Okay. Now let's work on how we can obtain the equations of motion in curvilinear motion. Basically, it's exactly the same for the rectangular coordinate system like linear motion. The only difference is you have a different coordinate so that your expression for the acceleration is different. Okay. So remember in chapter two we learned how we can express the acceleration for rectangular coordinates, simply ax and ay, and when you are using tn or nt coordinate your normal direction or acceleration is expressed by v square over rho, and tangential is just v dot and if you're using the polar coordinate, there are how many, four acceleration components, two for the r components which is r double dot minus r theta dot square and two components for the theta direction, which is 2 r dot theta dot plus r Theta double dot. So only difference is you have a different expression for the a. However, the rest of the steps are exactly the same. So you draw the free body diagram and express pictorial relationship between force and motion into mathematical form, that's really how you can obtain the equations of motion. Now, let's work on the example. Okay. There is a small block of mass m it's on the bar OA, and then OA, or with the distance r from the pivot point, and this bar is rotating counterclockwise with a constant angular velocity omega, and it's about to slip impending slip at theta is theta naught, and what's going to be the coefficient of static friction between the block at the bar? Okay, the first step is to find the coordinate which coordinate are you going to use? Yes, polar coordinate you have r and theta so I'm going to set the polar coordinate origin. Okay, now yeah, and then I will remind you there are four acceleration components two for the t and two for the theta, that you may have by now remember are double dot minus r theta dot square, 2r dot theta dot plus r theta double dot. Now draw the free body diagram, you free the block, how many contacts only one contact with the bar here. So you have horizontal and vertical forces and the gravitational forces that will generate the r directional acceleration and the theta directional acceleration. So this pictorial expression between the force and the motion can be transferred to the mathematical way, mathematical form. So r direction you have a friction and a gravitational sine component, and that will generate the acceleration in the r direction, and theta direction, you have two forces as well, like a normal force and the gravitational force, will give you the theta directional acceleration. Now, this is the end of getting the equations of motion. Now from the problem what it said was its impending slip, impending slip means for now you are not slipping yet. So those r is going to be constant values so r dot and r double turns out to be zero and also the omega is constant. So theta double dot turns out to be zero. So this gave you the informations about the normal force, normal force now turns out to be mg cosine theta, so if you plug that into the upper equations of motion by the friction is going to be mu n, then you will end up having the relationship or the static friction coefficients in terms of the parameter that you know already, what's been given already. And note that this r omega square here, that's what we call this a centripetal acceleration. Okay. Now I said about the coordinate as a polar in the beginning, but can you use tn and t coordinate as well? Yes, of course. So if I set the tn coordinates, so note that the t direction is parallel to the velocity, right? So since it's rotating circular, so the velocity at this point, since there is no r directional motion, is perpendicular to the distance from the moment arm from O and A, when the box- so if this is a t direction and then this is n direction. So acceleration for the t an is p square over rho, t is just v dot. Now draw the free body diagram, as I said, the free body diagram doesn't matter which coordinate you chose, so same gravitational friction and normal force. Now those force components will be splitted into t and n direction. So you have equations of motion in n direction which is friction and mg sine theta is going to be mv squared over r, mv squared over rho, which is r, and t directional equations of motion, two forces will give you on mv dot here, but the since omega is constant the v dot is zero, so you ended up having the same equation. So you plug in the end information to the top equations of motion, then you will end up having equations of the mu s. Okay. So exactly equivalent to the equations of motion while you have obtained in using the polar coordinate. So we would conclude that the coordinate doesn't matter when you obtain the equations of motion. So when you draw the free body diagram, it doesn't really matter which coordinate you chose and the resultant acceleration could be either expressed by the polar or the tn or even Cartesian coordinates. So you can obtain the equations of motion based on your coordinate notation and those equations are all equivalent. So in this video we learn how we can formulate the equations of motion for the curvilinear motion and you should note that everything is same except the expression for the acceleration, depending on where you set the- which coordinate you chose to describe the motion.
