Mass Distribution with Calculus - Video Tutorials & Practice Problems
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1
concept
Using Calculus to Solve Mass Distribution Problems
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10m
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So we've got that, we're gonna use calculus to calculate the gravitational force for non spherical distributions of mass. So let's recap everything that we've learned about Newton's law of gravity whenever we had point masses like M one and M two and we had the distance involved, the gravitational force was just given by Newton's law of gravity. Now, even when we had planets, like in the second example, over here, we kind of just pretended that these planets or these objects could be treated as points and all the mass were just concentrated at the center. So that's big M and this guy over here is sort of just like little M and as long as we had the center of mass distance, which was that little R, it basically worked the exact same way except instead of M ones and M twos, we just had big MS and little MS. So we just use Newton's law of gravity and we just replaced it with the appropriate variables. The difference is that when we have a non spherical mass distribution, we can't just imagine that all of the masses is concentrated at the center here. It doesn't work that way. So we need a new technique to solve these problems. And one of the things that we can do is we can sort of pretend that this rod of mass and big end that I have here can be thought of or broken up into tiny little pieces of, of, of rectangles or mass right here. And the smaller that they become, they become a differential mass. So I'm gonna call that D M and we can treat that DM as a point mass. So that means that this point mass DM generates a tiny amount of force in this direction. And that's gonna be DF and it's due to that center of mass distance of little R right here. So if you wanted the equation for that DF, we can actually figure that out using Newton's law of gravity. It's gonna be the exact same as this equation over here. So it's gonna be G but instead of big M, we're actually gonna use the differential mass. So that's gonna be DM times M little M out here divided by R squared where that's the center of mass distance. But remember that this is only one tiny piece of the force from one tiny piece of the mass. So to solve for this total force, we have to basically go along the rod and we have to add up all of the tiny amounts of forces that are generated from all these tiny amounts of masses. So think back to calculus, what do we call it when we added up a bunch of tiny infinitesimal things we call that an integral. So the way that we're going to solve these problems is we have to integrate along whatever mass that we're given. So that means that the total amount of force is going to be given as the integral of DF which is just going to be the integral of G times DM M over R squared. So one of the things you're always gonna see this in your textbooks written like this. But one of the things that we can do with this equation is we can actually pull out the big G and the little M as constants to the outside of the integral, right? Because those things don't change. So the way the more useful form that we're always going to use here at clutch when we solve these problems is we're gonna use the GM times the integral of DM over R square. So we're always going to start from this equation when we're solving these problems. All right guys, that's basically it. So we actually have a list of steps that we're gonna need to solve any one of these mass distribution problems. But rather than telling you, I actually want to go ahead and show you. So we're gonna work out this example together. This is actually a very classic common type of problem that you'll see. Uh if you're doing this, this topic right here. We've got a hollow ring. It's got a massive M, it's got some distances of that radius. We have a distance in which a little mass over here is sitting. And we need to find out what the gravitational force is. So the first thing that we do in all of these problems is we write out the equation for F which is right over here. So we have F is equal to and then we have GM times the integral of DM over R squared. So that's the first step. OK. So the second step is we have to pick, let's see, it says, pick two D MS and we have to write an expression for R. So I've got this DM that's gonna be over here. And then I'm gonna, I'm gonna pick another DM like this over here. And then we know that these D MS act like point masses and they produce forces on this mass over here. They're gonna point in this direction like that, that's gonna be one DF. And then this guy is gonna be in this direction and that's gonna be another DF. OK. So now we actually have to figure what the RS are the center of mass distances. So that is this right here, these pieces right there and then I've got one over there. So this is gonna be R OK. So we have to use uh write an expression for F, using the problem's geometry. That means is that we're just going to use the uh length variables that are given to us R and D. Now, this little R, if you think about it is actually just the hypotenuse of this triangle that we've made here. So we can actually use the Pythagorean theorem to come up with that expression for a little R. It's actually gonna be the square roots of R squared plus D squared. OK. Now, we're just gonna save this for later. We're not actually gonna start plugging it in yet because what happens is if we start plugging this in, we're