While practising my skill at determining moments of inertia, I encountered the problem:
Find the moment of area about the x-axis. Good luck... The answer which is provided by Wikipedia is
$$I_x = \left(\frac{\pi}{8} -\frac{8}{9\pi}\right)r^4$$
In an attempt at acquiring this result, I fell back on the definition of the moment of area, given by
$$I_x =\int_{\Omega} y^2 dA = \int_{\Omega} R^3 \sin^2 \theta dRd\theta = ...$$
Looking at $\Omega$, I realize that limits of integration are hard, and quicker to fall back on parallel axis theorem, which states that
$$ I_x =I_0^x +s^2 \cdot A$$
A new system of coordinates I set at the centre of the circle. The moment of the area we are going to compute through the centroid which is not at the centre of the circle. We know that $I_0^x=\frac{\pi}{8}r^4$, which is easily obtainable from the definition. The coordinates of the centroid I derive from$$x_C = \frac{S_y}{A}, \\ y_C = \frac{S_x}{A}$$They turn out to be equal to each other and equal to $\frac{4R}{3\pi}$, so$$I_x = \frac{\pi}{8}r^4 + \frac{8}{9\pi}r^4 = \left(\frac{\pi}{8} +\frac{8}{9\pi}\right)r^4$$As you may see, my answer differs from the given one. How can that be? Can this be associated with the orientation of moments like torque? I do not know and would like to pose that question to you if it is allowed.
1 Answer
$\begingroup$The answer is provided by Andrei who commented (made a comment) on my question. As he stated, the problem is of incorrect usage of the parallel axis theorem, probably because I abused the definition of the moment of area. According to it, we set the origin at the centroid and take any small area of the object into consideration, denoted by $dA$. Next, we calculate the distances from the point to the axes and obtain the expression $r^2 dA$, which can be split into $(x^2 + y^2)dA$, since $r^2 = x^2 + y^2$. This is how we define the moment of area and compute this by$$I_z =\int_{\Omega} r^2dA = \int_{\Omega} x^2 dA + \int_{\Omega} y^2 dA = I_x +I_y$$where $x,y$ are distances to the axes of rotation. Suppose that we wish to compute the moment of area through another point, not just a centroid. Here comes the parallel axis theorem in handy, which states that$$I_A = I_C +s^2 A$$Here A stands for any point, and C for centroid, so that a new system of coordinates is set at point A, and the old one is at C; s is the distance from A to C in the sense that we compute this as a distance from one x axis (new one) to another (old one). Now look at the picture, shown in the description of my question. Problem is that we are assigned to determine the moment of area through the centroid, which is $I_C$. In order to do the task, we have to make use of the parallel axis theorem (because direct integration is complicated and should be avoided if there is another way to solve the problem). So, we know the moment of area $I_A$. Therefore, according to the formula above, we get $I_C$. However, we are not going to rush into completing the task.
Note that the problem occurred because when I was going to use the theorem and find $I_C$, I had to use the formula of the known moment of area $I = \frac{\pi r^4}{8}$, which, in its turn, was defined as well as in the definition; in other words, I knew the moment of area for the circle and to find it for the semi-circle, I just divided the expression in the formula by 2, which was forbidden to execute. In the end, I disregarded the fact that the semi-circle has another centroid other than the circle's one.
So,$$I_C = I_A - s^2 A =(\frac{\pi}{8} - \frac{8}{9\pi})r^4$$
$\endgroup$
