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Current time:0:00Total duration:13:56

AP.PHYS:

CHA‑4.D (EU)

, CHA‑4.D.1 (EK)

, CHA‑4.D.1.1 (LO)

, INT‑3.F (EU)

, INT‑3.F.2 (EK)

, INT‑3.F.2.1 (LO)

all right so we know how to find the torque now but who cares what good is torque what what good is it going to do for us well here's what it can do we know from Newton's second law that the acceleration is proportional to the force what we would like to have is some sort of rotational analog of this formula something that would tell us all right we'll get a certain amount of angular acceleration for a certain amount of torque and you could probably guess that this angular acceleration is going to have probably something with torque on top because torque is going to cause something to angular ly accelerate and then on the bottom maybe its mass maybe it isn't that's what we need here if we had this formula this rotational analog of Newton's second law then by knowing the torque we could figure out what the angular acceleration is just like up here by knowing force we could tell what the regular acceleration is so that's what I want to do in this video I want to derive this rotational analog of Newton's second law for an object that's rotating in a circle like this cube all and not just rotating in a circle something that's angular ly accelerating so it'd be speeding up in its rotation or be slowing down in its rotation so let's do this let's derive this formula so that if we know the torque we could determine the angular acceleration just like we determine regular acceleration by knowing the force and Newton's second law so how do we do this in order to have an angular acceleration we're going to need a force that's tangential to the circle so in order to angular ly accelerate something you need a force that's tangential because this force is going to cause a torque so let's say this is the force causing the torque we know how to find it now remember torque is R times F times sine theta but let's make it simple let's say the angle is 90 so that sine theta will end up being 1 because sine of 90 is 1 and let's make it simple 2 in this way let's say this force is the net force let's say there's only one force on this object and it's this force here well we know that the net force has to be equal to the mass of the object times the acceleration of the object and you're probably like big whoop we already knew this what's new here well remember we want to relate torque to the angular acceleration so let's write down the torque formula how do you find the torque from force remember that the torque from a force is going to be equal to the force exerted that torque times are the distance from the axis to the point where the force is applied now in this case that's the entire radius because we applied this force all the way at the edge if this force was applied inward somewhere it would be only that distance from the axis to the point where the forces but we applied it at the very edge this V F times the entire radius and then there's also a sine of the angle between F and R but the angle between F and R is 90 degrees here and the sine of 90 degrees is just 1 so we can get rid of that so this is simple the torque exerted by this force F is going to be F times R what do we do with this we'll look it down here we've already got an F down here if you're creative you might be like well let's just multiply both sides by r down here that way we'll get torque into this formula in other words if I multiply the left side by R I'll get R times F and now that's going to equal R times the right hand side is going to be R times M times the acceleration and this was good look at now we have R times F that's just the torque torque as R times F or F times R so I've got torque equals R times M times the acceleration but that's no good remember over here we want a formula that relates torque to angular acceleration not a formula that relates torque to regular acceleration so what could I replace regular acceleration with in order to get angular acceleration maybe you remember when we talked about angular motion variables the tangential acceleration is always going to equal the distance from the axis to that object that's got the tangential acceleration multiplied by the angular acceleration alpha so this is the relationship between alpha and the tangential acceleration is this tangential acceleration it is because this was the tangential force so since we took the tangential force that's going to be proportional to the tangential acceleration these are both tangential here and these forces are all tangential that means I can rewrite the tangential acceleration as R times alpha and that's what I'm going to do I'm going to rewrite this side as R times alpha because o our alpha is the tangential acceleration so this whole term right here was just tangential acceleration and now look what we got we've got torque is going to be equal to R times M times R times alpha I can combine the two R's and just write this as M times R squared times alpha the angular acceleration and now we're closed if I wanted a form of Newton's second law I could leave it like this or I can put it in this form over here and just solve for alpha and get the alpha the angular acceleration of this mass is going to equal the torque exerted on that mass divided by this weird term this M the mass times R squared and this is what we were looking for this is what we're looking for over here I'm going to write it in this box the rotational analog of Newton's second law for rotation is this torque divided by this term here this M R squared what is that well it's serving the same role that mass did for regular acceleration and the regular Newton's second law and remember this mass was proportional to the inertia of an object it told you how hard it was to get that object accelerating how sluggish an object is how resistive it is to being accelerated that's what this term down here is going to be people usually call this the moment of inertia but that's got to be the most complicated name for any physics idea I've ever heard of I don't even know what this means moment of inertia that just sounds strange it's represented with the letter I and it's serving the same role it's in this denominator just like masses and it's serving the same role it's serving as the rotational inertia of the system in question so in other words something with a big rotational inertia is going to be sluggish to angular acceleration just like something with a big regular inertia is sluggish to regular acceleration so if this ball and we can see what it depends on look at for a ball on the end of the string the moment of inertia for a ball on the end of the string was just M R squared this was the denominator this was the term serving as the rotational inertia for this mass on a string and what that means is if you had a bigger mass or if the radius were big this object would be harder to angular ly accelerate so it'd be difficult to get this thing going and start speeding it up but on the other hand if the mass were small or the radius were small it'd be much easier to angular ly accelerate you could whip it