gonna have to write this a bunch of times. It's just gonna get really annoying. OK. So we have this second step right here. So we have the expression for R. Now let's take a look at this third step here. 33 says we have to break uh the integral into its X and Y components. So in other words, what happens is that F, which is the integral of DF actually gets split into two things. The FX components is gonna be the integral of all the D FX S and the FY component is gonna be the integral of all the dfys. Now, where do these components actually come from? Well, remember that these dfs actually are two dimensional vectors, right? They point in opposite directions or different directions. So we have to use vector decomposition to break them up. So what I'm gonna do is I'm gonna sort of draw an angle right here relative to the x axis. That's my angle theta. And what happens is now I can break this DF into its components. So I've got D FX and then I've got dfy over here. And by the way, the same exact thing happens for this DF vector. So I have another one of these components that's gonna be here and then another one of these components that's gonna be over here. So dfy. So what we can see here is that my D FX components are always gonna be sort of pointing to the left and they're always going to be together and they're gonna be adding together. Whereas these dfys here are always going to be equal and opposite. So because these things are equal and opposite, what happens is that their components of DFY will always end up canceling out. So for any mass here, for any little point, I can think of the mirror, opposite point on the opposite side of the ring that's directly symmetrical and those Y components will always end up canceling out. So that means that this integral just goes away and I don't have to deal with it anymore. Now remember that if I have this angle right here, I can write the D FX. Uh so now that I've actually sort of split this and canceled out the uh components, I have to expand this into sines and cosines So remember that this D FX components can be written with this angle as the integral. Uh And actually, I'm gonna have GM integral of DM over R squared. And this is actually gonna be the cosine of that angle right here. So now what I have to do is I have to expand this into signing cosine, which I've done. Now, I have to rewrite this cosine in terms of the side links that are given because what happens is I'm integrating DM and I've got this R variable here. So I don't want this cosine sitting in here. So what I have to do is I have to relate it using the triangle. Now remember that this cosine angle here is always the adjacent over the hypotenuse. So given this triangle right here in which I have the D as the adjacent side. This and I've got the R as my hypotenuse. This is actually just gonna be D over R. So we're actually just gonna replace that cosine of theta in there for D over R. So that means that for four, my equation becomes FX is equal to or we could just write F uh is equal to the integral. Uh Well, it's actually, it's gonna be GM, the integral of DM over R squared. And then this cosine of theta actually just becomes D over R. So actually what this becomes is GM integral DM uh D divided by RQ. OK. So that is step four. So we're done with that. So now what we have to do is for step five, we have to plug in the expression for R from step two and then pull all the constants out of the integral. OK. So we've got the F is equal to and now you've got GM, now we have the integral uh and then we have DM. And then what happens is we're gonna plug in our expression for R. Now remember that expression for R is this guy over here. So actually what this R cubed becomes is it becomes R squared plus D squared. And then, because this is a square roots, which really means that this is uh to the one half power, this actually becomes to the three halves power over here. And then we have this D, now what we have to do is we have to pull all of these letters that are constant outside of the integral. But if you take a look at this, remember that D is an in, is it just a constant, right? This is just a constant horizontal distance up here. So that is uh that just gets pulled out. And then this R squared and then D squared, those are both constants as well because they are capital letters. Remember that this radius of the ring never changes and the D the distance also never changes. So in fact, everything out of here is just a constant, all of these things here are constants and they get pulled out. So these are constants. All right. Now, we're done with that. So we pulled all the constants out of the integral and then we're gonna rewrite this. So this actually ends up being uh F is equal to. And then we've got G MD divided by now, we've got R squared plus D squared to the three halves power OK. And then we've got to just write that last little remaining integral piece. So you have the integral of DM. OK. So now what happens is if we look at step six A, if we're only left with DM, the integral DM, then the integral of DM here is just M, right. So uh if we're just integrating our differential DM, that's just basically the whole entire mass. So then what happens is that we're done, right? So we just replace that with big M and we're done here. So that means that the force is equal to GMM. And that's gonna be D divided by R squared plus D squared to the three halves power. And that is actually our force. We don't have to do the integral because we already did it in this step. OK, guys. So let me know if you guys have any questions and I'll see you guys in the next one.
