all right this is going to be a review of every topic on the AP Physics e mechanics exam uh the first topic is kinematics uh so if the AP be given you a position as a function of time guess what you're probably going to take a derivative of it because the derivative of the position equals the velocity so if I take a derivative of this I'd get 6 T ^2 - 6 that would be my velocity function right here and if you take a derivative of the Velocity function you get the acceleration so if I take it if the a P be throwing you a velocity as a function of time take a derivative of it that would get you your acceleration over here you can keep taking derivatives that give you the jerk the snap the crackle the pop so on Lock and Drop but you probably don't want to do that if the AP be giving you a acceleration as a function of time you're probably going to integrate it because the integral of acceleration is going to be the change in velocity you know if you just separate this differential equation over here and you integrate you'll get this it's the change in velocity though be careful it's not the velocity it's the difference in velocity between a limit one and limit two whatever those are and so if I integrate both sides here I integrate my acceleration DT I'm going to get 12 T ^2 / 2 that's just 6t ^2 plus a constant what's that constant that constant would be the initial velocity which for our case was -6 so you'd have to you'd have to be given that information you can't just figure it out uh without knowing that beforehand uh what do you do if you have a velocity graph I said you take a derivative to get the acceleration but you could also take an integral uh integrate the velocity graph to get the change in position how far that thing went so if I integrate this with respect to time I'm going to get 6 T cubed over 3 - 6t plus a constant that's just going to be my position as a function of time so I mean there's a geometric interpretation uh the derivative of the position is just the SL so the slope of the position graph equals the velocity the slope of the Velocity graph equals the acceleration the area underneath the acceleration graph equals the change in velocity and the area underneath the velocity graph is the change in position so you know you can do this for any function if you get acceleration velocity or position as a function of time you can integrate take a derivative if you want what's the verb of derivate no derivative take a derivative and you get your vales uh what if the velocity is constant this one's easy I mean this is about the easiest formula you get if the velocity is constant distance equals rate time time do you ever have to use can you ever use such an easy formula on the AP Physics C exam yeah if you got a 2d kinematics problem something flies off a cliff like that in the horizontal direction there's no acceleration in the horizontal Direction there CU gravity just be pulling down gravity doesn't pull right or left that means the distance in the X Direction Just equals velocity X time T whatever that velocity was in the X Direction stays the same the whole way so you could be using that what if acceleration what if well what if velocity is not constant but what if acceleration is constant then you can use the kinematic formulas I mean you could derive these using calculus but they're given to you on the formula sheet if you need them uh this you'd have to use for the y direction because in the y direction the velocity keeps on changing so these are all for the y direction if you want you know acceleration in the y direction is always 9.8 uh but you could typically treat that as just -10 uh if you want to find a value uh what if the acceleration is not constant well okay then you got to use some calculus here do something sometimes you use conservation of energy to get your value maybe you can use conservation of momentum or you just straight up derive it from Newton's Second Law and knowing what you need to do what is Newton's second law well it's this one right here it's the fact that the acceleration is going to equal the net force per Mass uh divided by the mass the inertia of the system uh you know you can do simple problems on this or you can do some harder ones one that the AP likes to do is air resistance so air resistance is typically negative some constant times the velocity or the velocity squared if you want to represent it in that way but often times it's just velocity what do I do with this we can throw it in here so let's say we have some car with a mass m and it's coasting to a stop so it's not pushing on the gas you got your gravity you got your normal force those are just going to balance but you got this resistive Force backward that's going to slow this car down to a stop I mean the graph's going to look like this it started with some speed and it's going to slow down to a stop like that how would you get this fun function velocity is a function of time you'd use this formula here uh you'd say that the acceleration would be BV that's the force backward and it's in the opposite direction of V that's what's important it doesn't matter if it points left or right here the negative here doesn't mean left it means opposite direction of V so if this car is moving right negative right is left if this car is moving left negative left is right so you always need that negative there even if the car was going to the left uh you divide by m this is a function of v this isn't constant um I can't plug this into a kinematic formula V is not a constant here what do I do I turn this into a differential equation how do I do that I write a as DV DT so I'm going to write a as dvdt how do I know to write it as dvdt because my force is a function of v so I'm going do equals negative BV over M now I need to separate this equation I got V and T are my variables and I got a v over here B and M are constants B and M can go anywhere I need all the v's on one side all the t's on the other so I'm going to divide both sides by V and then I'm going to multiply both sides by DT so I'll get b/ M DT I've separated so separate then I integrate from0 to T I'm going to assume I start my clock at Zero from V initial not from zero this car had some initial speed whatever it was to v i integrate both sides on the left hand side I get a log a natural log so I get natural log of V minus natural log of V KN equals B over M on the right hand side I just get T from 0 to T uh and then I'm going to write this difference of logs as the log of the ratio V over V going to equal NE B over Mt I'm going to take e to the both sides because that gets rid of my log so e cancels that I get e to the that I'm going to bring it back up here sorry a little messy I'm going get V over V KN = e to theb over Mt I solve this for V that V that I'm getting is v as a function of time and it is going to be v e to the b/ M * time that's why this is a negative exponential it dies off like e to the negative uh what if I was falling through the air now I've got another Force here Gravity mg how do I do it I do it the same way I plug in up here but I'm going to have dvdt dvdt now I have two forces here I'm going to treat down as positive because the treating the direction of motion as positive typically makes things easier so Down's going to be positive mg minus BV notice I'm still putting the minus here it points up why well I'm falling down positive but that doesn't matter opposite the direction of the Velocity even if I called up positive I'd want to call this negative BV so just be careful there it's always got to be negative because it's opposite directional velocity divide by m i just do the same move I separate I integrate and then I forgot to celebrate integrate separate integrate celebrate I celebrate I got my answer right over here so here we got to separate First Step DV divided by I have to move this whole thing these guys move as a family if you try to separate these it's bad don't separate families what are you a monster take this thing family moves together over here mg minus BV equals DT over M you separate these guys you are never getting this thing to fully separate here now all my V's are on one side my T's are on the other I'm going to integrate uh I'm going to integrate from 0 to T this time let's say this guy started at rest V initial zero so I'm going to integrate from Zer to V KN oh not to V not to V final this is the V final at this time t uh left hand side's going to give me well I do youu you know like U substitution 1 /b if you do that log of mg minus BV on the left equals and then I get uh T over M on the right and be careful from zero to V and you might be like oh