Mass Distribution with Calculus - Online Tutor, Practice Problems & Exam Prep

Calculus is essential for calculating gravitational forces in non-spherical mass distributions. Using Newton's law of gravitation, we treat differential mass elements (dm) as point masses, generating tiny forces (dF) that we integrate to find total force (F). The equation is expressed as F^2 = GM⁢r^2 + d^2. This approach allows for accurate calculations in gravitational problems involving complex shapes.

Using Calculus to Solve Mass Distribution Problems

We've got that. We're going to 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 m₁ and m₂, 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 kinda just pretended that these planets or these objects could be treated as points, and all the mass was 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 is that little r, it basically worked the exact same way. So instead of m₁s and m₂s, we just had big M's and little m's. So we just used 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 the masses are 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 M, big M that I have here, can be thought of or broken up into tiny little pieces of rectangles or mass right here. And the smaller they become, they become a differential mass. So I'm going to call that dm, 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 going to be df, and it's due to that center of mass distance of little r right here. df=g•dm•m/r2 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 add 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•dm•m/r2. 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 more useful form that we're always going to use here at Clutch when we solve these problems is we're going to use the g•m•∫−∞dm/r2. So we're always going to start from this equation when we're solving these problems.

Alright, guys. That's basically it. So we actually have a list of steps that we're going to 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 going to work out this example together. This is actually a very classic common type of problem that you'll see if you're doing this topic right here. So we've got a hollow ring. It's got a mass m. It's got some distances over that radius. We have a distance 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 g•m•∫−∞dm/r2. So that's the first step. Okay? So the second step is we have to pick Let's see. It says pick 2 DMs and we have to write an expression for r. So I've got this DM that's going to be over here and then I'm going to pick another DM like this over here and then we know that these DM's act like point masses and they produce forces on this mass over here. They're going to point in this direction like that. That's going to be 1 df and then this guy is going to be in this direction and that's going to be another df. Okay? So, now we actually have to figure out what the r's are, the center of mass distance is. So, that is this right here, these pieces right there, and then I've got one over there. So, this is going to be r. Okay. So, we have to use, write an expression for f using the problem's geometry. That means is that we're just going to use the 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 little r. It's actually going to be the square roots of r² plus d². Okay? Now, we're just going to save this for later. We're not actually going to start plugging it in yet because what happens is if we start plugging this in, we're going to have to write this a bunch of times. It's going to get really annoying. Okay? So we have this second step right here, so we have the expression for r. Now, let's take a look at this third spep here. Step 3 says we have to break, 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 2 things. The f x component is going to be the integral of all the DFXs, and the f y component is going to be the integral of all the DFY's. Now, where do these components actually come from? Well, remember that these D F's actually are 2-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 going to do is I'm going to, 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 dfx 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 going to be here and then another one of these components that's going to be over here, so dfy. So what we can see here is that my dfx components are always going to be sort of pointing to the left, and they're always going to be together, and they're going to be adding together. Whereas these dfy's 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 as 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 dfx. So now that I've actually sort of split this and canceled out the components, I have to expand this into sines and cosines. So remember that this dfx components can be written with this angle as the integral. Actually, I'm going to have g•m•∫−∞dm/r2, and this is actually going to be the cosine of that angle right here. So, now what I have to do is I have to expand this into sine and cosine, which I've done. Now I have to rewrite this cosine in terms of the side lengths 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 going to be d / r. So we're actually just going to replace that cosine of θ in there for d / r. So that means that for 4, my equation becomes f equation becomes f x is equal to or we could just write f, is equal to the integral, which actually is going to be g•m•∫−∞dm•d/r3. Okay? So that is step 4. So we're done with that. So now what we have to do is for step 5, we have to plug in the expression for R from step 2, and then pull all the constants out of the integral. Okay. So we've got f=g•m•∫−∞dm. Now we have the integral, and then we have dm. And then what happens is we're going to plug in our expression for r. Now remember that expression for r is this guy over here. So, actually, what this r³ becomes is it becomes r2+d2. And then because this is square roots, which really means that this is, to the one-half power, this actually becomes to the 3/2 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 constants outside of the integral. But if you take a look at this, remember that d is an integral is just a constant. Right? This is just a constant horizontal distance up here. So that is, that just gets pulled out. And then this r² and then d², 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. Alright. Now, we're done with that, so we pulled all the constants out of the integral, and then we're going to rewrite this. So, this actually ends up being, f is equal to, and then we've got g•m•d/rd3. Okay? And then we've got to just write that last little remaining integral piece. So you have the integral of DM. Okay. So now what happens is if we look at step 6 a, if we're only left with DM, the integral of dm, then the integral of dm here is just m. Right? So 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 G•M•M•d/r+d23/½, and that is actually our force. So we don't have to do the integral because we already did it in this step. Okay, guys? So let me know if you guys have any questions, and I'll see you guys in the next one.