2
example
Gravitational Force of Rod Parallel to Axis
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11m
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All right guys, we're gonna work this one out together as well. So we have the gravitational force between a rod of mass M and length L and this mass over here. So we're solving for the gravitational force. But this is a mass distribution, remember because this is non spherical. So we're just gonna go ahead and stick to the steps. The first thing that we're gonna have to do. And by the way, the first part of this problem is just setting up the integral expression. The first thing we do is just write out the formula for F, right. So we've got F is equal to, we've got GM. And then it's gonna be the integral of DM over R squared. So boom, that's the first step. You always start with this equation. Now, for the second step, it says that we're supposed to pick two D MS along our sort of shape or object as we're gonna pick this DM over here. So this is gonna be DM and we're gonna get this one over here. This is also gonna be DM cool. So we need to figure out an expression for little R, which is the center mass distance. So between this DM and this mass over here, this is R and between this mass and this mass, this is R. So clearly, we can see that R is actually changing throughout the shape, right? So let's come up with an expression that relates the variables that are given to us in the problem. That's L and D. So if you think about this, this whole entire distance over here is actually L plus D. But the thing is we're not taking all of that distance, remember, because we're actually not including this little piece right here. So that means we're really not including sort of that piece. And if we're looking at this DM over here, then this distance over here sort of ignores all of this distance uh that I've highlighted in blue. So it's clear that we're not actually going all the way from L plus D, we're going to subtract something. And because this sort of distance here is variable that we're not using, I'm gonna assign it a variable and we're gonna be in the X direction. So I'm just going to say that this is X. So that means that this little R distance here is actually L plus D minus X. It's the whole entire thing minus this little piece right here. OK. So that means we have R equals L plus D minus X and that is step two, we're just gonna save this for later. We're not just gonna start plugging it in yet because they're gonna have to keep writing it over and over again. OK. So step three says we're gonna break DF into its X and Y components so that we only do that if the dxs or dfs actually have components to them. But if you take a look here, this mass is always just going to be pulled straight to the left no matter what. Right, because everything is sort of on the axis of the X axis, right. So this DF doesn't have any components. So that means we can kind of just get rid of this whole step here. We don't really have to do it because there are no components and we just get F is equal to and then GM and then integral of DM over R squared. So it's just gonna be the exact same thing. So now this fourth step over here, so we're done with that is we're going to expand blah blah blah into the X and Y components, but we don't have any. So we don't have to do that, right? So now we're actually gonna skip steps three and four and we're just gonna go straight into step five, which is, we're actually gonna plug in our expression for R into the integral and then we're gonna take out all the constants, right. So that means that we're gonna do F equals and now we've got GM integral of DM divided by now, I can start plugging in my R, right? So now I have RL plus D minus X and then that's squared. So can we pull out any constants? Well, remember that L and D are constants, right? These are actually just links that are given to us. The problem is we can't pull them out because all of these constants here are sort of enveloped in this term that involves X. So this guy right here is actually the variable. And so because these two distance here are attached to this variable, I can't remove them from the integral, right. So basically, what happens is I'm already done with that step. So there are no constants to be pulled out of the integral. So now let's take a look at steps six, right. Step six has two different possibilities. If we're just left with the integral of DM, we're gonna replace it. But as we've seen, we're not gonna pull out in any, any constants. And so we're not just left with the integral of DM. We actually have this term here on the bottom, which means that we're not gonna go with six A, we're actually gonna go with six B, which is otherwise we're gonna change the differential of DM, right this guy to match the changing variable. So this right here is our differential and because it's DM and I have a variable that's changing, that's X. These two things don't talk to each other. They're not the same thing. So what I have to do is