but the zero limit yields zero no it doesn't you're going to plug zero into V you are not going to get zero so be careful you're going to get netive 1 let's just move this negtive 1 overb over it's not going to hurt anybody let's move this over get out of the way so 1 overb is going to go over to the right I'm going to plug these in here I'm going to get log of mg minus BV minus log of just mg because I'm plugging in zero for my V the zero limit did not give zero be careful there equals I move my negative B over I'm get negative B over M * T all right I'm going to write this difference of logs is the ratio of logs I'm going to solve solve boring algebra I'm going to end up getting that v as a function of time you know try it on your own if you want is going to be I guess it's mg over B * 1 - e to the b/ m * T this is going to look like this starts at Z at tal 0 1 - 1 is going to give zero and it builds up to some ASM toote what is this ASM toote it's the terminal velocity that's the fastest you're going to get so at some point at some point this air resistance equals the force of gravity and at that point you stop speeding up you don't stop you just continue on with a constant velocity this is that constant velocity it's called the terminal velocity you can get it easier just ask hey when will this equal 0 what equals z when mg equals BV IE when mg over b equal VT so that's solve for this velocity here and it just gives you what the terminal velocity is without having to go through all this mess right here so that's differential equations you know separate integrate celebrate uh sometimes there's seemingly easier Force problems you're like oh this one's easy there's no like differential equations here yeah but let's it's not necessarily easy you got a 4 kogam Mass on top of a 6 kgr mass and there's enough friction here that they just move together right they don't slip right there but it's frictionless on this floor right here so you're going to pull on the 4 kgam Mass with 20 newtons and this whole thing's going to slide to the right and I want to find what's the force on the 4 kgr Mass exerted by the 6 kgr Mass what do I do there's all kinds of masses like what Mass do I plug in you got to keep this straight you got to make sure you're always talking about the same thing you can't be you can't be plugged in acceleration of the six 6 kilogram Mass force on the four mass of the 10 you have to be a you have to be acceleration of six force on six Mass on six but the first thing we're going to do since since these slide together anytime two masses accelerate at the same rate you can treat them as one big Mass so I'm just going to treat this as acceleration of the whole thing equals net force on the whole thing over mass of the whole thing I'm just going to be like well okay fine the force on the whole thing is just 20 you know there's no other external forces like there's normal and gravity but those Balance out there's no other horizontal forces there's internal forces but those aren't you know aren't going to affect the acceleration of the whole system only external forces affect the acceleration of the whole system 20 newtons is the net external Force over 10 kilograms is the mass that gives me two meters perss squared this is useful because this is the acceleration of both of these masses now I can focus on either one You' be like which one it doesn't matter you can focus on either one actually we can do the four let's do the four so the four is going to have an acceleration of two we know that 2 m/s squared I'm using Newton Second Law uh equals so I got 20 to the right minus whatever that force is exerted by the six on the four and that's going to go leftward the six is going to exert a force on the four to the left if you don't believe me I'll prove it to you in a minute divided by well the mass is 4 kg so 4 kg I solve this for f I'm going to get 8 over here equals 20 minus f f in other words is going to be 12 Newtons that's the force on the four by the six what if I had picked the six same thing I could just be like okay fine 2 m/s that's squared sorry that's going to be the acceleration of the six because everything's accelerating at the same rate equals well there's only one force on the six the four is dragging the six with it some Force F it's going to be equal and opposite to this one you know Newton's third law says that force on 4X 6 equals force on 6x4 and that's it there's like gravity and normal but these just cancel out there's no more horizontal forces this is frictionless so I got that Force F I divide by what six that's the mass of the six I multiply look at what I get 12 Newtons they're the same because Newton's third law says force on 4X 6 has to be equal and opposite to force on 6 by4 I was just finding the magnitude here I wasn't really worried you know I threw in this negative that took care of the sign there so really they are equal and opposite I just got the same value because I peeled off that negative and treated this as the absolute value of the force all right uh how does this get harder I mean you they might throw some inclines at you over here what if you got a block on an incline well you got to remember that the component of gravity that goes down along the incline is always mg sin Theta and you can remember this cuz like it's slides down the ramp sometimes people put mg cosine Theta for this but you got to remember like mg that's the noise it makes as it slides down the ramp so I got some four I'm going to treat G as 10 uh because you can do that on the multiple choice and I'm pretty sure free response they don't care 4 * 10 sine of 30 well that's just a half so I get that the force down the ramp here is just 20 newtons uh what about the normal force if I wanted the normal force that's also mg but this one's cosine of theta as long as you're talking about this Theta right here of the the ramp's angle with respect to the horizontal so that's going to be 4 * 10 * cosine of 30 cosine of 30 isun over2 so that's going to be like 40 over 2 * < tk3 we can call that 20 < tk3 newtons okay what do we do with that well what if they give you some coefficient of static coefficient of kinetic if you remember the maximum force of static friction is always going to equal mu FN this is the maximum force of static friction this isn't the force of static friction I'll show you what I mean let's say we just plugged in values mu s is < tk3 okay and then mg cosine Theta we just said is 20 < tk3 newtons so this is going to be 20 < tk3 this is going to be 3 * 20 is 60 newtons that's not going to be the if if this question was posed like this and they said what's the static frictional force don't be telling them 60 newtons you kidding me 60 newtons you're telling me 60 newtons backward 20 newtons forward this box would slide up the ramp it obviously doesn't slide up the ramp because static friction is just going to do the minimal amount it needs to do so this is just how big it could be but the value it's going to be right here if it's if this is not sliding if it's at rest there this is going to be 20 it only exerts the force back that it needs to it could keep doing that until you get to 60 you know if I start pushing forward here I can add another 40 Newtons before this thing would budge that's what this number tells you so this this msfn tells you the maximum static frictional force now mu kfn this is the kinetic friction if you're sliding this is the amount of kinetic friction so let's say this was < tk3 over 4 that that was our coefficient of kinetic friction I multiply that by 20 < tk3 I'm going get 60 over 4 that's going to be 15 Newtons okay let's say I give this a little kick like I just Boop I give it a kick so it overcomes static friction as it slides down the ramp now you could be like well my acceleration parallel to the ramp because Newton Second Law works for every direction even this parallel this parallel Direction down the ramp this is going to equal the force down the ramp which is 20 newtons minus my coefficient or my ktic friction which is 50 Newtons back up the ramp divided by my mass of four I would get 54s m/s squ again it wouldn't do this on its own because this maximum static friction force is bigger uh than this Force down the ramp but if you give it a little Boop so you force it to start sliding then you can go into kinetic friction mode find the acceleration that would be the acceleration down the ramp what's the acceleration perpendicular to the ramp zero this thing doesn't come flying off or going into the ramp so the net force on a block on a ramp is always whatever the force is in the parallel Direction okay how does