around like crazy but if the mass were very big or the radius were big this moment of inertia term would get much bigger this is the moment of inertia for a mass on the end of a string and that's what the eye is here so you can think about it as the rotational inertia that's a much better name for it people are coming around and realizing that you should just call it this because that's what it really is this moment of inertia is kind of a historical term that's stuck around not a very good one rotational inertia is much more descriptive of what this eye really is and we should note the units of this moment of inertia since its mass times radius squared the units are going to be kilogram meters squared these are the units of moment of inertia and this is the formula if you just have a point mass and by that I just mean a mass where all of the mass is traveling at the same radius in a circle it doesn't have to be tied to a string this could be the moon going around the earth but as long as all of the mass is at the same radius and traveling around in the circle release mostly at the same radius let's let's assume this little radius of the sphere is really small compared to this radius of the string if that's the case we're basically all the mass is traveling around in a circle at the same radius this would be the formula to find the moment of inertia so how does this ever get harder how'd it where do you have to look out for well we only considered one force you can imagine maybe there's many forces on this object maybe there's some other force this way well in that case you just have the net force here to make sure it's M times a and you just have to make sure you use the net torque up here so this formula will still work if you have multiple torques on this object or the system you just have to use the net to work up here you add up all of the torques where torques trying to rotate it one way would be positive and torque trying to rotate it the other direction would be negative so you'd have to make sure signs are correct up here and what about rotational inertia what if your object isn't as simple as a single mass so what do you do then let's look at that let's take this formula here copy that let's get rid of all of this and let's say you had this crazy problem you had three masses now and one force on this mass - was 20 Newtons downward and one force was upward 50 Newtons on this mass one and they're all separated by 3 meters and can rotate we're stepping it up as complicated can rotate in a circle but we can do it we can do it with the formula we just derived let's use that this is going to be useful let's say the question is what's the angular acceleration for these masses and this particular setup of forces we're going to use this formula for Newton's second law in angular form will say that the angular acceleration if that's what we want is going to equal the net torque how do we find the net torque now there's two forces well it's not that bad you just find the torque from each one individually and you add them up just like you would do with any net vector find each individually and add them up but it's not going to be 50 minus 20 these are torques we've got a plug torque in up here not force this is going to be a torque and until you multiply that force by an R it's just the force so don't try to just stick this 50 up in here it needs to get multiplied by an R what are be careful you might think three meters but no the R is always from the axis of rotation which is the center all the way to where the force is applied so the torque from this 50 is going to be nine meters times the 50 Newtons now we've got a torque it's not a torque until you multiply that force by an R so that was the torque from the 50 Newtons how about the torque from the 20 Newtons you might be like alright I got it now it's going to be 20 Newtons but I can't just put 20 right we got to multiply by an r it's got to be 20 Newtons times and it's not 3 it's always distance from the axis so it's from the center all the way to where this 20 Newtons was applied and that's going to be 6 meters and sometimes when people get here they're just so happy they remember the are they just do plus and without thinking about it but they're going to get it wrong you can't do that look at this 50 Newtons was trying to rotate the system counterclockwise right the 50 Newtons trying to rotate it this way the 20 Newton is trying to rotate it that way there in each other these are opposite signs of torque so I've got to make sure I represent that up here I'm going to represent this 20 Newton torque as a negative torque and that's the convention we usually pick counterclockwise is usually positive and clockwise is usually negative but no matter what convention you pick they've got to have different signs in here so be careful there so that's our net torque up here how do we find the rotational and inertia or the moment of inertia well we know from the previous example the moment of inertia of a point mass that is a mass going in a circle where all of the mass is going at that particular radius it's just mr squared but now we've got three masses so you might think this is hard but it's not that hard all we have to do is say that the total moment of inertia is going to be the sum of all the individual moments of inertia so we just add up all the individual moments of inertia in other words this is just going to be the moment of inertia of mass one if that's one kilogram that's going to be one kilogram times R squared that's what this means you take all the masses M one times R one squared plus M 2 times R two squared plus M 3 times R 3 squared you keep going if you had more you just add them all up and that would give you the total moment of inertia for a system of masses so if we do them one at a time this 1 kilogram times the R for that one would be 9 meters that's distance from the axis to the mass that'd be 9 meters squared plus all right mass 2 if that's 2 kilograms and that's going to be times 6 squared and now we keep going we take this 3 kilogram mass and we also add its contribution to the moment of inertia or the rotational inertia and that'd be 3 kilograms times it's only 3 meters from the axis squared so times 3 meters squared and if we add all this up and pull all this into the calculator we'll get that the alpha the angular acceleration is going to be one point eight three radians per second squared so that's the rate at which this object would start accelerating if it started from rest it would start to speed up in this direction and start speeding up over and over and over if these forces main contained the torque that they were exerting so recapping just like Newton's second law relates forces to acceleration this angular version of Newton's second law relates torques to angular acceleration and on the bottom of this denominator is into the mass it's the rotational inertia that tells you how difficult it's going to be to angular ly accelerate an object then you can find the moment of inertia of a point mass as M R squared and you can find the moment of inertia of a collection of point masses by adding up all the contributions from each individual mass