unlike the, unlike we did for the ring, I actually have to match this changing variable uh using the density right here. So this is something we haven't talked about just yet. So what happens is we're gonna get this DM and we can relate this DM to the density, which is LAMBDA. And we're gonna write a different expression for that involving the mass and the length of the objects. And by the way, you're gonna see all of these expressions. This is only for one dimensional shapes. Whereas we're gonna use these equations for two dimensional shapes and then these for three dimensional shapes. So this would be like a flat sheet or surface or something like that. And this would actually be a three dimensional object like a cube or a sphere or something like that. So we're dealing with just a horizontal rod, which is just a one dimensional we can imagine is very thin. So we're actually gonna use the one dimensional expressions. So what I need to do is I need to relate DM uh to the mass and the length that are given to me. And so that's actually gonna be this expression right here. So DM is equal to M over L times DX and then I just substitute it in for that expression. So now what happens is this F equation becomes GM and now I've got the integral of M over LDX, right? Because that's this piece right here, divided by L plus D minus X. And now I've got to square that. So now if you've looked, if you've taken a look at what happened, what's happened here, we've gotten AD X and we have this X variable. So now these two things talk to each other, they are the same sort of variable. So I can actually go ahead and carry out the integration. The last thing we have to do is we have to pull out the constants, right? We, we wanna be pulling out constants from the integral. So that M and L because they're cap capital letters can be pulled out. So we've not got that F is equal to and we've got GMM over L and now we've got the integral of DX over L plus D minus X squared. OK. So that is step six B and we're done with this. So now the last two things we need to do are we just need to determine the limits of integration. So that's based on the actual shape of the objects. So what happens is as we're integrating along the X direction because we integrate this way, we need to go from some starting point to some final points. The starting point is actually gonna be right here. So what I'm gonna do is I'm gonna call that X equals zero, right? We can sort of choose that to be the zero points. Now we get to the end of the rod and that's gonna be over here. Now, that distance is X equals L. So that's gonna be right here. So that's actually where we're integrating from zero to L. So those are gonna be our limits of integration. So that means that F is equal to and now I'm just gonna reverse M and M. So I've got GMM over L and now the integral of zero to L and we've got DX over L plus D minus X squared. OK. So if you, if your professor only wants you to set up the integral, sometimes you might just have to only do this in problems. Then we're done here. This is the answer. This is the integral expression with the correct limits and all of that good stuff, right? So this is the expression. So if your professor actually wants you to carry out the integral, then stick around because we're actually going to do that in part B. But if you only are asked to set up the integral, then this is the answer. So in part B, now we're actually going to evaluate the integral, which is step eight. So we're gonna use integration techniques to integrate this expression. Now, what one thing we can do is I've got an X over here in the bottom and I've got AD X that's gonna be on top, but I have this whole expression right here. So one thing I can do is I can just use a U substitution. So just think back to U substitutions, we're just gonna call you this whole entire expression right here. L plus D minus X. And so we take the derivative with respect to, to you to X and with these two things are consonants, the L and the D. So that means that those go away or they evaluate to zero. And when I take the derivative, I just get negative DX, right? It only affects this piece right here. So uh I don't want DU, I actually want DX because I have this DX right here. So one of the things I can do is say that negative Du is equal to DX. So this is the reason this is important is because every time I see L plus D minus X in this expression, which uh which is over here, I'm gonna substitute with U every time I see DX, I'm gonna substitute that as well. So what happens is this piece right here gets substituted and this piece right here gets replaced as well. So what our new integral becomes is our new integral becomes F equals GMM over L. And now we've got the integral of DU divided by and now we've got oh or sorry, negative Du, right. So don't forget the minus sign and then we've got U squared. OK. That's done on the bottom right here. So uh this integral can actually be written in a different way. Uh Because this integral looks a little nasty. Um So