this get harder well people don't like centripetal they get confused by this it's the same formula it's still Newton's third law it's just the acceleration we plug in is this the centripetal acceleration which is always directed into the circle in toward the center of the circle for any object going in a circle your cental acceleration points into the circle and this is it turns out is v^2 over R so you just take this plug it into there and suddenly you've got Newton's second law for a centripetal motion problem what forces do I plug in only forces that are like into you know radial I'm going to put radial forces forces that are like into or out of the ramp into is going to be positive out of is going to be negative what if it's parallel you don't put that in you don't put any tangent forces in here only forces in toward the center of the circle or out that's why I call them radial so like down here you'd have normal force on this car you'd have gravity right here how would you use this formula you'd have something like V ^2 over R is going to be all right forces into are positive so I got to put FN minus mg / the mass and then I can solve for what do I want to solve for how about up here I got gravity straight down still this time I have normal down this this Force up here is down like this this is force uh surfaces can only push this can't pull so the force from the roof here would be pushing down on the car now they're both positive because you might be like well they're both negative CU they're downward no they're both positive because they're inward toward the center of the circle so I'd have FN positive plus mg over M what if the car's on the side of the wall now FN points this way gravity's down you know I got some other Force this way if I want maybe he's on the brakes or something doesn't matter these tangential forces don't matter all I'd have here is v^2 over R would be forces into that's still positive FN goes into over M and that'd be my centripetal for that case so be careful into forces are positive out of forces are negative and tangental forces don't get plugged in here at least if you have v^2 r on the left hand side what if you got a like a orbits problem same issue you're going to put v^2 over R over here it's going to equal the net force over the mass it's just the force of gravity on an object you know between planets and whatnot the most General Newton's universal law of gravity is FG is going to be big G at least the magnitude of this Force Big G mass of one of the objects mass of the other object Earth and Sun divid by the distance between them squared that's what you would plug into here but you might be confused you might be like all right fine I'll do that and mass of the Earth mass of the sun distance squared you might be like but what is which mass is this is this Mass down here the mass of the Earth or the mass of the Sun well it depends do you want to find the velocity of the Earth or the velocity of the sun it's probably velocity of the Earth I mean Sun ain't going nowhere Earth's the one that's going to have the speed well if this is the velocity of the Earth then this is the acceleration of the Earth this needs to be the force on the earth and that needs to be the mass of the Earth so this needs to be mass of Earth that means mass of Earth cancels and the speed of Earth does not depend it's kind of counterintuitive does not depend on the mass of the Earth what it depends on is the mass of the sun because the mass of the Earth cancels out and it's like duh yeah because the mass of the Sun is what's like pulling this thing toward it change the mass of the Sun you'll change the speed of the earth change the mass of the Earth you won't change the speed of the earth you know everything out here if you had another mini Earth baby Earth if it had the same that it would need the exact same speed the Earth does to orbit at the same radius it wouldn't need any more or any less because that speed of the earth only depends on oh oh be careful this R is also d That's the radius of the circle is also D so I can cancel one of these and you realize oh shoot I can get the velocity of the Earth or anything at that radius it's going to depend on G Universal constant mass of the Sun the thing that's orbiting over the distance orbiting from the Sun but does not depend on the mass of the orbiting object uh energy and work is cool let you save some time sometimes people get kind of confused this is a big nasty equation right here it's like what is this talking about this is the work done by a force forces do work how do they do work they do work if there's a force over a distance like if you apply a force on something and it doesn't move you're not doing any work but if you apply a force on something and it does move you will do work you'll do work and that work will oh my goodness this thing doesn't work yay okay you'll do work and that work will equal the integral of f do DX what is a dotproduct we can always rewrite this as magnitude of one of these times magnitude of the other times the cosine of the angle in between these two things so this angle here is angle Teddy wants to do some work I know buddy I want to do work too f * DX so this is the angle between the force and the direction of motion so consider gravity let's just make it real here this is the force of gravity it points down obviously if this ball also moves down uh that's the direction of motion these would be in the same direction this Theta would be zero cosine of Z would be one that makes sense this is how you do the most work you want to do the most work you push in the direction of motion this would be positive work um what if I move this ball upward if I move it upward DX goes up forces is down that's 180° so now this is 180° cosine of 180 isn't one it's negative 1 you're doing negative work grav sorry you if you gravity gravity is doing negative work if this ball moves up and gravity's down gravity would be doing negative work trying to take away energy if this ball moved horizontally that'd be 90° forces at 90 degrees don't do any work because cosine of 90 is just going to be zero and so uh this is how you well you might be like what the heck is this integral here for well gravity is nice gravity just be staying the same the whole time I could pull F out I would just get the the work done is the magnitude of the force times how far I went guess what MGH is no surprise that's the potential energy but what if my Force wasn't constant well maybe my force is like a function of X maybe my Force gets bigger and bigger or smaller and smaller then you just plug that force in here integrate it that gives you the total work done you might be like well why do I care about work you care about work because this number tells you how much change in energy if something does work on something else it's adding or taking away energy if it did positive work it adds energy do negative work takes away energy um you be like well why is this equal to negative Delta U well think about it let's say the mass goes down we just said the work done would be positive right for Gravity to be mg times how far it moved down okay great that's the change in energy of the ball so you got to be careful this over here is worked on by gravity this is force of gravity this isn't change of energy of gravity if you're doing work you're giving that energy to someone else the ball gained energy what kind of energy gained kinetic energy so the kinetic energy would be what's equal to MGH this MGH turns into kinetic energy ball gained energy here but that means gravity lost energy obviously if a ball moves down you have less MGH less potential energy than you did before that's why there's this negative here if the ball gained 10 Jews Jewels is the unit of energy uh if the ball gain 10 Jews by Falling downward well then the change in potential energy I move this negative over the change in potential energy would be -10 Jews because if a ball moves down it gains kinetic it only does that because it loses potential energy and so another way to geograph or geographically yeah go on a map another way of geometrically thinking about this Force integral is that if you get a force function as a function of position while the area here would be the work done why do you care about that because it's the change in energy so it could let you figure out like oh how fast is the ball going I can find that area that's the work done so it's probably like you know maybe the object's sped up that's how this energy manifested I can plug it into this formula that this area is 1 12 mv^ S one last thing on this you might be like dang this guy