Hi. In this video, I'm going to talk about ME question regarding the integral of the equations of motion in polar coordinate. So from the AP course MA, we apply the integral over the displacement and time to obtain the work and energy in impulse-momentum relationship, and when you are using the polar coordinate, there's one thing you have to consider to the characteristics of polar coordinate. Let's talk about it more in detail. Let's start with the impulse-momentum relationship in Cartesian coordinate. I know that this is a vector form. The v is a vector including x and y components and same for the f, but can I just apply impulse momentum components by components, per say, like x in x directional linear momentum is going to be obtained by the addition of any linear impulse in x direction. For example, if I have equations of motion in x and y direction, can I just separately take the integral for the x components and y components, and that will give me the linear impulse in x and y direction? Yes, since your derivative of the v_x is going to be giving you the x direction acceleration, I can apply impulse momentum in x direction and y direction separately. How about the polar coordinate? Again, the same linear impulse-momentum relationship, those are vector form, and can I just apply this relationship separate for the r directions, and Theta direction. For example, I have F_r direction is going to be F equals m_a, and take the time integral and will then give me the r directional linear momentum and Theta directional linear momentum. To answer this question, I'm going to talk about how those acceleration in r direction and velocity in r direction are related with. Here, you have a v, we usually have r component and the Theta component, and acceleration is obtained by taking a derivative of r direction of velocity and Theta direction of velocity. R direction of velocity derivative here over r double dot ER term and when you take the derivative of the ER vector in polar coordinate, which is non-zero, it is going to be Theta dot e Theta. So your a_r is going to be contributed by the derivative of the v_r as well as derivative of b Theta. Since you have a Theta derivative, the Theta term, you have some e Theta components as well as e_r term which is centripetal acceleration. Therefore, this doesn't hold. So your derivative of the velocity in r directions gives both e_r components and e Theta components. Acceleration in r direction, acceleration in Theta direction, respectively. So you cannot simply apply impulse-momentum relationship in r and Theta direction components separately. Then how you can do the integral correctly from the F equals m_a. If you're handling the m_a part, by taking the derivative of the velocity over time, you can simply obtain the total velocity components like this, the linear momentum change. When you take the integral of the force term, which is F equals m_a, you have a components by r directions and Theta direction altogether. Here, note that this, two r dot Theta dot, is coming from derivative of the v_r term and derivative of the Theta term. So I'm going to separate it out as r dot d Theta d_t, and d_r d_t and Theta dot. So if I just rewrite everything in terms of derivative term like, r double dot is going to be t r dot d_t, and then multiply it by the d_t term here, and split it out everything, every component. I can simplify the a term as t_r dot r Theta dot d Theta, and here r dot t Theta, Theta dot d_r and r_d Theta dot e Theta term. Now those a_r d_t, which is this term, is not going to be v_r, that's the message that I'm going to deliver here. So since those a_r was contributed by the derivative v_r and v Theta, let's combine this green underlined term together here and orange underlined term together, separately. I have green underlined term to the front and orange underlined term to the separate parenthesis. What is this? When you remind yourself for the derivative, the v_r and v Theta, you have a derivative r dot and e_r, that will give you this d_r dot and e_r term and r dot d Theta e Theta term. So the front part is going to be what? d e_r term. The second part is in the similar manner, it's going to be d_e Theta term, right? So what you can have is m dv_r and d_v Theta. Note that here, when you are taking the step from this line and this line, you have a mixture of the contribution for the Theta term and the Theta directional components and then r directional components, therefore, you will end up having total linear momentum derivative. Note that your a_r is not going to be directly v_r, those are combined by the v Theta term in a Theta term. This is the example that I derived and I have this results and let me know which part I've made a mistake. I_ r is going to be F r d_t, which is m_a r d_t correct, a_r is going to be r double dot minus r Theta dot square, which has been rewritten as a numerator and denominator form, further simplification here. I have d_r dot r Theta dot d Theta. It can have an integral term here. This has been left and it's going to be m delta v_r. From here to here, it is not correct because this one is right, this is delta v_r, but I still have this term left. Then this should be handled. So this equation, this step has an error. So how about the impulse in Theta direction? Theta direction is timed integral of the f Theta. That's going to be f equals ma, which is Theta directional acceleration. I can also cause simplification. So that ended up having this point. So here 2 m_r dot d Theta is r dot Theta term integral and m_r d Theta dot I have m_r Theta dot, which is going to be m delta v Theta. Where I made a mistake here when I take an integral here, that is from here to here, I assuming that everything here is independent to Theta I handle this as a constant value however, this r dot is not an independent variable to the Theta and r is also a function of Theta dot. There are relationship, there is over the Theta and Theta dot. Therefore, you cannot directly now take the integral from here to here. That therefore the total