remember we can actually sort of evaluate this or we can rewrite this as a negative U to the minus two power times DU. Remember the minus sign just means that it goes in the bottom exponents, right? Or the the denominator flips. So how do we solve these integrals? Remember if this is just a normal integral. So I'm gonna scroll down um the integral of U to the N power DU is just equal to, we raise the power by one and then divide by that new exponents. So that means that this integral here F which is GMM over L is gonna be uh let's see. So when we evaluate this integral, we're gonna raise the power by one. So it's gonna be negative U to the negative one power and now we have to divide by the new exponent. So we're gonna divide by negative one. Well, hopefully you guys should see that these negatives will cancel. So we're gonna cancel out those and we just get you to the negative one power. So in other words, what happens is we just end up with GMM over L times U. The problem is we still have to evaluate this at the limits but we have to do is we have to change this. So we have to turn this back into X. So turn this back into X whenever you do a U substitution, you always have to change it back into X. So uh let's see, we, what we actually get is we get G big M little M over L and then we get L plus D minus X and there's only one power and now we actually evaluated the limits which is X equals zero to L. So for our final answer, sort of write down here, what we get is we get GMM over L and now parentheses one over when we evaluate this top limit over here, we're gonna get L plus D minus L and then we're gonna get minus one over L plus D minus zero. All right. So what happens is these LS will cancel out over here and then we're just left with this expression. So that means that for your final answer, your force is equal to GMM over L and now we have one over D minus one over L plus D and that's actually the whole entire integral. So that's that part B, all right guys, that's how you solve these problems. Let me know if you guys have any questions.
3
Problem
Problem
a) Set up the integral for the gravitational force between a rod with a density λ and length 2L, and a mass m at an arbitrary distance D directly above the midpoint of the rod.
b) Evaluate the integral.
A
B
C
D
4
example
Gravitational Force from a Solid Disk
Video duration:
13m
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Video transcript
All right guys. So we have another example here, we're gonna be calculating what the gravitational force is between a solid disk and a small mass that's here on the outside. Now, I just want to warn you, this example can be pretty challenging. So just make sure that you absolutely need to know how to do this. Maybe it's for exam or homework or something like that because the chances are you might most likely won't see it. OK. The other thing I wanna mention is that we're actually gonna use this hint to think about this disk as being made of very thin rings here. And we're gonna add together all of the contributions of force from these rings as we go out towards the outer edge of the disk. The reason this is really helpful is because it is we're gonna be able to use a lot of the steps that we use for the ring problem to solve this disc. OK. So with those two things being said, let's jump into the first step, which is that F is equal to GM the integral of DM over R squared. So that's step one. Now step two says that we have to construct two D MS and then relate it to the center of mass distance. When we did this for a ring, we said that the, uh the R distance was this guy right here. And we just use the relationship between this triangle that we've made. So we've got this triangle that we've made here and this little R distance was the square roots of D squared plus something. Now, what is that distance? What is essentially this vertical distance right here? Well, let's use the hints, this sort of thin concentric ring that I've formed has a radius of R prime. So not big R, we're gonna use that R prime variable instead. Now I know this can be pretty confusing because we have a little R and we have a little R prime. So what I want you guys to remember from this video is that this little R is not equal to R prime. These are two separate variables. Little R is the center of mass distance and this R prime is the radius of this sort of thin ring that I'm talking about right here. OK. So that means I have the expression for little R that's equal to R prime squared plus D squared. And that is step two. Now, step three says we basically have to break up the forces into their X and Y components, right? So we said that these rings produce these forces that pointed in this direction like that, that was my DF and we, we separated them into their X and Y components. That was D FX. And this was DFY. Now for the ring, we actually found out that there was always a mirror opposite DM that would cancel out that dfy. And the exact same thing is gonna happen here. So there's no DFY contributions because