find-the-indicated-moment-of-inertia-or-radius-of-gyration-find-the-radius-of-gyration-of-a-plate-13

Question

Find the indicated moment of inertia or radius of gyration.
Find the radius of gyration of a plate covering the region bounded by $x=2, x=4, y=0,$ and $y=4,$ with respect to the $y$ -axis.

Stanley Enemuo · Accepted Answer

well given their wash up too equal to four And there was born the zero and the X. Up to it&#x27;s B. And x. sub one. It&#x27;s a. Also equal to two. No. To calculate the moment of inertia we have K times the integral of for X squared With the limits being two and 4. That&#x27;s equal to Forte Tom&#x27;s excellent. Third over three. I give this 22,000. I give this 224 k. Over three and then the mass of the plate stealing by the former. So we have K times the integral of four X. Dx. There are limits being 24 and we get a K. Now on the rages. Our sub Y is equal to the moment of inertia over the mets. And that gives us 3.06.

Eric Mockensturm · Answer

we have to find the area moment and radio driving radius of gyration of this cross section again. But now about the y axis with the y axis goes up the center here. This one&#x27;s easy. Because this why access passes through the central rate of all of our pieces of our geometry here. So we just have, you know, 1 12 HB cubed or, um, basically all on all of these cases where this is the, um, base or the Here&#x27;s Yeah, here&#x27;s the height h on this is being. And so if we plug in those values from the what we get from our from the dimension from the book, you get that this is zero point, as you are, too. 0.2 05 centimeters to the fourth. This one up here is your 0.691 centimeters for the fourth. And this one here is five point, um, 53 centimeters to the fourth, and we can see that this one is quite a bit larger because again, it has a lot of area away from this x axis. So we at we had all these up and we get that the area moment about the Y axis is six point to 4 to 6 centimeters to the fourth. The area we have from before and so we can get the radius of gyration about the Y Axis is 0.887 senators.

Eric Mockensturm · Answer

now, I asked to find the area moment and radio preparation about the Y axis for this shape. And we have all the dimensions. And so, um, the Y axis goes through the central rate of this piece and the central of this piece. So we just get these simple expressions here, and we wind up with I about why this access for this member is 10.67 inches to the fourth. Um, for this member, it&#x27;s 5.33 interest of the fourth. They looked the same here, but I&#x27;ve drawn it out of scale. And in fact, this is much thicker than this. So this makes sense. And then these guys, basically we need the take the area moment about about why of its about to its central right. And we get this. And then we need to take punk, apply the area times the distance from the why access to the centrally and that I called the three and we can get that from the schematic in the book. And we also know that this one is gonna be the same because of symmetry about this, and so that these two each one has a area moment of 15.25 inches to the fourth. So our total area moment about the why access is, um this Plus this plus two times this and we get 46.5 inches to the fourth. Our area is again 18 interests squared, and our radius of gyration becomes 1.6 or seven inches.

Eric Mockensturm · Answer

who asked to determine the moment of the second area moment of the and the radius of gyration of the shaded area. With respect to the X axis, Mr are shaded area looks something like like what I&#x27;ve sketched here and what we&#x27;re gonna do is break it up into into three parts. So the first part, it will be the center beam here or the center section. So this part here and then to would be this upper flan and three will be this lower flan so we can base, weaken, take, find the area moments for each section and then add them up to find for the total. And we want to do it about, um, this X axis here, which is in the center of this vertical member. So the, um, formula for the area moment of a rectangular shape about its central aid, which this is that, um, is 1/3 be one h one cube. So be one. Is this thickness And H is this distance here, and we&#x27;re giving those in the problem. And those there they are. And now for two, we can take the area moment about its center of mass. All right. It&#x27;s, um, sense. Right? And then we need to shift it using the parallel axis deer. I&#x27;m from here to here. And so that distance ties. So then we need the area of this, which is B to H two and then this distance here. Um, and, um, that&#x27;s deep. Well, call that D to so we know, we know a, um that&#x27;s totally And so here&#x27;s the values for, um, for here. And then we have the third part, which is this part here. And so we have the center. We have the center, the area moment about the central right, and then plus the area times this distance from the central to our X axis. I call that the three. And these were all given in the problem. Calculate the total area of this cross section and that we get 8.16 centimeters squared. Um, and then we can calculate the area moment about X, and we just need to add all these things up. So this is I x one I x two and I x three. And they seem reasonable because, you know, this is gonna pretty high area moment about excuse. It&#x27;s tall this one because it&#x27;s far away from the X axis and this one is bigger than this one and because it&#x27;s much wider. So that looks reasonable. And if we had everything up, we get 39.0 centimeters to the fourth. And then if we want to find the radius of gyration, we just need to take the square root of this divided by this and we get 2.19 centimeters.

Eric Mockensturm · Answer

you can ask to find the area moment and the radius A generation of the shaded region with respect to the X axis. Um, and our our shape looks something like this where X axis goes through the center of this member here and were given all the dimensions that we need in the in the problem. And so I divided it into four regions, 12 in blue and 12 and red. Now we need to find the area moment of each of those regions about this X axis, and we can just start doing that, you know, we can see that the, um in red here, the X axis goes through the central right, So we just get the area. Moment, for that part is simply this expression here that will be used many times. And then for the second Red Member, we have it&#x27;s area moment about its center of mass. Plus, it&#x27;s area times the distance from this, uh, century the center it of this to the central to this and that I called to and again we can get all of these dimensions from the schematic in the in the book and then for um, the radius, Uh, the area moment. About of X one, that Red First Member or the Blue First Member. It turns out that the X axis passes through its central aid and same with this one and that since these two are the same they have they&#x27;re gonna have the same area moment about that excess. And so that&#x27;s just, um 1 12 b one h one where we used blue for a different be one on H one that appear. And we have those values. And we can, um, plug everything in and we get this thing here is to point, um, 67 inches to the fourth. This here is 25 point three inches to the fourth and weaken. This is much bigger than this, because this is farther away from the X axis. Um, and then these two guys both have area moment of nine inches to the four. So we gotta multiplying by two. So we have two of them, and then if we add everything up, we get 46 inches to the fourth, then the radius of gyration is easy, because we can calculate the area of all these rectangles and had them up when we get 18 inches squared, and then we take the square root of this divided by this and we get 1.6 inches.

Eric Mockensturm · Answer

in this problem were asked to determine the second moment of area and the radius of duration of the shaded area shown with respect to the Y axis and our region are area of interest is bounded by these two functions. Here we know that that why equals a the both equal be so we can get K one in terms of A and B and K two in terms of A and B the area where the integral we need to do to get the second moment of area about the Y axis is X squared D A and a. Our region goes from y equals. Why one dysfunction to y equals y to this function and X goes from zero to a We can crank through those in a girls and we get that i sub y equals 3/35 times a cube. Be now the find the radius of gyration. We need to calculate the area and instead of the second moment of area, we just have the area itself for the first moment covered the is your moment of area and we have a one here instead of an X instead of X squared and we can do this integral. And we get that the area is a B over three. The radius Our generation is the square root of I over a and that gives us three a all over square root of 35.

Problem

Find the indicated moment of inertia or radius of…

Problem 8 Easy Difficulty

Find the indicated moment of inertia or radius of gyration.
Find the radius of gyration of a plate covering the region bounded by $x=2, x=4, y=0,$ and $y=4,$ with respect to the $y$ -axis.

Answer

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Kayleah T.

Caleb E.

Michael J.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 26

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Caleb E.

Michael J.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Basic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Caleb E.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Caleb E.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus