so we're going to talk about torque that's why I put this one here down from Family Guy you know it grinds my gears when the T Cell strength of toilet paper is lower than the torque required to unroll it I E when it pulls off a piece of paper pulls off too fast okay so let's talk about torque itself what do we mean by this torque is a little bit tough because it's it's it's not an exact Ono one um version of Something linear so I'm going to say it's the rotational equivalent of force but only kind of okay it's not exactly so all right but what is it well it's not exactly a force it's a rotational kind of equivalent so watch if I've got some kind of pivot point so this point right here and I've got some kind of stick going out well then depending on what I do with this thing so if I apply for example a force this way or if the force is straight down or something then there's going to be some kind of you know lever action here that's going to uh be formed so it's not exactly a force uh but we'll see it's close to that and it depends on the r here and we're going to use a symbol a Greek symbol called too for torque okay it's going to be called ta here and the equation is going to go like this this is on your data booklet so torque equals f * R * s of angle Theta so this you don't have to memorize it's in your data booklet okay so let's try to figure out what are the different uh units here well torque we'll leave that one for a second here f is going to be the applied force so in this case this would be a force either going down at an angle or something um R is the distance from the axis of rotation so it it depends on how far away this whole thing is how long it is then Theta is the angle between the applied force and R and that uh that Theta is measured in degrees it is not the same as angular displacement remember that one that's measured in uh radian so just be aware of that so basically depends on where your angle is and a nice exam tip here is this very very often uh the center of mass is going to help for toao and Alpha so in other words so if you're considering some sort of stick like this well where is all the mass the mass isn't all at the end it's you know everywhere but what we do then a nice trick is we just consider where is the center of the mass and just deal with that as a point that's moving around and of course if the angle is the Theta is 90° in other words if the force the applied force is straight down for example here compared to this horizontal here on this on this R here um well then because Theta is the angle between uh the applied force and R if theta equals 90° well what's the s of 90 s of 90 is just 1 so what does that mean that means the equation for too just becomes to = f r * 1 which is just to equals F and this is the very very common end result it's quite rare actually on exams that you're going to have to deal with the sin Theta but just in case you need it uh you have the angle but this is the more useful one you're going to end up probably using tal f * r a lot more so let's go back and look at Newton's first law if you remember it from linear terms says that an object in motion stays in motion unless otherwise uh unless it's acted upon by like an external unbalanced or net force well there's a rotational version as well and it says that hey it's going to keep rotating at a constant speed as long as there's no resultant or net torque so that's what I'm just trying to put down here so if there's no resultant net torque it means there's no acceleration it just keeps spinning at a constant speed just like in linear terms you know if there's no net force then there's no acceleration it goes at a constant speed and keep in mind so we have something that we call rotational equilibrium that's just if there's no net torque all right let's do Newton's Second Law Revisited so do you remember the equation fals ma that's one of the formulations of Newton's second law that the net force equals mass time acceleration well we have a version here now remember I said that force is kind like torque so that would be T instead of M do you remember what we put for M we put in moment of inertia and instead of a we say Alpha so this is also an equation from your data booklet so let's remind ourselves what are all the letters here we've got to is torque I is moment of inertia and Alpha is angular acceleration this one for example is radians per uh second squared so that's an important one there okay uh what about moment of inertia well if you remember it's Mr s or some Sigma of that so it's going to be uh kilog time me squared and torque now of course you could do it in these units here but remember torque was the same thing as f * R so it's Newtons uh time meters it's a newton meters here and here's a nice little tip here for exams is that a net torque means it accelerates I'm just repeating this it's really important okay so if it's a net torque it's going to have an angular acceleration I put this one here because when you're forever alone and your physics book trolls you look calculate the total torque acting on this oh God this complicated looking heart let's figure out the torque Direction remember I said torque isn't exactly like force here's where it's not exactly like force so let's consider something that's rotating in this case in this direction like this we've got R and we've got an applied force this way what we do is we use our right hand rule in my case it's my right hand here and what do you do put your fingers so start your fingertips in the direction of R and then curl them in the direction of f in my case up and then curl them to the left and then my thumb is going to be pointing in the direction of the torque so in this case right here tow that's the torque and this one here technically then is going to be the direction of rotation you can see I mean that's what's going to happen here um so we're going from R to F so in this case right here if I was going to draw it this way torque goes that way it's a bit weird I can't imagine they're asking you very much about this other than just yeah if something is rotating which way is the torque well I guess it just use your right hand rule and torque is your thumb there we go so you're much more likely going to be needing to calculate the value of the torque that's going to be really important or you're going to be asked you know to find the angular acceleration which came from the torque so you'll see that the torque is going to be useful for calculating um so let's actually do an example where we have to calculate something so let's look at a pretty tricky question but if we can solve this we can solve