there's always gonna be equal and opposite symmetry. So that is gonna be step three, we're actually gonna get rid of this fy component and only the FX survives only these D FX s will add together. So that's gonna be the integral of D FX which is the integral of DF and now we have to relate it to either sine or cosine. So remember when we had this angle here that we sort of drew relative to the X axis, this was theta which means that this angle right here is actually going to be cosine of the, of theta because we need to relate it to the parallel component to this or the horizontal components. So we're gonna be using cosine. Now for step four, we actually needed to construct an expression for cosine by relating it to the sides of the triangle that we built. So this triangle here, the opposite the adjacent side, which is what we need for cosine is equal to D and the hypotenuse is O is equal to R. So we just had that expression. And now we can just plug this in for the cosine of the angle here. So that means that this just becomes F equals and we have GM integral of DM over R squared times D over R. So we can actually combine those two terms and say that this is equal to GM integral of D MD over R cubed, right. So that is step four. Now in step five, we just have to take this expression that we made for little R and just plug it into this equation. So we've got F is equal to GM and we've got integral of D MD over. Now we have R prime squared plus D squared to the three halves power, right? So we have three powers of that because this is actually equal to one. This is a, a one and one half power this square root right here. So we c it becomes three halves cool. So that's step five. And we actually have to pull out these uh intr these constants that appear. So this D is a constant, it's gonna pop out to the outside. This D is sort of involved with the changing variable with this R prime. So I can't pull it out. So that means that F just becomes now, I have G MD integral DM over uh R prime squared plus D squared to the three halves power. OK. So now we're done with step five. So now let's take a look at step six. So if we're only left with an integral of DM, then we replace it. And we're done. The problem is we still have this expression on the bottom here. So that means that we're not going with six A, we're gonna take a look at six B instead. OK. So six B says we have to change this differential which is DM and then relate it to the changing variable of integration, which is this R prime right here, right? So when we did this for the ring, we just related this DM by using the one dimensional shape equation. So for a ring or a rod or something like this, we were able to relate it to these equations, but this is actually a two dimensional shape. So we need to use these equations, not these ones. So we need to do is say that DM is related to the total mass over the total area times the differential area. Now, the problem with this expression is that this differential area D A here doesn't match up and doesn't get us closer to matching it up with this changing variable of Dr prime. So we're gonna need another expression here, right. So the A, so if you can think about this, the total area of a disk is equal to pi times big R squared, right. So this is the total area of this disk if I wanted the differential area, which is the area of this tiny, little like sort of ring or these thin concentric rings, I have to take the derivative of the area with respect to little R or just with respect to R. So this actually becomes too pi and then instead of big R, I'm gonna replace it with little R prime because that's our changing variable times Dr prime, that's the differential. So these expressions here are we gonna be used to sort of relate this D A. So now what happens is my F can become G MD and now this is gonna get, this is gonna give the integral of M over A. And by the way, these are the total masses and total areas times two pi R prime, Dr prime divided by R prime squared plus D squared to the three halves power man, that's a mouthful. So now what happens is I've got my differential Dr prime related to the changing variables, which are, those are the other R primes there. Now everything sort of talks to each other. These are all the same variables so I can carry out the integral. But the last thing I have to do is I always have to pull out the constants, right? It's always gonna be easier for me to pull out these constants towards the outside. And then I'm just gonna replace this area expression here uh with this pi R squared. So that means that my expression just becomes F is equal to. Now, I've got G uh M MD and then I have a two pi now I have divided by pi times big R squared. And so what happens is I have P that cancel and now I have the integral of RDR divided by uh R prime squared plus D squared to the three halves power. OK. So I'm almost done here, almost done. We just have to determine what our limits of integration are so that we always get the limits of integration by relating it to where