talking about this if you have work the net workor IE work done by all the forces it doesn't just equal any change in energy it definitely equals the change in kinetic energy that's it you add up all the forces on an object find the work done by all of it that's change in kinetic energy work done by gravity might change your potential energy or your kinetic energ who knows spring energy but networ by every force on an object has to change just the kinetic energy of that object okay uh what if you had a spring well it's kind of interesting interesting because like the spring Force we know what the spring force is it's like KX or at least the magnitude of it so say I plug that in up here okay fine work done is integral KX DX well shoot guess what I get I get 12 KX s it's no surprise that that's the formula for the potential energy from the spring you might be like why is it positive here well this spring lost this energy like I really should have been plugging in a negative here or whatever lost this energy if I want to find you know the if I want to get rid of this negative this would just be you know positive 12 KX s but that's how you get the formula for Springs and you can do kinetic you know you can do conservation of energy just like normal on the AP Physics C exam I started off with 1 12 KX s what did it turn into probably 12 mv^2 you know energy will be conserved this energy of the system will equal zero the change will equal zero if there's no external work so as long as you put everything in your system everything interacting spring box in my system well then you say well changing the total energy zero in other words the energy I had initially equals the energy I had finally you just be set potentials equal to kinetics or vice versa and you can solve for what you want to solve for uh you could also rearrange this work formula look at this work we said is integral all let me write it this way let me write negative Delta or yeah negative Delta U is equal to integral f. DX and let me move that negative over I don't want the negative here I want the negative over here boom there we go okay think about what this saying this is saying hey you want to change in potential energy well then add up a bunch of that's what the integral means integral means add up a bunch of f. DXs when you're like what if I only have one little bit of f. DX if you only have a little bit of FDX then that just changes this by an infinitesimal amount you only change this U by a little amount if you're not adding up a bunch just a little bit you still have that negative now I rearrange this and I would get forces negative Dux what does that mean well if you had a potential energy graph this means the slope of U as a function of X the slope here well negative that slope would be the force on the object so right here I'd have a positive slope that's a negative Force you might be like that's weird how can I think of that well look if I put a ball here which way does it roll leftward duh negative Direction well right over here look at I well look at my slope over here I got a negative slope of my U negative slope of U times this negative out front is going to be a positive force yeah because if I put a ball here rolls in the posit you know goes in the positive direction so negative Direction positive direction what if I put a ball right here well there's no slope du DX is zero that makes sense how much force would I have making me go right there if I put a ball right there none this is an equilibrium position and this is in stable equilibrium so this a point where the force is zero um but and if I displace it a little to the right it rolls back to the left and if I displace it a little bit to the left it rolls back to the right stable equilibrium right here this up here is unstable equilibrium look the force is still zero my slope is still zero here but if I displace my ball a little bit to the right it's going to like Cascade that way if I displace it a little to the left Cascade that way this is like sitting on the head of a pencil this isn't going to stay steady like if you did it perfectly I guess so but this is unstable equilibrium right here even though the force is zero there's a stable equilibrium force is zero and that's kind of a quick way to figure out oh I put the ball here it rolls downhill now the ball's not going up and down this is supposed to be like a ond example the ball's only going left and right but you can kind of think oh ball rolls downhill therefore ball goes rightward ball rolls a downhill therefore ball goes leftward Ball's at a perfect position to have no Force so just sits there and you can check if this is right does this formula actually work well this is the potential energy uh gravitationally the universal law of gravitational potential energy what if we took a derivative of it let's and in fact negative derivative so this is the uh gravitational potential energy let's take a negative DD well it wouldn't be X here the relevant variables R so DDR take a derivative of this this negative cancels that negative I take a derivative of this mess I'm going to get G that's a constant M1 M2 over derivative of 1/ R would be uh 1 r^ 2 and that is indeed the force of gravity would be would be negative this you might be like negative I don't remember seeing that negative but remember Big G mm over R squ gravitational force the negatives there it just says it's attractive so if you see a negative force it just means attractive Force when it's a radial Force like this so the negative is always there we usually just write this as absolute value so we get rid of that negative but there's technically a negative there because it's attractive if there was a positive there it would be repulsive um okay how about momentum and impulse momentum and impulse like dang I want a cool formula like work and energy does well I'll give you one here we go look at the momentum which is defined to be MV and it's a vector so it has a direction unlike kinetic energy kinetic energy has no direction it's a scalar but momentum is a vector momentum rightward is different from momentum leftward um momentum has a cool formula like work instead of work though it's impulse impulse is not the change in energy it's the change in momentum and it's equal to integral of f not DX but integral F DT so if you find the area under FDT so if they give you you got to look what is this graph you know I mean you got to check is this X or t if it's t area here gives you the change in momentum so in other words like MV final minus MV initial so that's useful uh and that we call that the impulse the impulse is the change of momentum or integral FDT what if f is constant well then it's easy then it's just like f * delta T you can pull that out of the integral and you could rearrange this formula just like we did a minute ago instead of saying that Delta p is integral FDT rearrange it you get that f is dpdt uh what does this mean well if you get a momentum graph the slope of this momentum graph is the force so in this case the slope is positive that means there's a force on this in the positive direction it's not like ball rolling downhill anymore that was potential energy up here uh the slope is zero so the force at that point would be zero and so on it's just there's no negative sign um why do we care about momentum because if your momentum is if your system is closed your momentum's conserved in other words you could just do momentum initial equals momentum final this is why we liked energy as well you could just say conservation of energy or conservational momentum just be remember these are vectors negative counts so if I want to find the final speed of the masses I can be like okay momentum is MV so my initial momentum in the system is 6 * pos3 plus 2 * -1 this thing is a vector you got to treat it like a negative if it's negative equals let's say they stick together okay well then you have 6 Plus 2 total mass time unknown V I'd have 18 - 2 is 16 = 8 * V my V final of this whole system would be 2 m/s what if I wanted to find the impulse you're like damn I don't know what impulse means but remember impulse is the change of momentum or integral FDT FDT DT but I don't need to do that if I want on each Mass let's say I want to find the impulse on the six then I find the change of momentum on the six I'd say momentum final of the six minus momentum initial of just the six not the whole system equals U momentum final of the 6 would be 6 * final is 2 m/s minus 6 * initial was three so I get 12 - 18 is -6 and the units is going to be either Newton seconds or kilogram m/ second what if we did this for the two well you'd do