okay everybody let's going with our discussion of rotational motion um we're g to last time we talked about rotational Dynamics when things can accelerate um we're going to now talk about this special case where we have what we call rotational equilibrium um so for rotational equilibrium that basically means that something is stationary we often look at stationary things but it could just be um no acceleration so um conditions for our rotational equilibrial so sorry um say that the acceleration and angular acceleration are going to be zero or you can think of it as the sum of forces is zero and the sum of torqus is zero this is how we're going to utilize that what we're going to utilize in order to figure out uh stuff about rotational equilibrium so this is largely for us things that are stationary that could rotate but aren't um okay so because you're things in rotational equilibrium turns out the the axis of rotation um is a bit arbitrary can't write that in top um so was it that mean for us it means we can really pick a convenient Pivot Point a p convenient axis of rotation um as opposed to one that's physically like the the hinge of a door definitely the axis of rotation but if we consider some other things we can set our axis to be somewhere that's just convenient um so that's that's something we're going we are definitely going to exploit um so if we look at hockey sticks um if we push so the on the left we have equilibrium okay um we're going to push equal on both sides and we're also going to push um since we pushed at the same position um it's not going to rotate either those torqus are even but if we push at the top and the bottom we push equal side to side which means the net force is zero but because the the forces the torqus aren't lining up that hockey stick can rotate so non so you get non equilibrium because your torqus are nonzero or the net torque is non zero um so example so excuse me I started this video and immediately start yawning um so in order one common torque we're going to have can I take a step back for a second um is gravitational Force the torque due to gravitational force right because everything's got weight and we go oh no where is that weight that gravitational force actually acting so we're going to let because the gravitation Force acts all along the thing that you're looking at right um we're going to set the gravitation Force to act at center of gravity so it's as if all the mass is concentrated at the center of gravity we treat that as the place that the gravitational force is acting um so it's it's a way to model the gravitational torque so the center of gravity is the point at which the application of the total gravitational force would have the same effect on the rotation as the sum of gravitational force of my two those pieces is basically what it just said center of gravity is this spot where everything lines up oh an example of this let me move my head out of the way um so I I have little my little tablet pen here if I find haha the center of gravity is is above my finger I put the the center of gravity on the pivot point this case my finger is the pivot point and so I can balance the pen on my finger right that's that's my thumb in the background um if I slide it off so you can find the center of gravity of a thing by balancing it on a on a small object so how can we calculate the center of gravity great question um using this guy these two equations so center of gravity it's the sum over all the pieces right of the piece M times uh X which is the position the X position um on some sort of coordinate system uh over the total so you could write this like M1 X1 plus M2 X2 plus y yada yada over M total now give you the position of the center of gravity in the X Direction similar for the y direction um this is written for Point particles a set of Point particles in some sort of space um however we don't often have a set of Point particles we've got you know like a ruler uh a pen uh a seesaw board we're often going to assume that our objects are homogeneous that the mass is evenly spread out the density is the same throughout um we're also going to assume that the center of gravity is basically the geometrical Center because the mass is evenly spread out now is this always true no in real life that's not always quite true true um if we take a meter stick and you try to balance the meter stick you would G like geometrical Center is 50 cm but depending on how it's made how much how dented it is you might be off a couple centimeters either way so we're going to assume in our theoretical calculations Center gravity is in the geometrical center of the thing so if you have a long wooden board it's in the middle of the board that's can be useful um so I'll probably set up a center of gravity calculation for class um I don't have one right here but we'll we'll do it later so let's employ uh conservation momentum rotational equilibrium to the Seesaw this seesaw so a board of mass m Big M serves as a seesaw for two children child a is on the left child B is on the right um child a has a mass and sits so far from the pivot point right you got a little it's sitting on a little triangle to Pivot uh so his center of gravity is 2 and a half from the Pivot Point at what distance X should the P should child bead sit um she also has a mass um in order to balance the Seesaw so by balancing it you have rotational equilibrium assuming your board is uniform and centered over the pivot Okay so we've got a lot all kinds of good stuff going on here I'm going take it real slow we can see all the pieces um and try to remember all the all the steps so I got all kinds of good stuff I want to write down what I know first all the numbers I know and go from there so I have the mass of the board uh two kilograms that seems kind of light that's okay uh the mass of child a is 30 uh the distance of child a to the pivot point right which is what the distance they're telling me it's 2 and2 m um RB is X is what the mass of child B is 25 kilograms um assume the board is uniform so the center of gravity is in the middle and then um it's centered over the pivot point so that the mass that's not how I want to write that um the gravitational force of the board is over the pivot point that will become critical later okay so because it is balanced we have equilibrium so our sum of forces should match and our sum of torqus should match okay oh we're going to start summing forces we're going to sum torqus we need what I call an extended free body diagram so a free body diagram we had all the forces just smashed onto um a single dot we said which way are they going because forces don't actually care where they're applied they just care which way torqus care about