Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by extremely light rods (Fig. E9.28). Find the moment of inertia of the system about an axis

(c) that passes through the centers of the upper left and lower right spheres and through point O.

Video transcript

Hey, everyone in this problem, we have a rectangle that measures 30 cm by 50 cm. We have point masses weighing 420g fitted on each of the four corners. We're told to assume that the connecting rods between the point masses have negligible mass. And we're asked to find the moment of inertia of the system about an axis perpendicular to the square passing through one, one of the corners and to the center of the short edge at point P. Alright, so let's go ahead and think about what is the moment of inertia. Now, the moment of inertia for point masses recall is going to be given by the sum over J of the mass M J times a radius R J squared where J is the number of point masses that we have in our system. Okay. So for our system, we have four corners, we have four point masses. And so this is going to be equal to M one R one squared plus M two R two squared was M three, our three squared plus M four, our four squared. Now the mass M 123 and four, we know are all the same and we know that there are 420 g of the mass M that's equal to 420 g. Okay, let's go ahead and write this in our standard unit of kilograms because we can see that the answer choices for the moment of inertia are kilogram meter squared. And so we're gonna want our mass in kilograms. So this is going to be equal to 420 g times one kg per g. Okay. So to go from grams to kilograms, we divide by 1000 Here, the unit of Graham Cancell and we're left with 0. kg. Alright, Let's start with Part one. Okay. Now that we know in general what we're looking for and .1 is asking us to find the moment of inertia if the axis is one of the corners. So let's label our corners, top left is going to be corner one, top right is going to be corner to bottom right is going to be corner three and bottom left is going to be corner four And let's choose this first corner corner one to be our access, Okay. four per one of this question. So this is gonna be I'm gonna call it see that's gonna be our axis. All right, so we know the mass, okay. Mass of all the corners is the same and our is gonna give us a distance from that access two the point mass. So our one is going to be the distance from Point Mass one, which is his top left corner to the axis. Now, we've chosen the access to be this corner. And so the moment of inertia or sorry, the distance R one is just gonna be zero m. All right, what about our two R two is going to be the distance from 0.2 To our axis, which is at .1. So that's going to be the distance of the wit which is 50 cm. So our two is going to be 50 cm. And again, we want to convert this to our standard unit of m. So we multiply by one m per 100 centimeters. And so we get that R two is going to be 0.5 m. Okay to go from centimeters to meters. We effectively divide by 100. Alright. What about our three, two or 3 here? Well, R three is down in this bottom corner and we want to find the distance to this top corner. Okay. We know the length With a width is cm, you know that the height is 30 cm. And so we can use some triangle math, we can use our Pythagorean theorem to find this hypotenuse which is going to be our three. So doing that are three squared, OK. By the Pythagorean theorem, it's going to be equal to 30 centimeters squared Plus 50 cm squared, our three squared is equal to 3400 centimeter squared. Alright. And now we have centimeters squared, we want to convert this to meters squared. So in order to do that, we have 30,400 cm, squ, we're gonna multiply for every one m squared, We have 10,000 centimeters squared. So we divide by 10,000, we get that are three squared is equal to 0. meters squared. Now you might be tempted to take the square root here and you can go ahead and do that. However, if you look at our equation, our equation for moment of inertia has R squared R one squared, R two squared R three squared. So we can actually just leave this as our three squared instead of taking the square root and then having to square it again. So let's just leave it as our three squared. And the last distance we have to look at is our four and our four is going to be the distance from point mass four to our axis, which is point mass one right now. Okay. So we have the distance from 4 to 1. You know, it's going to be that width On the small side of our rectangle which is 30 cm. and again converting two m, we divide by 100, The unit of centimeter cancels and we have 0. m. Alright, so we know our mass M, we know all of our distances R one R two R three and R four, we can use our moment of inertia equation M one R one squared plus M two, R two squared plus M three, R three squared plus M four R four squared in order to calculator moment of inertia. So our moment of inertia And when we have our access at the corner, I see is going to be the mass, 0.4, two kg times the sum of the distances squared. So we get zero m squared and we've just Packard out that mass because it's the same for all of those terms plus our two squared, 0.5 m squared plus our three squared, which is 0.34 m squared. Okay, if you do decide to leave our three squared a 0.34 m squared instead of taking the square root and squaring it in the equation, just be sure that you don't accidentally square it again in the equation. Okay. We have .34 m squared and then we have 0. m all squared. Alright. So if we work out what is inside of our brackets there, we are going to get 0. meters squared. And when we multiply these two together, 0.42 kg times 0.68 m squared, we're going to get zero point 2856 and the unit is kilogram meters squared. Okay. And so that is our moment of inertia. If we take the axis Perpendicular to one of the Corners. And if we look at our answer choices, if this was on a test, we would see that that corresponds with answer choice C. But let's go ahead and do part two to make sure that that answer works out as well. And just to see how we would approach the problem. Alright. So part two is asking us to have the access at the center of the short edge at point P. Alright, so what is our one going to be if her access is at point P? Well, our one is going to be the distance from .1 two point P. We know that that's half the distance of this smaller side, the side is 30 cm. And so the distance there is going to be 15 cm, 30 divided by two. So our one is going to be 30 centimeters Divided by two which gives us 15 cm. And again, we're converting this two m, we multiplied by one m per 100 centimeters or effectively dividing by 100. We get that this is 0.15 m. All right. What about are too, let's go back to our diagram. Now we have our two, okay. And we can see the distance from point mass to to P, we can form a triangle again. But because P is in the middle, we actually know that this is going to be the same as the distance 2.3. Okay. So we can calculate R two and R three together. We know that R two is equal to R three. And we can find it by looking at this triangle, no, The bottom is going to be 50 cm. The side is going to be 15cm okay up to the distance of P and so Her hypotenuse will be our two or 3 using Pythagorean theorem. We have the R two squared is going to be equal to 15 centimeters squared plus 50 centimeters squared. Okay. Now, we've already converted both of these values into meters previously. So this is going to be 0.15 m squared Plus 0.5 m square. Okay, we can look back at our previous calculations to see those conversions. And this is going to give us an R two squared value of 0.27, 25 m squared. We're gonna leave it as our two squared. Okay because in our equation, we have our two squared. So no need to take the square root and then square it later. And because R two is equal to R three, R two squared is going to be equal to R three squared. Okay. So we have our one value R two value in our, our three value. Now we need to do our four. Well, it turns out that our four is going to be equal to R one. If we look back at our diagram we can see that the distance from this midpoint p of the side separating corner one and four is going to be the same. So we have the distance from 40.1 to P cm. And so the distance from .42 p is also 15 cm. Alright. So we have all of our distances, we have our masses. Let's go ahead and work out our moment of inertia and I where the access is at point P is going to be equal to the mass, which is the same in every corner times of some of these radius is squared, 0.15 m all squared plus 0.27 to five m squared. Okay. That one, we left as our two squared. So we don't need to square it again. Same with our three squared, 0.27 to five m squared. And finally plus 0.15 m all squared. This is going to give us 0.4, two kg times 0.59 m squared for a final value of 0. kg meters squared. Okay. And that is going to be our moment of inertia if we take it the axis at point P. Alright. So if we look back at our answer choices, we see that we found a moment of inertia if we took our access perpendicular To the square and passing through one of the corners as 0.286. And if we took it about the center of this short edge at Point P, it was 0.248. And so our answer is going to be C thanks everyone for watching. I hope this video helped see you in the next one.