Gravitational Force of Rod Parallel to Axis

Alright, guys. We're going to 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. We're solving for the gravitational force, but this is a mass distribution, remember, because this is non-spherical. So, we're just going to go ahead and stick to the steps. The first thing that we're going to have to do, and by the way, the first part of this problem is just set 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 \( g \times m \), and then it's going to be the integral of \( \frac{{dm}}{{r^2}} \). So boom, that's the first step. You always start with this equation.

Now for the second step, it says we're supposed to pick 2 \( DMs \) along our sort of shape or object, and so we're going to pick this \( DM \) over here. So this is going to be \( DM \), and we're going to get this one over here. This is also going to 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 \( l + d \). 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 not actually 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, that I've highlighted in blue. So, it's clear that we're not actually going all the way from \( l + d \). We're going to subtract something. And because this sort of distance here is variable that we're not using, I'm going to assign it a variable, and we're going to 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 \( l - x \). It's the whole entire thing, minus this little piece right here.

For step 3, it says we're going to 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 \( g \times m \), and then integral of \( \frac{{dm}}{{r^2}} \). So it's just going to be the exact same thing.

Now, step 4, we're going to expand 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 going to skip steps 3 and 4, and we're just going to go straight into step 5, which is we're actually going to plug in our expression for \( R \) into the integral, and then we're going to take out all the constants. Right? So that means that we're going to do \( f \) equals, and now we've got \( g \times m \) integral of \( \frac{{dm}}{{(l - x)^2}} \). So can we pull out any constants? Well, remember that \( l \) and \( d \) are constants. Right? These are actually just lengths 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 step 6. Right? Step 6 has 2 different possibilities. If we're just left with the integral of \( DM \), we're going to replace it. But as we've seen, we're not going to pull out 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 going to go with 6a, we're actually going to go with 6b. Which is, otherwise, we're going to 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, using the density right here. So this is something we haven't talked about just yet.

So, what happens is we're going to get this \( DM \), and we can relate this \( DM \) to the density, which is \( \lambda \), and we can write a different expression for that involving the mass and the length of the object. And by the way, you're going to see all of these expressions. This is only for 1-dimensional shapes, whereas we're going to use these equations for 2-dimensional shapes, and then these for 3-dimensional shapes. So this would be like a flat sheet or a 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 one dimensional. We can imagine it's very thin. So we're actually going to use the one-dimensional expressions. So, what I need to do is I need to relate \( DM \), to the mass and the length that are given to me. And so that's actually going to be this expression right here. So \( DM \) is equal to \( \frac{{m}}{{l}} \times dx \), and then I just substitute it in for that expression.

So now, what happens is this \( f \) equation becomes \( g \times m \), and now I've got the integral of \( \frac{{m}}{{l \times dx}} \), right, because that's this piece right here, divided by \( (l - x)^2 \). So now, if you've looked if you've taken a look what have happened here, we've gotten a \( dx \) and we have this \( x \) variable. So now these two things talk to each other. They're 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 want to be pulling out constants from the integral. So that \( m \) and \( l \), because they're capital letters, can be pulled out. So we've not got that \( f \) is equal to, and we've got \( g \times m \times m \), over \( l \), and now we've got the integral of \( \frac{{dx}}{{(l + d - x)^2}} \).

Okay? So, that is step 6b, 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 point. The starting point is actually going to be right here. So, what I'm going to do is I'm going to call that \( x = 0 \). Right? We can sort of choose that to be the 0 points. Now, we get to the end of the rod and that's going to be over here. Now, that distance is \( x = l \), so that's going to be right here. So that's actually where we're integrating from, \( 0 \) to \( l \). So those are going to be our limits of integration. So that means that \( f \) is equal to and now I'm just going to reverse \( m \) and \( m \). So I've got \( g \times m \times m \), over \( l \), and now the integral of \( 0 \) to \( l \), and we've got \( \frac{{dx}}{{ (l + d - x)^2 }} \).

