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Determine the moment of inertia for the shaded area about the $y$ axis.

$I_{y}=1.22 \mathrm{m}^{4}$

Physics 101 Mechanics

Chapter 10

Moments of Inertia

Motion Along a Straight Line

Motion in 2d or 3d

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Simon Fraser University

University of Sheffield

McMaster University

Lectures

04:34

In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

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Determine the moment of in…

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Determine the moment of i…

in this question, we have to evaluate the moment of inertia off this shaded area around the Y axis. So this figure, we rotate around the Y axis like this. Now, what is the moment of inertia? Appreciated area. For that we can use this expression for the moment off inertia around the Y axis. So to use this expression, we have to determine what is the element off area D A. We have to choose an element off area that is parallel to the axis off rotation so that we avoid using the parallel access. The're, um So my element off area will be parallel to the Y axis before it is something like that. So vertical line, another vertical line. And here we have our element off area. Okay, so this is what the expression calls d A. So this is the element off area. As you can see the area off, that element is given by the following. There are two possibilities. We have one possibility here. So we have a lift that is given by the X on. We have a height that is given simply by why the reform, the expression for our element off area is the following d A. Is equals to why the ex reform the integral for the moment of inertia is moment off ownership even by the integral off X squared. Why the X? Now we have to make a choice. We either serve this equation here for why and substitute why here and go with the integration or we change the variables to why, so that we got a d. Y here instead, off the X and so on. I will choose to just solve this equation for why and then plug in tow the integral. So when this equation for why is as difficult as taking a square it so why is given by the square root off one minus? Remember that 0.5 is the same as one over to. So I will write X over to and that's it. Off course we have plus or minus here. But since we know that all values are positive, we go with the positive route. So the negative route we'll have no use for us. So let us disconcert er it. Okay, then the moment off inertia is given by the integral from zero up to two off X squared times the square it off one minus x divided by two the X Now all we have to do is solve this integral. We begin with the following substitution. So let us define a new variable you that is equals to one minus X divided by two. So the differential element the U is equal to minus the X over to so that we know that the X is minus two times you. The reform our integral is given by integral from U equals to one up to you equals to zero. Notice that you is equals to one minus X over two. So when X is equals to zero, we get u equals to one on. This is the result I'm writing here and when X is equal to chew, we got you equals to one minus two divided by two which is one so one minus 10 This is what I'm writing here. Now we proceed and substitute x So what is X as a function off You serving the definition off you for X, we get the following X divided by two is equals to one minus you so X is equal to two times one minus you. Therefore, X squared is four times one minus. You squared and then we have the square it off you. That multiplies the X, which is given by minus two D you okay? Now let me factor out the constants. So we have four times minus two. So we have a factor off. Minus eight times the integral from 1 to 0. Off one minus. You squared times you 2 1/2, which is just the square it off you. Now we expand this expression to get the following minus eight times the integral from 1 to 0 off. One minus. You squared. It is the square off the first. So one plus the square off the second you squared minus two times the first times. The second That's it times you 21 over to do you. Now we multiply you 21 over to buy each off those factors. So we distribute the factors off you to get the following. So now we have minus eight times the integral from 1 to 0 off the first factor, which is one times you 2 1/2. So we have you to one over to do you plus the integral from 1 to 0 off you squared times you 2 1/2, which is you to five over to U minus two times the integral from 1 to 0 off you times you toe 1/2. Which is you to treat over to the U on these results in the following minus eight times you 2 3/2, divided by three over to plus you to 7/2. Divided by 7/2 minus two times you 2 5/2, divided by five Over to on. This is to be evaluated at one and zero on this results. In the following, we have minus eight times. So when we evaluate this at zero business just zero. Now we have minus the value off that when evaluated at one. You 2 3/2. Evaluated at one is just 12 3/2. So it's just one here. One divided by 3/2 is to over three. So the result is minus two or three and then we have the next term again, evaluating at 00 So everything we have is the minus it evaluated at one which results into over seven and then we have the final term, which is zero when evaluated at zero. So we're left with plus two times that term evaluated at one which is again one and then we divided by five or two, resulting into divided by five. Then this is minus eight times minus 2/3, minus 2/7, plus 4/5 on this results in 128 divided by 105 which is equals to 1.22 m to the fourth. And this is an approximate result. This is the answer to this question.

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