I know that the moment of inertia of the ring through the Diameter is $I_{x}=I_{y} = mr^2/2$.
But I cannot get this formula using the integral. Give me some hints how to do this. Thanks a lot!
[Math] Moment of inertia of the ring through the diameter
integration
Related Solutions
You are rotating the cylinder about an axis perpendicular to the axis of symmetry. In this case, polars may not be your best bet.
Align the $x$ axis with the axis of rotation. For each point on the axis of rotation, which extands between $x \in [-a,a]$, there is a corresponding rectangular cross-section of the cylinder. Each point of the cross section is at a distance $D$ from that point of the axis of rotation. That rectangle has dimensions $2 \sqrt{a^2-x^2} \times h$. A point within that rectangle has distance from that point on the axis $D$ equal to $\sqrt{y^2+z^2}$. The get the element of inertia from this rectangle, integrate $y^2+z^2$ over the rectangular area. Then integrate the collection of elements of inertia from rectangles like this over $x$.
Setting this up: the element of inertia $dI(x)$ is an integral over that rectangular area:
$$dI(x) = \rho dx \int_0^h dz \: \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} dy \: (y^2+z^2) = \rho dx \frac{2 h}{3} \sqrt{a^2-x^2} (a^2+h^2-x^2)$$
Once you determine this, integrate over $x$ to get the total moment of inertia:
$$I = \rho \frac{2 h}{3} \int_{-a}^a dx \: \sqrt{a^2-x^2} (a^2+h^2-x^2) = \frac{1}{12} M (3 a^2+4 h^2)$$
I leave the details of evaluating the integrals to the reader.
If the shape is located and oriented in a "good" way relative to the $x$- and $y$-axes, the answer is yes, you can derive the second moment around any axis in the plane given only $I_x$ and $I_y.$
More generally, for an arbitrary shape at an arbitrary location and orientation, you will also need the product moment of inertia, $$I_{xy} = -\iint_A xy \, dx dy.$$ It's also conventional in this context to write $I_{xx}$ and $I_{yy}$ instead of $I_x$ and $I_y.$ Using this notation, then for a shape lying entirely in the $x,y$-plane with the centroid of the shape at $(0,0),$ the moment of inertia around an axis given by a unit vector $\hat n$ can be computed (using matrix notation) by the use of the inertia tensor, $I,$ as follows: $$I_{\hat n} = \hat n^T I \hat n = \hat n^T \begin{pmatrix} I_{xx} & I_{xy} & 0 \\ I_{xy} & I_{yy} & 0 \\ 0 & 0 & I_{xx} + I_{yy} \end{pmatrix} \hat n. $$ Note that $\hat n$ can be, but is not required to be, a vector in the $x,y$-plane. You can use the parallel axis theorem to get the moment of inertia around an axis parallel to $\hat n$ that does not pass through the shape's centroid.
This is a special case of a more general formula for the moment of inertia of a three-dimensional object around the axis through the centroid of the object in the direction of the vector $\hat n$: $$I_{\hat n} = \hat n^T \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix} \hat n. $$ (These notes offer this notation; also see this page and the page that follows it, or these lecture notes and the notes for the following lecture.)
For an object entirely in the $x,y$-plane, the $z$-coordinate of any point is zero, hence $I_{xz} = I_{yz} = 0,$ and we can use the perpendicular axis theorem to substitute for $I_{zz}.$
You can simplify the calculation further, however, by finding the principal axes of your object. This is a set of orthogonal axes such that when the axis of rotation is described by a vector $\hat n$ relative to those axes, the moment of inertia is simply $$I_{\hat n} = \hat n^T \begin{pmatrix} I_{xx} & 0 & 0 \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{pmatrix} \hat n $$ when the principal axes are taken as the $x$-, $y$-, and $z$-axes. That is, if you place your object on your coordinate plane so that its principal axes coincide with your $x$- and $y$-axes, or conversely choose $x$- and $y$-axes that correspond with the object's principal axes, then you can derive the moment of inertia around any axis merely by knowing its moments of inertia around the two axes. But that assumes you are able to find the principal axes in the first place, which (for an irregular body) may involve computing $I_{xy}$ in some initially chosen coordinate system and then finding the eigenvectors of the inertia tensor in that coordinate system, which indicate the principal axes.
A possibly useful fact: if your planar object has any bilateral symmetry, the axis of symmetry is one of your principal axes. That's easy to confirm by setting one of your coordinate axes to the axis of symmetry and computing $I_{xy}$ in that coordinate system.
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Best Answer
$I=\int r'^2 dm$
$dm={{M}\over{2\pi}}d\theta$
$r'=r\cos\theta$
$I=\int_0^{2\pi} r^2\cos^2\theta {{M}\over{2\pi}}d\theta$
$I={{Mr^2}\over{2\pi}}\int_0^{2\pi} \cos^2\theta d\theta$
$I={{Mr^2}\over{2\pi}}[{{\theta}\over{2}}+{{\sin 2\theta}\over{4}}]\biggl|_{\theta=0}^{2\pi}$
$I={{Mr^2}\over{2\pi}}[(\pi+0)-(0+0)]$
$I={{Mr^2}\over{2}}$