final answer is not correct. Same for the angular impulse momentum relationship. It's all vector form. Can I just do it component by component? It's our components separately. I don't think so. So whenever you have a r direction equations of motion, Theta direction equations of motion. If you took the r cross product and the time integral, you can obtain the equality of two here. However, those are not directly give you the r component angular momentum of the r component in angular one the Theta components. No, again that's because you have a velocity derivative gives both contribution for the r direction acceleration and Theta directional acceleration as well. Very similar way what we have done for the linear momentum relationship. Take the derivative of this one, or this one is simple. You have a term canceled out. You have a total velocity cross product by the r. So you can have, you can have r cross-product to be Theta. But if you have F equals ma integral, note that you have acceleration components are coming from both Theta components and the r components of the velocity derivative. So if you just make it simple, if you simplify the equation at the stage here, you should swap some of the components for the e_r components in the Theta components by originally originated from the v_r term and originated from the v Theta term. So from the derivative of the v_r and v Theta, you can tell that the first part, which has been contributed to the acceleration in a_r and acceleration in Theta respectively will be combined form. So that what you can have is v_r and v Theta angular momentum formulation. Note that from when you're taking the step from here to here some of the a_r components are split it out as of v Theta component and some of the a Theta components are mixed in the v_r components. How about the work and energy, Is that the same? Yes. Total work done is the work done by the r direction and Theta direction. That's correct because when you have a defined definition work done is f and then dot-product of d_r. Those force could be splited out and r components should be also splitted as an e_r and e Theta and dot-product means you have a multiplication for the e_r and e_r Theta r components separately and Theta components separately. So you are up to the point here. You have a work done by the r and Theta, that's going to be same as kinetic energy change, which is m_r dot square, and r, has been moved here, Theta dot square. But that does necessarily means that your r direction or work done is going to be same as m_r dot square and your Theta direction integral is going to be same as r Omega square. So you have equations of motion in r direction and Theta direction and you have an integral over the displacement for the forces, you will have kinetic energy change. But that does not necessarily independent of v_r and v Theta term here, no. That's because again you have acceleration is a r-direction acceleration are contributed by both v_r and v Theta. Then how you can integrate correctly, you have a force F equals ma. If you have an acceleration integral, it's easy. You have a V m_v and total d_v, is the total velocity vector. So you have a norm square, which will give you the r dot square and r Omega square. For F equals ma, you will have a force work done by the r-directional forces and Theta directional forces. That's correct and you will have an acceleration and then you have r vector. Now split it out and e_r vector in the Theta vector. So you do the dot product by component by component. So you have all the r vector components and you have all the Theta component vector. Now if you have another formulate the r double dot as a numerator and a denominator form, d_r dot d_t for simplification later you have all four terms here and here, here is a trick. You have r dot Theta dot d Theta, which is d_r, d_t, d Theta, d_t and d Theta. If you combine this one as a Theta dot square, you have r dot d_r dot r Theta dot squared d_r 2_r Theta dot squared d_r and r square Theta d Theta. This one is Theta dot. Now, instead of just calculating the practices separately, you have to rearrange them as r dot d_r separate. Then this green line term should be calculated together. Then those orange lines underlined terms all together as a blue line. What this blue line stands for. If you take a derivative of r Omega total square as a derivative, this is the one. So this is the 1.5 derivative of r Omega total square. So what you can have is the kinetic energy in r direction and kinetic energy in the Theta direction again. So your work done by the r-direction and Theta directions are all merged into the total sum of r direction and Theta direction or velocity square not separately contributes to term by r directional work done is m_r dot square and Theta directional work done to the r Omega dot r Omega squared term. So let's check where I made a mistake. Work done in the r direction, d_r. m_a r d_r, which is d_v r d_t d_r and this one is going to be v total v and I have a dot products, only the v_r term has been left. So I have m_r dot square term. Seems like correct where no mistake. Yes here. Since a_r has been contributed by both the derivative of v_r and the v Theta. So this step is wrong. How about the other approach? W_r is going to be FR ma So acceleration has been written and rearrange it. I just take a derivative here, 1.5 2 r dot square and here m_r squared Theta dot squared. What has gone wrong? Yes, here. When you take the derivative, that means when you take an integral, note that your integral parameters are independent or dependent to the integrating term d_r. Again, Omega is a function of r, so Omega cannot be handled as a constant. So this step is wrong. In this ME question, we've just gone through the integral step of the equations of motion in polar coordinate to obtain the work energy and linear impulse momentum and angular impulse momentum. Note that when you take a polar coordinate, try to handle the whole integral as a vector form, not separate for the r component and Theta components.