Okay. So if your professor only wants you to set up the integrals, 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 8. So we're going to use integration techniques to integrate this expression. Now, one thing we can do is I've got an \( X \) over here in the bottom, and I've got a \( DX \) that's going to 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 going to call \( u \) this whole entire expression right here, \( l + d - x \). And so we take the derivative with respect to to to \( x \), and with these two things are constants, the \( l \) and the \( d \). So that means that those go away or they evaluate to 0. And when I take the derivative, I just get negative \( dx \). Right? It only affects this piece right here. So, 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 \( l - x \) in this expression, which, which is over here, I'm going to substitute with \( u \). Every time I see \( dx \), \( x \), I'm going to 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 \( f = g \times m \times m \), 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 \( u^2 \). Okay? That's on the bottom right here. So, this integral can actually be written in a different way, because this integral looks a little nasty. So remember we can actually sort of evaluate this or we can rewrite this as, negative \( u \) to the minus 2 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? Well, remember, if this is just a normal integral, so I'm going to scroll down. The integral of \( u \) to the \( n \) power \( du \) is just equal to we raise the power by 1 and then divide by that new exponent. So that means that this integral here \( f \), which is \( g \times m \times m \), over \( l \), is going to be, let's see. So when we evaluate this integral, we're going to raise the power by 1, so it's going to be negative \( u \) to the negative one power. And now we have to divide by the new exponent, so we're going to divide by negative one. Well, hopefully, you guys should see these negatives cancel, so we're going to cancel out those, and we just get \( u \) to the negative one power. So in other words, what happens is we just end up with \( g \times m \times m \), over \( l \), times \( u \). The problem is we still have to evaluate this at the limits, but what we have to do is we have to change this. So we have to turn this back into \( x \). So, let's see. We what we actually get is we get \( g \), big \( m \), little \( m \), over \( l \), and then we get \( l + d - x \). And there's only one power. And now we actually evaluated the limits, which is \( x = 0 \) to \( l \). So for our final answer, sort of right down here, what we get is we get \( g \times m \times m \), over \( l \), and now parenthesis, one over when we evaluate this top limit over here, we're going to get \( l + d - l \). And then we're going to get minus one over \( l + d - 0 \). Alright. So what happens is these \( l's \) 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 \( G \times M \times M \), over \( L \), and now we have one over \( d \) minus one over \( l + d \). And that's actually the whole entire integral. So that's that part b. Alright, guys. That's how you solve these problems. Let me know if you guys have any questions.

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.

Gravitational Force from a Solid Disk

Alright, guys. So we have another example here. We're going to 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 an exam or homework, or something like that, because the chances are you might most likely won't see it. Okay. The other thing I want to mention is that we're actually going to use this hint to think about this disk as being made of very thin rings here, and we're going to add together all 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 we're going to be able to use a lot of the steps that we use for the ring problem to solve this disk. Okay? So with those two things being said, let's jump into the first step, which is that \( F = G M \int \frac{dM}{r^2} \). So, that's step 1. Now step 2 says that we have to construct two DMs and then relate it to the center of mass distance. When we did this for a ring, we said that 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 root of \( d^2 + r'^2 \). 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 \( r' \). So not big R. We're going to use that \( r' \) variable instead. Now, I know this can be pretty confusing because we have a little r and we have a little \( r' \). So what I want you guys to remember from this video is that this little r is not equal to \( r' \). These are two separate variables. Little r is the center of mass distance, and this \( r' \) is the radius of this sort of thin ring that I'm talking about right here. Okay? So, that means I have the expression for little r, that's equal to \( \sqrt{r'^2 + d^2} \), and that is step 2. Now, step 3 says we have to basically 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 always separated them into their x and y components. That was \( dF_x \) and this was \( dF_y \). Now, for the ring, we actually found out that there was always a mirror opposite dm that would cancel out that \( dF_y \). And the exact same thing is going to happen here. So there's no \( dF_y \) contributions because there's always going to be equal and opposite symmetry. So, that is going to be step 3. We're actually going to get rid of this \( F_y \) component, and only the \( F_x \) survives. Only these \( dF_x \)'s will add together. So that's going to be the integral of \( dF_x \), 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 theta because we need to relate it to the parallel component to this or the horizontal components. So we're going to be using cosine. Now for step 4, we actually need to construct an expression for cosine by relating it to the sides of the triangle that we built. So this triangle here, the adjacent side, which is what we need for cosine, is equal to d, and the hypotenuse 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 \( F = G M \int \frac{dM d}{r^3} \). So, we can actually combine those two terms and say that this is equal to \( G M \int \frac{dM d}{r^3} \). Right? So that is step 4. Now in step 5, we just have to take this expression that we made for little r and just plug it into this equation. So we've got \( F = G M \int \frac{dM d}{(r'^2 + d^2)^{3/2}} \). Right? So, we have three powers of that because this is actually equal to 1, this is a one and one-half power, the square root right here. So we cube it, it becomes \( 3/2 \). Cool. That's step 5. And we actually have to pull out these, integral these constants that appear. So this d is a constant, it's going to pop out to the outside. This d is sort of involved with the changing variable with this \( r' \), so I can't pull it out. So that means that \( F \) just becomes, Now, I have \( G M d \int \frac{dM}{(r'^2 + d^2)^{3/2}} \). Okay? So now we're done with step 5. So now let's take a look at step 6. 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 6 a, we're going to take a look at 6 b instead. Okay. So 6 b says we have to change this differential, which is dM, and then relate it to the changing variable of integration, which is this \( r' \) right here. Right? So when we did this for the ring, we just related this dM by using the one-dimensional shape equations. 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 dA here doesn't match up and doesn't get us closer to matching up with this changing variable of \( d r' \). So we're going to need another expression here. Right? So, the A So if you can think about this, the total area of a disk is equal to \( \pi \times r^2 \). Right?