the where so we extend the fre body diagram to show where the forces are applied and get a get a sketch of what's going on so um I'm going to use the board as the base for because all these forces are acting on the board right so um I have a kid pushing down FG of kid a right child B is pushing down over here I'm also going to show a little pivot point because that'll be used that's important later too um the gravitational force of the board is centered on the pivot so it's at the Pivot Point these are not to scale right the board's only two kilograms he should have a much shorter line now if we look at this we go ooh all of my forces are down in order to balance my forces is something has to go up well the board's sitting on the pivot point so the pivot point is providing some sort of normal force upwards the Pivot Point pushes up so okay we want to know so we've labeled let's see okay so we can use our extended free diagram to write the sum of Tores you can also write the sum of forces on it from it in this case we don't need the sum of forces because we don't actually care what until the normal force and we know all the other ones so I'm just going to go straight to the sum of torqus so my sum of torqus is zero right that's a condition um I'm going to say that the torque um positive is out of the page right positive is up positive is Right positive is out up out yeah um okay so I need add the torqu so Torque from a and you can just you can look at your your free body diagram and say okay I got a torque for each of those forces we're going to write a torque for each four of all four of those forces and go from there so then I have the torque for the normal the torque for the board the torque for the child B right okay so torque we remember is r F sin Theta all these forces are acting perpendicular to the board so all of our signs um are one so we're going to drop those all of our signs are sign of 90 so become one so we just have RFS um so we have r a fa one um RN n FN one right all these ones are for the Signs R board F board one and then RB FB one okay so let's figure out we we know all this we know some stuff for the two kids let's look at the distances for the normal force and the and the board so where do I put it put them under the Ling okay so if the the the normal force is caused by the PIV by the pivot and so it's sitting on the pivot how far apart is the pivot point and the force give it give a contemplation is zero why is that because the force is on the pivot point there is no distance between the pivot point and the force so there is no lever arm so those torqus are zero similarly for the force from the board because our gravitational force is on the pivot point there is no distance so that torque is zero so the normal force and the gravitational force of the board aren't contributing to the torqus because they have no torque it's like if you push on the hinge of the door it's not going to move the door because you're on the hinge there is no le arm okay so we're down to oops oops I push it but I pushed the button hold on there we go um R A F A Plus RB FB okay it's been a while since I've done this let's we forgot I forgot um we also have to consider our directions so um the middle two torqus zero they have no Direction so we have to consider the direction of A and B so we can add and subtract appropriately so for a the lever arm is to the right right he goes over this way so we go right and down a has a torque that's coming out at me he is Plus for B the lever arm is to the right the force is down the torque is in oops so he is negative oops that's not where I wanted to put that that's right I can put it there too we need that negative okay so um I can add RB FB to the other side right because you're subtract you have torque a minus torque B add torque B to the other side um we want X which is RB um ra a and the gravitational force force a is gravitational you have the mass G um and then you're going to divide FB which is mass time G G's are canceling so we're going to have 2 and2 m time 30 kg over the 25 kg it's going to be three right right three so the lighter child has to sit further from the Pivot Point in order to create the same torque as the heavier ch um okay recap what do we do uh first we figure out all the things we do right just wrote our numbers down you draw the extended free body diagram off the sketch off the off the picture Center every body diagram um draw all the different forces you have um from there you can figure you're going to write your sumat torqus which is going to equal zero because it's an equilibrium it's balanced um so we add up all the torqus that come from you have a you have the option the possibility of a Torque from every single Force um the two forces that are on the Pivot Point produce no torque because there is no distance there's no lever arm um we had to consider the directions of our two torqus that that existed that remained not zero um we can add subtract appropriately and then you do some math to sub in the fact that the forces are MGS so now you figure out how far that little girl has to sit from the Pivot Point excuse me okay so oh I gave myself space um I have another one we're going to do that guy in class why because I think we get so much more out of it when we sit together um so one thing I do want to talk about before we go from this video um is the fact that the forces that act in a rotational equilibrium they don't actually have to be perpendicular to your L arm both a seesaw and and this this guy that's sitting here on a board um all those forces are perpendicular but the the they don't have to be so your s Theta does not or you can have have can be sin Theta doesn't equal one so you would have to include the S Theta bit in your sum of torqus so those torques consider the angle between the lever arm and the force um you can also consider situations that are not in rotational equilibrium which we did uh in the previous video um so something sliding as it goes around um a case that I've seen a lot for INE but not uh not all perpendicular forces is like you have a ladder on the corner right you say oh if I know some stuff about the situation like here's my little ladder pretend I can draw um I don't about the situation but like the wall is pushing here this guy's pushing here and like so there's they're not perpendicular to the L arm which is the ladder it's a good time um so know that uh so oftentimes the sin Theta is one but not always so still think about it still consider it okay I'm going to leave this here um make sure to do your checkpoint let me know what questions you have what more other types of examples you want to see um I from there so do your checkpoint I'll see you in class