we start and we stop in our integration. So in other words, we're adding up a lot, all of these rings starting from the center of the disk. And then we're gonna go all the way out until we hit the edge. So in other words, we're going from our prime equals zero, which is the center of the disc all the way to the edge of the disk where it becomes big R. These are actually the limits of integration we're gonna use. So that means our equation becomes F equals GM MD. Uh There's actually a two this here over R squared and now we have the integral from zero to R and then we're gonna have RDR over R prime squared plus D squared to the three halves power. OK. So just for those of you who only had to set up the integral, you're done, this is the last step that you have to do all you had to do is just set up that integral. Now, if you actually want to carry it out, we're gonna need some integration techniques. So that's what we're gonna do in step eight. So if you all you were here for was just setting up the integral, you're done, you can just move on to the next video. So for step eight, now what we have to do is we have to basically do a U substitution because we have an R prime and we have AD R that's on the top here. So I've got the U is gonna be R prime squared plus D squared. So if I want DU, I just have to take that derivative with respect to R prime. So that's just gonna be two R prime times Dr prime. Uh But I want this RDR prime because that's what exists on the top of this numerator here. And so what I'm gonna do is I'm just gonna move the two over to the other side and I get DU over two is equal to RDR. So now what happens is I'm just gonna replace this RDR with this expression and I'm gonna replace this guy with a U. That's right here. So my integral just becomes F is equal to. Actually, I'm gonna wait, I'm gonna sort of hold off on all of this stuff. And I'm gonna just say the integral uh this just becomes the integral of DU over two. And then we have U to the three halves power which is equal to uh this is equal to U to the three halves power. Uh So to the negative three halves power over two times DU. OK. Um Cool. So let's actually carry out this integral. And what ends up happening is we get that this integral just becomes um we're gonna raise the power by one which is gonna be U to the negative one half power, Then we have to divide by the new exponent that we got. So there's a two on the bottom here and we're gonna multiply it or as it is divided by negative one half, right? That, that goes to the bottom there. OK. So now what we have to do is we can cancel some stuff out. Uh And now we can basically substitute our expression for F back inside of this. I just don't want to keep writing it over and over again. So we've got two GM MD over big R squared. And now what happens is this two will cancel out with this two right here. And then we have to turn this U back into that R prime, right? So we always have to relate this back to our integration variable. So this F actually becomes, we have GM MD over R squared. When we divide by a negative one half, it becomes a negative two. So that's gonna go on the outside here. That's negative two or it's gonna go on the top. And this is U to the negative one half power, which actually means it's on the bottom here and it's a square root. But remember we have to replace this U with that R prime. So it's actually gonna be the square root of R prime squared plus D squared. So this is actually what our integral sort of evaluates to. So now that we've changed it back, now, we have to evaluate this at R prime equals zero and big R, OK. So uh we're actually gonna do that sort of, let's see if I can create some room. Uh So we're actually gonna do that all the way over here. Um So I've got that F is equal to and then I've got GMM uh then I've got D over R squared and then I have negative two over and I have the square roots of, and I'm gonna have big R squared plus D squared and I have minus negative two. Actually, this is gonna be minus negative two over the square roots of zero plus D squared. So now what happens is I get the zero that cancels out and I got two negative signs that turn it into a positive sign. So what happens is now, I can sort of rewrite this as F equals and have GM MD over R squared. And I can actually pull out this two that exists. So I'm gonna pull out this two. I'm gonna replace it over here. So this is gonna be F equals two. And now I have, this is going to be a one over. And this is actually gonna be this term right here is gonna just boil down to ad because the square root of D squared is D and then I have minus, which is gonna be this term one over. And this is gonna be the square roots of R squared plus D squared. And that is actually the answer. So I ran it a little bit of room there on the bottom. But this is your final answer for the gravitational force of that disc. All right. So let me know if you guys have any questions and I'll see you guys in the next one.
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