Determine the moment of inertia for the shaded area about...

Key Concepts

Axis of Rotation

The axis of rotation specifies the line about which the moment of inertia is calculated. In problems concerning the y-axis, distances in the integration are measured perpendicular to this axis. The selection of this axis influences the limits of integration and the form of the distance function within the integral.

Differential Element of Area

In the calculation of area moments of inertia, the shape is divided into infinitesimally small parts, each represented as a differential area element (dA). The moment contribution of each element is determined by multiplying dA by the square of its distance from the specified axis, and integrating these contributions gives the total moment of inertia.

Moment of Inertia

This is a physical quantity that measures the resistance of a body or a shape to angular acceleration about a given axis. It is calculated by integrating or summing the products of each infinitesimal mass or area element with the square of its distance from the axis of rotation, reflecting how distribution of mass or area influences rotational dynamics.

Integration

Integration is the fundamental calculus tool used to compute the moment of inertia by summing the contributions from all differential elements within the shape. In the context of area calculations, it involves setting up an integral that accumulates the squared distances of differential area elements relative to the axis of rotation.

Transcript

00:01 In this question, we have to evaluate the moment of inertia of the shaded area around the y -axis.

00:06 So this figure will rotate around the y -axis like this.

00:11 Now what is the moment of inertia for the shaded area? for that, we can use this expression for the moment of inertia around the y -axis.

00:20 So to use this expression, we have to determine what is the element of area da.

00:26 We have to choose an element of area that is parallel to the axis of rotation so that we avoid using the parallel axis theorem.

00:34 So my element of area will be parallel to the y -axis.

00:39 Therefore, it is something like that.

00:43 So a vertical line, another vertical line, and here we have our element of area.

00:51 Okay, so this is what the expression calls da.

00:55 So this is the element of area.

00:57 As you can see, the area of the area of the area of the element is given by the following.

01:02 There are two possibilities.

01:04 We have one possibility here, so we have a width that is given by dx and we have a height that is given simply by y.

01:14 Therefore, the expression for our element of area is the following d -a is equals to y -d -x.

01:22 Therefore, the integral for the moment of inertia is the moment of inertia given by the integral of x squared y d x.

01:34 Now we have to make a choice.

01:36 We either serve this equation here for y and substitute y here and go with the integration, or we change the variables to y so that we get a d y here instead of a d x and so on.

01:49 I will choose to just solve this equation for y and then plug into the integral.

01:55 So when this equation for y is as difficult as taking a square root, so y is given by the square root of one minus.

02:04 Remember that 0 .5 is the same as 1 over 2.

02:07 So i will write x over 2 and that's it.

02:11 Of course we have plus or minus here, but since we know that all values are positive, we go with the positive root.

02:18 So the negative root will have no use for us.

02:23 So let us disconsiderate.

02:24 Okay? then the moment of inertia is given by the interiors.

02:30 From 0 up to 2 of x squared times the square root of 1 minus x divided by 2 d x.

02:40 Now all we have to do is solve this integral.

02:42 We begin with the following substitution.

02:45 So let us define a new variable u that is equals to 1 minus x divided by 2.

02:51 So the differential element the u is equal to minus the x over 2 so that we know that the x is minus two times the u.

03:01 Therefore, our integral is given by an integral from u equals to 1 up to u equals to 0...

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