find-the-indicated-moment-of-inertia-or-radius-of-gyration-find-the-radius-of-gyration-of-a-plate-15

Question

Find the indicated moment of inertia or radius of gyration.
Find the radius of gyration of a plate covering the first-quadrant region bounded by $y^{2}=1-x$ with respect to the $x$ -axis.

Stanley Enemuo · Accepted Answer

the moment of inertia. I said along the X coordinate this K times the integral of Y squared minus Y to the fourth. Dy With the limits beams during one. This be hey integral of Y squared his wife to the third of three minus the integral of negative one to the fourth. This negative watch. Yeah, massage With limits being giant. one that gives us to take over 15. Now we can find the mass of the solid using the same form as previously discussed. We have the integral of one minus Y squared. You are Again using the same limits which is there in one. This gives us K two K of the three. And then we could find our which is the square root of the the moment of inertia over the man. And that zero points for 47.

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

who asked to determine the second moment of area and the redistrict freight of generation of the shaded area expected the X axis. So the shaded area is bounded by K one X squared and que two squared of axe. We know that at X equals a. They both equal be pan, so you can get K one and K two in terms of A and B. The integral we need to do to get the second moment of area about the X axis is to integrate y squared over the area, and the area goes from y equals y one to y two on X course from zero to a you can crank through that and we get that I set backs equals 3 35th a be cute. That should be a cute and we get that the area. If we instead of putting, why here we just put one so that will give us the area and we get that the area is a B over three, and then the release gyration about the X axis is square root of I Cybex all over a and that gives us three B all over a squared or 35

Eric Mockensturm · Answer

we&#x27;ll ask to determine the second moment of area and the radius sub duration of the shaded area shown with respect to the X axis. The area is bordered by why because K X square we know that at X equals two a. Why is aged okay? Was h over for a square theater? Go we need to do is integrating y squared over the area and that area goes from y equals zero to y of X and X equals zero r A to a And if we crank through those in a girl&#x27;s with not pretty but we can do it We get that ice of X equals 127 over 1344 eight times eight cube. And if we Catholic the numerically we get 0.9458 cube the area Can you if we just plug in one here we get the area and do these in a girls and we get that that is 7/12 h and then the radius of gyration. And this was respect to X access is, um, square root of this divided by this. So we get that that turns out to be square root of 1 27 all over 28 times H, or 0.40 to H.

Eric Mockensturm · Answer

were asked to find the second moment of area in the radius of gyration of the shaded area shown with respect to the X axis. And that area is bounded by why want equal see, signed K X and Y two equals M X plus B. We know we have four conditions on these functions. That why one at X equals A S H by two at X equals A is 24. Why one and why two at two. A. Or both. Zero. So that tells us we can get these for Constance here. Um, in terms of a terms of h. So we get that M equals minus two h over a B equals for h C equals eight and K equals pi over two A. To get the second moment of area about the X axis we need to integrate. Why squared over the area of interest? And that area goes from why was why one toe y two and X equals a to a and we can crank through these ugly in a girls. And when we do that, we get that ice of X equals to a cube all over nine pi times three pi minus two. If we calculate that numerically we get 0.525 a cube, we can find the area by integrating one over the area of interest. And that area is a H times one minus two over pie. And then the radius situation is just the square root of ice of X all over a and that is K X, and that gives us a value of zero of 1.202 times h.

Problem

Find the indicated moment of inertia or radius of…

Problem 12 Easy Difficulty

Find the indicated moment of inertia or radius of gyration.
Find the radius of gyration of a plate covering the first-quadrant region bounded by $y^{2}=1-x$ with respect to the $x$ -axis.

Answer

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Allyn J. Washington, Richard S. Evans

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

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus