and put the limit. So you have what you have is 1x 2 m v2² minus v1 squ. So this is the change in uh uh change in the kindinetic energy because half m1 squared half mv2 squared is going to be the kindinetic energy. So the work done is going to give you the change in the kindinetic energy when you apply a force and the particle is going from one position to the another position. So what it actually one thing what it's giving is the change in the kinetic energy. The second important thing is like when the force through which you are doing certain work if if if this is the force and the system is going to be conservative. So what do you mean by the conservative force system you have gone through this gravitational force and the coolum force these two forces are the conservative forces. So because the work done by these forces are going to be path independent. So when uh when you apply the conservative force and the object is moving from one position to another position, it's not going to depend upon path. So, so whatever whatever uh the path you choose from going from this position to this position or this position to this position or you go like you go from here and here and here and like this and then you go this whatever the path you choose going from this location one to location two and if the and if you calculate the work done and if the work done is not depending on a path it depending on only the initial and the final position.

[00:12:30]Okay, this is one and two. is the coordinates of initial and final position. So this is going to be the conservative force and because it depends upon the initial position. So if you do the work done if you calculate the work done for uh for a closed curve. So when whenever you do uh circle around this integration this this means uh the integration around one to two position is going to be closed curve. So going you starting from this you're going to the second position and then you coming to the this position one and two going from here and then you coming to here. So when going from here to here 1 to two and two to one that that is going to be a closed path and if you calculate the work done going to closed path it comings out to be zero then this force is going to be a conservative force. So one thing is that the work done is giving you the change in the kinetic energy. Second thing is that when the force is conservative then uh the work done is path independent and the work done along the closed curve is going to be zero. So if the force is conservative so we can write this force as a gradient of some scalar function of position. So I should say that this is this is a vector like we normally don't put a vector sign over this uh dell operator because this is supposed to be a vector operator. So this is known as the gradient. This is known as the gradient and this vr is my scalar potential. So if the force is conservative force that is not depending work done not depending on a path the work done is closed along then this force is arising because of certain scalar potential okay scalar function of the position. So that's this is what let me let me tell you what is this gradient is all about. So this gradient is basically uh an operator a vector operator which is defined as I * del x + j * del y + k * del z. So I ji you know this is these are the these are the position unit vector along x y and z.

[00:14:46]This is the unit vector along I, unit vector along Y and unit vector along Z-axis. These are the JK vector. And this is the partial derivative of the function whatever whatever the function I apply here. And this is going to be at these positions. So this is going to be the partial derivative. So del X, dy and dz are these are the partial derivatives.

[00:15:16]partial derivative operators with respect with respect to x y and z. So, so uh what do you know what is partial derivative? Let me quickly let me quickly summarize this uh partial derivative and what is total derivative. So suppose you have a function this is some function x and y and it depends upon uh for example let's let's try to start with this is 2x² + 3 y. So this is a function which depends upon x and y. If I do the partial derivatives that is d del x of this function f. So this is going to be d del x of x² plus d del x of 3 y. So what is the answer to this? This is because I'm doing with respect to x. So this is going to be x. This is 2 2's are 4x and plus because there is no x and in this there is no x only y dependence is there and I'm doing with respect to x. This is going to be zero.

[00:16:30]This is the partial derivative. What is the difference between this partial?

[00:16:33]What you do is like when you do a full derivative or total derivative with respect to x of this function. This is what you write this is ddx of x² plus ddx of 3 y. This is not going to be same as the partial derivative because here what we have is 4x and then the you have ddx sorry uh you're going to have ddx of y. So there is a difference between here when you do the partial derivative and you do the total derivative. Why? Because when you're doing the total derivative this y this y is going to be dependent on x could depend upon x and here we don't consider we we consider x and y as independent quantity. So in this f uh I can have x depends upon y and y depend upon x. This is when we do uh the total derivative. But here in this case of partial derivative uh x and y are independent quantity. x is not depending on y and y is not depending on x. So this is what do I mean by this partial derivative with x y and z. So uh now come to uh the point that we are talking about uh the conservative forces uh and then if the forces are conservative which are the work done by these conservative forces are uh path independent and this is zero along a closed curve so that can be written as uh this uh negative of uh the gradient of this scalar function which is going to be the potential. So now if this is the scenario then if uh the work done we can define the work done then this is going from from position one to position two this is my f dot this is my f dot ds so in place of f because this f is conservative now we considering this f to be conservative so I can write it as this is minus this is dv and ds remains same okay so now this ds dot uh gradient of this potential v is going to be uh the total derivative of the potential. So you might be wondering why how this this is uh this is going to write. So let me let me tell you one thing that uh when I have this x and y this is my function which depends upon x and y when I'm doing the partial derivative of this this is d f by d x and this is uh d f of d y two partial derivative I can have for this but when I try to write the total derivative of this df by dx I can have and df by uh dy now I've already told you what do we mean by uh partial derivative and the total derivative. But when I I try to write the total derivative of this function xy how you write this? This is the partial derivative of this function and how much the change in dx okay for this and then the partial derivative of this function with respect to y and how much is change in uh dy. So this is how we write uh the total derivative of a function in terms of the partial derivative and how much change is there in dx and dy. So if you remember that uh this f dot uh d s is basically when I write this is minus del v uh this is sorry this is a scalar thing and uh so d sv this is d is an operator so I have this d operator is i del d del x + j del y + k dz Z. Okay. So it is operating on this V. Okay. And then I have DS. DS is I can write this is I DX + J DY + K DZ. Okay. So when this this is the operating on this potential this vector V. So I have I d x of v plus j del d y of v plus k dz of this potential v.

[00:21:04]Okay. And then I'm going to have the dot product of this with the this dx j dy and k dz. Okay. So you know the dot products you have I do I is going to one J dot J is going to one and K dot K is going to one rest combinations are all going to be zero. So that's I do I do J is going to be zero and so on. So only this I and I component are going to survive. So what I have is dv by d x and then I have this dx and then plus this j and j component is always only going to survive. We have dy by dy and then I have dy and this is delv k and k component dot product and this is going to be dz and I have dz. So what if you if you compare from this equation that when you how you define the total derivative of function. So this is going to be total derivative of the potential because this is potential depends upon this r. So basically this is uh dv depends upon xyz.

[00:22:14]Okay. So partial differentiation with respect to x partial differentiation with respect to y partial differentiation with respect to zed and then what is the total change in x and y and z. So that's why uh that's why there uh I have written in this uh presentation. So this uh you see here. So this is what I have is [music]