anything so we have a uniform metal rod that means it's the same everywhere along the rod and it's set about some sort of pivot here and it's resting horizontally on a support that's frictional so here is a support so if you see it's kind of like my little pencil right here it's kind of sitting right here on the p on the support of course if I let the support go it's going to you know fall down so you know that's what we're going to be looking at in Part B but in part A we're just going to look at okay this right here is what's going on right now so it's it wants to spin or wants to rotate but this support is stopping it so the question would be what force does a support exert on the rod so let's take a look at this and see if we can figure out what's going on for starters uh let's look at this here and see about these applied forces because remember we often needing to know about um R for example the radius goes this way and the force because it's due to gravity is this way and because of that we have that the angle then between them is 90° that's going to be important because what does that mean do you remember the equation for torque torque goes um f * R * s of theta but if Theta is 90° s of 90 is 1 so that means what that means then that the torque is just going to be equal to f * R that's going to be an important first piece here we're going to be using so in other we can just multiply the force time R this distance out okay well the very fact that it's holding still tells us it must be in equilibrium okay so if it's in equilibrium there must be no net torque okay well what does that mean well that means that you know this this sort of torque caused by this sort of direction must be equal to the torque this way in other words the one caused by gravity you know that it wants to do is the same thing as it's like a normal force with this one here acting upwards they have to be equaling each other out or else this thing would be accelerating in some way so there must be no net torque okay well that means then I can say something like you know fub1 * R1 is going to be equal to F2 R2 in other words you know this if I deal with FS1 as just like the whole Rod right here this complete length of the rod and F2 R2 is going to be this one right here okay so this here I'll call this R2 I'll call this one here R1 okay well if that's the case then it should be straightforward but it's not quite because we have to use something really important here this Rod where is its mass the mass isn't at the end it's an actual physical Rod so in other words we have to use so for the rod we have to use the center of mass that's going to be really important here so we're going to use the center of mass of the rod so in other words it's not sitting at six it's actually technically sitting at three that's going to be the center of mass so let's deal with F1 R1 so F1 let's see it's going to be um the weight of the rod because you know f equals mg so we can say then this is going to be um the weight of the rod so 40 Newtons times R1 in other words times is 6 M all that over two and that's because of this Center of mass here so that's why so that I don't think was obvious maybe I'll make that one actually in um I'll make it in yellow here so you can say so the center of mass that's why we divided by two okay that's going to be the same thing as and by the way this was FS1 and that right there was R1 and then FS2 what's F2 that's actually what we're trying to find isn't it how that's the force we're looking at that so F2 * R2 what's that distance that's the distance out to the rod so that must be 5 m okay so FS2 * 5 and this is all of course divided by two well then if I want to do this then I just put it all together and get the answer so let's see here so I've got FS2 then equals um it's going to be this uh 40 * 6 well 4 * 6 is 24 so I add a Zer so that's going to be 240 uh divide that by two um and actually divide that by 10 that's because I'm dividing by the five here well that makes it 240 divided by 10 so that's just going to be 24 so the force uh uh that the support exerts on the rod then will be 24 Newtons okay so now we have a situation where the support is removed and now the rod begins to rotate about the pivot so in other words now it's going to start you know actually rotating like this and the moment of inertia of the rod about the pivot point is 30.1 kgr M squar so that's what's that letter moment of inertia that is I okay so now what well we want to know the initial angular acceleration so in other words we want to find Alpha now how can we deal with this well we've dealt already with a torque before remember from uh before at least so let's just find out what is the torque actually um of this Rod right here so what is the torque of the rod um or Torque experienced by the rod so this one right here if we do this one here remember again we're going to use this equation toal f * R and remember this whole length right here was 6 M so again we're going to say it's the same sort of thing we did before it's going to be the force which was 40 Newtons * 6 M that was F1 that was R1 but remember we divided by two and remember why we divided by two because of the center of mass okay that wasn't obvious so that's why I want to point it out here so if we do that one right there um well we're left with let's see here we've got 40 * 6 which is 25 40 / 2 so basically we get um this value then that the torque is going to equal 120 now keep in mind uh these are newton meters so this is important okay but now what so now that we found the torque now what well remember if we want the angular acceleration we have an equation for that don't we we have from Newton's Second Law instead of fals ma we have torque equals I Alpha so because of that then we can just solve for Alpha can't we we can just say hey Alpha is just going to be the torque divided by the moment of inertia okay well I know both of those don't I look I know everything here I need so the torque is going to be 120 and the moment of inertia is going to be 30.1 let's go ahead and calculate that so here's my trusty calculator I do a nice pretty fraction and I say hey what's 120 divided 30.1 and I end up with an answer of 39867 well that means my answer for Alpha then if I want it to two significant figures um which I think I was allowed from before then I would say this here is a 4.0 and this would be radians per second squared um if you want to do it to three significant figures of course then you could say it is um I just going to move this here or you could say it's uh you know 3.99 for example either way we have the initial angular acceleration