Ok so I'm extremely comfortable with calculating moment of inertia of continuous bodies but how do we do it for a system not continuous.
For example if 3 rods of mass $m$ and length $l$ are joined together to form an equilateral triangle what will be the moment of inertia about an axis passing through its centre of mass perpendicular to the plane.
i know that moment of inertia of each rod is $ml^2/12$ and c.o.m is at centroid?
also if 2 rods form a cross then to calculate the moment of inertia about its point of intersection would it be correct to sum up the individual moment of inertia of the rods form??
[Physics] Moment of inertia of rods
homework-and-exercisesmoment of inertia
Related Solutions
The missing piece you need is The Parallel Axis Theorem, which is incredibly useful for calculating moments of intertia in general. Let us know if you still have specific issues (e.g. how to implement this).
Edit:
The simple equations for moments of inertia for individual objects (e.g. $I = \frac{1}{3} m L^2$ for a rod about its end) are specifically with respect to a certain axis of rotation. The difficulty is when you need to know the moment of inertia about a different axis.
For example, the moment of inertia for $DE$ could easily be calculated if it was rotating about the line $CD$ --- it would just be $I_{DE,0} = mL^2/3$. But we want to know what it is about the line $AB$ instead --- so we use the Parallel Axis Theorem,
$$I = I_0 + mx^2$$
(for an offset distance $x$, which in this case is the length of the rod $BC$, so $x=L$. Thus, for the moment of Inertia of $DE$, about $AB$ is:
$$I_{DE} = mL^2/3 + mL^2 = 4mL^2/3$$
Note that this works because the original axis of rotation ($CD$) is parallel to the new axis of rotation ($AB$).
Look at the Fig.
The moment of Inertia towards the y' axes (green arrow) must be transformed to (x,y,z) this can obtain with this equation
$$\mathbf I_{xyz}=\mathbf R\,\mathbf I_{x'y'z'}\,\mathbf R^T$$
where $~\mathbf R~$ isthe transformation matrix between $~x'y'z'~$ and $~xyz~$ system and $~\mathbf I~$ is the inertia tensor
$$\mathbf R= \left[ \begin {array}{ccc} \cos \left( \alpha \right) &\sin \left( \alpha \right) &0\\-\sin \left( \alpha \right) & \cos \left( \alpha \right) &0\\ 0&0&1\end {array} \right] \quad,\alpha=\frac{\pi}{3}\\ \mathbf I_{x'y'z'}=\left[ \begin {array}{ccc} 0&0&0\\ 0&{\it I_{y'}}&0 \\ 0&0&0\end {array} \right]\quad\Rightarrow\\ \mathbf I_{xyz}=\left[ \begin {array}{ccc} \left( \sin \left( \alpha \right) \right) ^{2}{\it I_{y'}}&\sin \left( \alpha \right) {\it I_{y'}}\,\cos \left( \alpha \right) &0\\ \sin \left( \alpha \right) {\it I_{y'}}\,\cos \left( \alpha \right) & \left( \cos \left( \alpha \right) \right) ^{2}{\it I_{y'}}&0\\ 0&0&0 \end {array} \right] $$
thus $~I_x=\sin^2(\alpha)\,I_{y'}=\frac{3}{4}\,\underbrace{\frac{m\,L^2}{12}}_{I_{y'}}~$
for the 3 rods you obtain
$$I_\color{red}X=2\,\left(I_{x_1}+m\,H{_1}^2\right)+m\,H_3^2\\ H_1=\frac L2\,\sin(\pi/3)\\ H_3=L\,\sin(\pi/3)$$
$\Rightarrow~I_\color{red}X=\frac 54 m\,L^2$
Best Answer
The moment of inertia for a system of $n$ point masses, $m_i$, at distances $r_i$ from the pivot is simply:
$$ I = \sum m_i r_i^2 \tag{1} $$
We normally calculate $I$ by integration, i.e. we take each point mass to be an infinitesimal element of our continuous object and integrate to add up the moments of inertia of all those elements.
In your case let's call the three rods $A$, $B$ and $C$, then our initial equation (1) can be written as:
$$ I = \sum m_{Ai} r_{Ai}^2 + \sum m_{Bi} r_{Bi}^2 + \sum m_{Ci} r_{Ci}^2 $$
where all we've done is divide up our sum into the infinitesimal parts that belong to the three masses. But from equation (1) we know that $I_A = \sum m_{Ai} r_{Ai}^2$, and likewise for $B$ and $C$, so the total moment of inertia is just:
$$ I = I_A + I_B + I_C $$
So just calculate the separate moments of inertia for all the objects in your system then add them together. In your particular case the objects are identical so the total is just the moment of inetria of a single rod multiplied by three.