the same thing you'd say impulse on the two would end up being positive 6 these are going to be equal and opposite due to Newton's third law that's why the momentum's conserved if one of these gained moment momentum the other lost momentum in the same amount and you didn't change the momentum of the system unless there was an external Force only external forces can change the momentum of the whole system uh there was kinetic energy lost though these stuck together how could you find that you'd be like well what's my initial kinetic energy 12 6 * 3 squar mv^ squ plus 12 2 * sure you could write as negative 1 but it's going to get squared remember so this isn't a vector so it doesn't matter negatives kind of don't matter here just leave it off and it doesn't matter equals okay I get 3 * 9 is like 27 plus this is going to cancel here plus 1 is like 28 that's my initial Jewels before the Collision um what about after the Collision I'm going to have2 total mass is 8 * 2^ 2ar well one of these cancel one of these I'm going have 16 I lost a lot I lost 12 jewles of energy in this Collision okay so that's energy uh in a collision how about Center Mass well the center mass is the position where these Mass would balance like saying connect them with a rod I want to know where would I balance this thing well obviously it's going to be closer to the heavier mass in fact if this mass is three times the mass of the other mass then the center mass is three times closer to it so you just be like well doot doot uh it's going to be like over here somewhere so that if it's one away then this one would be three away how do I find the center of mass um I could just do there's a formula for it you could do X Center Mass is going to be M times I got to choose a position let's call this x equals z it doesn't matter where you call it you're just going to be measuring from this point if I call that x equal 0 I'm going to do 6 one of the masses times its position is zero so it's like mx1 plus the other one 2 * its position which is 16 M away look at that uh divided by the total mass so it's going to be 6 + 2 so this formula for Center Mass is really X Center Mass is like M1 X1 plus M2 X2 you just keep going divided by the total mass and what would I get here I'd get 2 * 16 uh 32 over 8 gives me 4 M it'd be you mean four like four meters where from where well from where you measured x equals 0 be four meters it'd be like right here you know that's where you'd balance it what's the speed of the center Mass it's the same formula look if you want to find the speed of the center Mass look how easy this is M1 V1 plus M2 V2 just replace these X's with V's you still divide by the total mass and that's the velocity of the center Mass so I could say 6 * 3 plus 2 * -1 I'm talking about velocities now so I got to throw that negative in there ID 6 + 2 I get 18 - 2 is 16 over 8 oh my goodness I get 2 m/ second um and you might be like what that looks familiar yeah CU look this is just like the total momentum here this would be what the speed is of the center Mass after the Collision if they Collide and stick together they're still going to be moving 2 m/ second we just found that a second ago you're like how come it didn't change they collided yeah only external forces can affect the speed of the center of mass force of 6 on two and force of two on six cannot affect the motion of their Center of mass only external forces can affect this so if this Center of mass was moving with 2 m/ second before the Collision it's going to be moving with 2 m/ second after the Collision which is kind of cool all right angular kinematics is just like regular kinematics we just got new characters instead of X we call it Theta this is instead of position this is the angular position like it's the angle you're at uh instead of V the velocity we call it the angular velocity that's Omega instead of a the acceleration this is angular acceleration Alpha and their their definitions are the same Omega the angular velocity is derivative of the position or angular position and the angular acceleration is the derivative of the angular velocity Theta typically has units of radians that means omega's in units of radians per second and that means Alpha's in units of radians per second squar and all the interpretations hold slope of Al slope of theta gives you Omega slope of Omega gives you Alpha area under Alpha gives you change in Omega area under Omega gives you change in Theta so it's all the same thing you know if if they give you one of these as a function of time you can take a appropriate derivative or integral to find whichever other thing you want um and again if the angular velocity is constant it's easy well the angle you go through is Omega time time just like distance equals rate time time angular distance equals angular rate time time if your angular acceleration is constant you get angular kinematic formulas they look like these they're just like the other ones they're only true if angular acceleration is constant if angular acceleration is not constant well you better be using conservation of energy angular momentum or just crafting your own angular version of Newton's Second Law which looks like this instead of a it's Alpha instead of forces it's torqu and instead of I or instead of M it's I what are all these things well I'll show you in just a minute but first let me be clear here um you can always relate these back let's say I know what angle I went through right here like this is the angle I went through it looks like about 60° or something like that we say I didn't want to know the angle I wanted to know how much distance I went through how much Arc Length I went through we usually call this s s if you remember is going to be R the radius times Theta as long as you're measuring this Theta in radians that's why we like radians so the arc length is R * Theta well we know Omega is D Theta DT let's smuggle an r in here because watch R Theta I just said is the distance so derivative of the distance with respect to time I know that's just velocity that means V the velocity is R Omega so R * Omega is the speed the regular speed like meters per second speed not like radians per second speed so you can always you can always take your angular variable kinematic variable multiply by an R get your regular variable so R * angular distance gives you regular distance R * angular velocity gives you regular velocity in fact in fact use this so often you guys should you should memorize it you know V is R Omega this is R * Omega looks like R Omega let's throw another R in here Alpha is derivative of Omega with respect to time so smuggle an r on both sides you know I can multiply both both sides by the same thing I don't change it but R Omega I just said is V but what is dvdt well that's just a so just like V equals R Omega a equals ralpha so I take R * my Alpha that gives me my acceleration but specifically my tangential acceleration not not my centripetal acceleration centripetal acceleration points inward we already know how to find that oh thanks adobi giv me messages equals v^2 R that's the centripetal acceleration that's always inward but the tangential acceleration is what you have if you're speeding up in your rotation you know if you do have an angular acceleration which is what you'll have if your rate of change of Omega is increasing then you'll have tangential acceleration and you can find it like this R * Alpha um sometimes people get confused they're like wait wait wait I've seen R Omega elsewhere when there's a ball rolling on the ground without slipping if it's rolling without slipping sometimes people use R Omega to find the speed of the center mass of the ball and that's true so R Omega does two things you can use R Omega to figure out hey what's the speed of a point on this rotating object about you know relative to its axis that you're rotating about and the further out you go the bigger the r is the faster this thing goes so if you're on a whirly ride the further out you go the faster will be G this thing's just rotating in place but if you have this thing on the ground and it's rot it's rolling across the ground without slipping now this romega also if you plug in the total R you know the the long the actual radius of the object you multiply by the Omega this thing spinning at you'll get the speed at which the center travels so in other words R Omega does two things it tells you the speed of an object about the center and if you're rolling without slipping it tells you the V the V of the center the center of mass will be moving with r omega as well so that's important if you're without slipping so in this formula you know Newton Second Law for angular Alpha is net torque per rational inertia you might be like what the heck is torque dude I don't know about torque well is the formula get on the formula sheet for the AP Physics C exam it's torque equals R cross F like dang what's a cross product well you can just call the magnitude of any cross product it's going to be equal to the magnitude of the first Vector times the magnitude of the second Vector times sign of the angle between the first vector and the second vector so the magnitude of the torque like don't be using a cross product you can draw this big old determinant and whatnot for this thing you don't need that maybe in college you need that or whatever but often you can just get by with finding the magnitude and then Direction separately so I got the magnitude of R magnitude of f sign of the angle between R and F what is r what is f we'll say you got some object this Rod here you got 20 newtons exerted on okay there's my f that's the force what is r r sometimes called the lever arm or whatever it's a point that goes from the axis the point you're rotating about goes from the axis all the way to where the force is applied that's r r goes from the axis to where the force is applied and this Theta right here is the angle between that R and that F so it's not this one that's bait don't be taking this bait that's this 30 would be the angle between R and horizontal I don't see horizontal in here I see R and F so you take R you take F you find the angle between those this would be 60° so you'd have to do okay my torque for this force would be R it's not 10 it's not four R is going to be this length right here which is 10 minus 4 so it' be 6 M these are bait too dude AP be throwing out bait 6 M * 20 newtons you know R * f * s of the angle between them which would be S of 60° that's how I get my torque you might be like but by getting rid of the cross product we don't know which direction it goes well yeah you do look the force goes this way guess which that goes counterclockwise like duh finding direction is easy for these easy problems you don't need to do the full cross product this is a torque of 6 * 20 * sin 60 it's in the counter clockwise Direction that's what you'd plug in up here You' just call one of these positive typically we call counterclockwise positive if there any if there were any torqus going the other way you just plug those in as negative up here in this numerator so how do you find the I you know there's this I in the bottom right down here like there's this I how do you find this well that's rotational inertia that gives you an idea of how hard it is to rotate something if you just have a point Mass rotating about the center it's not bad point masses have a rotational inertia so if this m is orbiting the whole mass is orb in about some Center R away then the rotational inertia is just going to be M * R 2 Mr squar over here what if you don't have that what if this Mass is like distributed it's rotating here and it's like a rod well then you got to do an integral like dang how do I do this integral they give you this formula but what they don't give you is well I'll show you you get integral R squ you're like DM so R is how far from the axis you're talking about DM you got to imagine like some infinitesimal M here physicists are simple we're like dang dude all I know is the rotational inertia for a point Mass someone comes up like hey I got this this Rod here and I'm like oh I'm going to make it into an infinite number of Point masses here's a point Mass it's right there it's infinitesimally thin and it's exactly a distance R away I'm like okay I know that this little piece of mass right here would have a would have a i of DM differentially small M so this would be a differentially small I times R 2 okay how do I find the total I you integrate both sides that's how I get this formula here but you're like what do I write as DM well so you got to come up with this idea of Lambda Lambda is the mass per length specifically it's like infinite so imagine I take this infinite decimal Mass right here I divide by the infinite tmal length of this region that gives me the mass per length right there that's DM per DL that's Lambda it's kilograms per meter um this is how I find DM I'm not going to ever leave this as DM really I'm going to write this DM as Lambda DL okay so I got Lambda * DL that's what you have to replace dm with they don't give you this on the exam you got to remember it equals let's say my Lambda is a constant like I have uniform density throughout here so I'm going to get Lambda integral R 2 DL you're like what do I DL but look DL is the same direction as Dr here so it's like simple they're the same direction I can call them the same variable Dr what do I integrate from well the closest Mass I have is zero away so I'm going go from zero here all the way to the length of this thing is l l away so I have Lambda * integral r^ 2 would be R cubed over 3 but evaluated between 0 and L so I'm going get Lambda * L Cub over 3 and you're like that's not what I remember the I of a rod being yeah yeah because look Lambda is the mass per length so this Lambda was uniform so I can just write it as total mass per total length so I know if it's uniform then the little bit of mass over this little bit of length should be the same as the total mass over the total length I know the total mass is M total length is L so let's plug that in U Mass per length is the Lambda L Cub over 3 this cancels one of those I'm going to get I is 1/33 ml 2 that is the rotational inertia of a rod about its end you might be like dang but what if what if I had the rod rotate about its Center okay fine just integrate from like negative L over2 to L over2 because the you know furthest Mass this way is goingon to be negative l over2 the furthest Mass this way is going to be L over2 you'd get 112th you just integrate from wherever you want to wherever you want these limits will give you the different factor out front here um and now you got it all together you can know how to find the I you know how to find the torque you put it all together that lets you figure out the angular acceleration but these are hard what if you have this like yo-yo it's a cylinder right let's say and it hangs from a string and it falls down it unwinds without slipping well shoot okay you got to draw your forces so I got a force right here gravity pulls from the Center mg I got another Force right here tension over here from the Rope that's it those are my two forces so here's what here's what rookies do they're like oh dude I'm a rookie look I'm gonna say first of all I'm going do like T minus mg dude rookie I want to torque up here not a force these are forces not torqu if you don't multiply by a f and multiply that F by an R it's not a it's not a torque so you got to multiply by R first of all and first and don't even call up positive call down positive it's going to be moving down positive we're going to call that the positive direction so see how much torque is there for this gravity zero look at where R is R is from the axis to where the force is applied this force is applied at the center R is from the center to the center that's zero this gravity applies no torque gravity doesn't cause things to rotate if you just throw them off a cliff this tension will cause it to rotate so this does exert a torque I'm going to have tension T time r r would be you know right here R that's R um angle between them is 90 degrees that's nice s of 90 is one and you're like well should I make this positive or negative well I'm calling down positive so you might think oh so let's call tension negative because it goes up no no no no no look we're not talking about whether this isn't tension plugged in here this is the torque of tension you might be like what's the difference well look the torque of tension here causes this thing to rotate this ways and if it rotates that ways it's going to cause the thing to move downward which is the direction we call positive so torqus that cause rotation that move the object in the direction you call positive you just call positive otherwise there'd be a relative negative sign you're going to have to deal with um they would give you the rotational iner of a cylinder you don't have to memorize these um you typically don't have to derive them either 12 Mr squ the rotational inertia of the cylinder you're like dang okay I can cancel that cancel that but I got two unknowns I don't know T I don't know Alpha yeah so you're going to have to do one more formula you're going to have to do a the acceleration is net force per mass now we do plug forces in here I'm going to call downward positive these are actual forces I don't have to multiply by R mg goes down call that positive minus t goes up divid M might be like dang but look now I got a and I got Alpha but remember a equals ralpha so I could just since a equal R Alpha that means Alpha equal a over R so I'm going to write this over here as a over R I canceled one r i cancel one R already I'm going to cancel this r with that R and I'm going to solve and get over here that t if I solve this for T move the 1/ 12 M over T is going to be 12 M A when I can plug that in right here I'm going to move this M over I'm going to get M A on the left equals mg minus a half Ma I get to cancel all my M's I'm going to move my half a over I get three Hales a equals g that means that a is 2/3 G dang these are hard problems but they're they're not that bad I mean you got to remember to multiply by R you got to remember to call the direction of motion positive and the direction of torque consistent with that motion positive otherwise you had a negative all screwed up here you got to remember a equals ralpha um there are easy ways to do this if you have kinetic energy then you know if you think about this in terms of energy it's a little easier so an object rotating can have rotational kinetic energy an object moving in a straight line can have translational kinetic energy and an object that's moving in a straight line and rotating so if a if the center of an object's moving and it's rotating about its Center then you have both regular translational kinetic energy and rotational kinetic energy why would you ever care about that cuz look how easy it is find the speed of this at this uh you know say it's a say it's a cylinder again 12 Mr s is the I of a cylinder find the speed of this at the bottom of the ramp you use Torx and you do that you could do it you saw all the mistakes you could possibly make though let's just do energy e initial equals e final let's put everything in our system the ball you know the ramp the Earth so what kind of energy does it start with it starts with MGH has a mass m let's say equals well it's going to end with 1 12 mv^ 2 it is moving the center is moving but it's also rotating so so I can't stop there I have to do plus a half I Omega 2 Okay so let's do MGH we're going to plug in a half mv^ 2 plus 1/2 be careful this this is the I the eye of a cylinder turns out is a half Mr squ so plug in whatever your relevant I is this is a half notice there's two halves now one half is normally there and then one half smuggled in due to the I times Omega squar you're like dang but I got V and I got Omega I only want one of those otherwise I have two variables yeah remember V is R Omega that means Omega = V / R plug V / R into Omega s right here and you'll get v^2 over R 2 and something magical happens R 2 cancels r s M's all go away we end up with GH is a half v^2 + 1/4 v^2 and half plus a 4 is what 34s is that right v^2 that means V is going to be the root if I solve this of 4/3 GH boom didn't have to do any torqus I just had to do conservation of energy you're like but what if they asked me for acceleration of this object fine just plug this into kinematic formula you know call this your displacement um you know V initial zero kinematic formula way to the acceleration if you need to um all right another way to just like we had momentum we could have angular momentum uh angular momentum is not MV angular momentum of the vector is R cross P you're like damn cross product again yeah but remember you can find the magnitude by just doing magnitude of this one magnitude of that one sign of the angle between them and we know what p is p is just MV so R * p is just m v * R might like well then why don't you write it as R MV because I like writing it like this look angular momentum is Mr RV I like remembering this as Mr V Mr V and you might like where's the S Theta well if r and v are at right angles then s of 90 would be one so if you just got a ball rotating a circle uh then that Theta is all nice and 90 if it's not that's fine I mean plug in the Theta if you need to so this would be like the angular momentum Mr V is the angular momentum of a ball you know orbiting around some point but what if the ball is orbiting around its Center well then the angular momentum is the rotational inertia of that thing times the Omega that it's spinning with so um in other words this I over here is Mr R 2 this eye over here of a sphere of a sphere turns out is like 25ths Mr 2 so if you're orbiting you know you use Mr V for your angular momentum if you're spinning you use I * Omega for your angular momentum you're like but I don't care about angular momentum well you should because just like momentum has a cool formula and energy has a cool formula angular momentum has a cool formula instead of work or impulse it's angular impulse is there a letter for this no I just picked K why I don't know because J is impulse so I just made K the angular impulse K is the angular impulse what is angular impulse it's the change in angular momentum how do you find it well you can find the change in angular momentum or you can integrate torque time time just like you can integrate Force time time to get the change in momentum integrate the torque times time get the change in angular momentum that means the area under a torque graph would give you the change in angular momentum IE like I Omega final minus I Omega initial um and then these are your formulas for angular momentum you could rearrange this formula if you want and you'd be like oh dude so the derivative of L with respect to time would be the torque yeah just rearrange this just like derivative of momentum with respect to time gives you force derivative of angular momentum with respect to time gives you the torque you might like I don't buy it I still don't care about angular momentum here's why you should care if you put everything in your system angular momentum is conserved that means L initial equals L final that means these problems get kind of easy so if I put all this stuff in my system then whatever the L initial is has to equal the L final so say I had some I had a cake down here spinning right I'm going to decorate it and I just blop I just plop this cake right on top of it this top one's not spinning but the bottom one is initially spinning so okay L is i o Omega initial so that's my initial uh what is this going to be well I of a cylinder is 1 12 M and the M down here is 2 m so 12 2 m * r^ 2 * Omega initial and this top one wasn't spinning initially so it has no angular momentum start equals well they they say they stick together these are cakes so they Boop they stick together they both start rotating with each other this is on like a little spinner well to end I have some total I right I final time Omega final I initial Omega initial I final is going to be 1/2 this bottom cake still has 2m R 2 12 2m R squ plus this top cake has rotational inertia 1/2 its mass is only M though M * R 2 and then I multiply by Omega final watch R's cancel out M's cancel out twos cancel out twos cancel out I have omega and IAL equals 1 plus half this thing isn't zero it's 1 * Omega final this is 3 that means Omega final is just going to be 2/3 Omega initial we got it without worrying about torque or anything like that so conservation of angular momentum is pretty nice to figure out final speeds of things that are spinning uh what if you had a point Mass hitting a rod still do angular momentum watch angular momentum initial equals angular momentum final we want to figure out what the Omega is afterward this is a point Mass though so I'm not going to use I Omega I mean I could if I want it' be dumb but I'm going to use Mr VM this is going to hit at a point L away so when it strikes it's l away Mr V equals this Rod had no spinning to start with no angular momentum this ball bounced bounced back and so it bounces back so it's going to be negative Mr V it's going the other way this ball effectively has clockwise rotation to start and afterward it's got like counterclockwise rotation so negative Mis but it's going half the speed cuz the velom speed V /2 plus so this is the angular momentum of the ball initially angular momentum of the ball finally plus I Omega of the rod afterward we can solve this for Omega we'd get Omega is what is this going to be three uh ml over we got a v over I so that's my Omega and I can do it with conservation of angular momentum you might be like but I want to do I want to do I Omega okay do it I want to do I Omega for L that's fine I for a point is m r 2 Omega remember is Omega but watch a magic trick watch this m r * R * Omega because r s is R * r r * Omega is V I still get Mr V it doesn't matter you try to use I Omega you're forced to do Mr V anyway so it still works out okay we're close I got two more little slides here hooks law is the force from a spring why is it negative well because if you displace this to the right the force is to the left and if you displace it to the left the force is to the right the Spring's a hater it's going to try to do again just going to work against whatever you do that's why there's a negative here uh this is going to cause this thing to simple harmonically oscillate if I pull this back and let go it's going to fly here then it's going to fly there it's going to fly here it's going to go back and forth forever if there's no friction or air resistance and it would look like this the position graph would look like the sign graph and there'd be Peak to Peak the peak to Peak would be the period and the period of a mass on a spring is 2 piun m over K how do you derive this formula well use Newton's Second Law we're going to do a is net force per mass and I'm going to say the net force is KX over M so I'm going to write this as a differential equation but I don't want to write a as dvdt this Force isn't a function of v this force is a function of X so I'm going to write a is second derivative of x with respect to t^2 because if you remember uh a is dvdt but v v itself is just dxdt so I can write this as second derivative of x with respect to t^2 = k/ M * X and uh you can't separate this so separate integrate celebrate doesn't work I can't separate the bottom here of the second order differential equation uh you way you solve this as you look at it you go oh look at that look at that gorgeous graph I remember I remember this equation if you have a second derivative of something equals negative a constant times that same thing this is the formula for a simple harmonic oscillator think about what it's saying it's saying hey can I think of a function that if I took two derivatives of it would equal negative a constant times that same function back I can that function would be X is a function of time would be some amplitude this graph is cosine so I'll just do cosine it could have been s though of you know some omega that's what we call this times T so you know in math class you might have called this b or something remember Omega is just 2 pi over the period the time it takes to go through a full cycle um let's see if this works let's take two derivatives of this one derivative of this thing one derivative of x would be I pull out an Omega I'd get EGA chain rule a Omega there' be a negative from the cosine so it' be negative Co or no not cosine derivative of cosine is s Omega T let's take another derivative two derivatives is going to be negative I pull out another Omega Omega squ a cosine Omega T and you're like oh look look two derivatives that's what this is second derivative of x with respect to T ^2 equals negative sum constant times this whole thing here is what we called X that is X so this is times x you're like oh dude that does work here as long as as long as Omega saral k/ M or in other words since Omega is 2 pi/ T you know this would have to equal so I'm I'm just going to call this omeg om GA is root K Over M that's what would go right here root K Over M that should equal 2 pi over the period in other words the period if I solve here would be 2 pi 1 /un K Over M orun M over K that's that's how you get it's not fun not pretty but that's how you get the period you prove you derive the period formula for a mass on a spring what if you had a pendulum last one we made it last slide here what if you had a pendulum well this this isn't from Force this is from torque so I say well I got a torque here Gravity is that way here's my Pivot right here Gravity is that way tension is this way I ask myself what's exerting torque here uh not the tension because look R goes from the pivot to where the force is that angle would be remember torque is RF sin Theta and Theta is between R and F so R goes that way torque goes back up that's 180 for the tension and S of 180 Z that the tensent exerts no torque but the gravity would exert a torque so if I imagine continuing my R out angle between R and F would be this same angle right here so this would be this same angle right there why do I care about any of that well look because I want to do alpha is net torque over I this is this is also going to be a simple harmonic oscillator as long as the angle small um I'll show you why in a minute and so oh you're like I know what to do instead of writing Alpha as D Omega DT I'm going to write it as second derivative of theta with respect to t^2 yep equals the only torque in here is gravity it's going to be Force time R which is the full length of this Rod times s of the angle between them and that angle here would be this angle there um if you don't like that you can think about it like oh look supplementary angle but s of supplementary angles or the same value so it doesn't matter it's going to be the same thing but I do have to throw a negative sign there's got to be a negative here just like negative KX this is negative cuz look this is displaced in the positive Theta Direction but my torque's trying to bring it the opposite way so my torque's op uh acting opposite direction of the angular displacement I divide by I the I of a point mass is ml s Mr s because I'm orbiting about this point right here and look I get to cancel M cancel M the mass of a pendulum cancels out I cancel l i cancel 1 L I get second derivative of theta with respect to t^2 um is going to be g g l sin th you're like wait so we're going to solve this by looking at it we look at it we go o ooh look at that and then you're like huh that's not a simple harmonic oscillator equation the simple harmonic oscillator equation is second derivative of some function equals negative a constant times that function not times s of that function okay here's where we get a little sketchy here physicists go well but hold on hold on they say if Theta is small and this is true for small Theta sin Theta is approximately Theta if you don't believe me plug it in like try plugging in and make sure you're in radians try plugging in sign of like 0.01 and you're going to get like 0.999 which is basically 0.01 it's off by like a smidgen so if Theta is small we pull a fast one we go oh this is a this is kind of equal to Theta and it's pretty close so the pendulum is only approximately a simple harmonic illat it will not be a perfect sign function it'll be close and it's close enough that it doesn't make much difference and if your instruments in the lab don't measure to that decimal place then it doesn't actually matter anyway so we say okay this is fine this is basically a simp harmonic oscillator for small angles and I can play the same game I say that this over here has to be Omega squar has to be G over L and remember Omega is always 2 pi over the period and therefore I could be like well then Omega is just root of this that equal to 2 pi over the period that means the period is going to be 2 Pi root flip this whole thing L over G and that's the period of a pendulum and if you don't like that you're like what physics is dumb dude why are they making approximations fine let me show you one that's not an approximation it's called the Toran pendulum say you got something on a string and you just like you rotate it right you twist the string you let go for a lot of cases this torque will just be negative beta some constant times how much angle you moved it through not sign of the angle just angle so if I double the twist I'll get double the torque trying to restore it for this case let me make some room here for this case you could be like all right fine second derivative of theta with respect to t^2 equals negative beta you know Theta over beta beta negative beta theta over what what are we looking at I the I of this thing so we're doing torque over I so we do negative beta over I * Theta now you look at this you're like that's a good looking simple harmonic oscill equation right there it's second derivative of theta equals some constant time Theta and you're like okay I know how to play this game this beta over I right here has to equal Omega squar that means the root of this thing has to equal 2 pi over the period and that means for this torsion pendulum the period would be 2 pi * the root of I guess whatever the I is divided by whatever the beta was um and that's how you would figure out the period of some Toran pendulum okay those were all the topics pretty much all the topics from the AP Physics C mechanics exam good luck