# Moment of Inertia (triangular plate)

I want to generalize the formula for the MOI of a triangular plate (sides $a,b,c$) about an axis passing through mid point of one sides and perpendicular to it's plane .

The mass of plate $M$ is uniformly distributed on it's area.

I can use parallel axis theorem if I know the MOI about an axis passing through COM of the plate, but I don't even know that. Also there is not symmetry in this, so taking small strips of masses to integrate will also not help.So, I was unable to start Please Help.

If you solve by integration then please help me get the integration term too.

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What have you tried? For example, you mentioned you're familiar with the parallel axis theorem, so edit the question to include how you tried to apply it and what step you got stuck on. –  David Z May 16 '13 at 7:40
@DavidZaslavsky: I can use parallel axis theorem if I know the MOI about an axis passing through COM of the plate, but I don't even know that. Also there is not symmetry in this, so taking small strips of masses to integrate will also not help.So, I was unable to start. –  ABC May 16 '13 at 7:44
That's the sort of thing you should put in the question. Show your evidence that you've made an effort. (Also, "commanding" people to "give the exact integration term and how we get it" may make others reluctant to answer. I'd suggest rephrasing that.) –  David Z May 16 '13 at 7:45
Aah! a downvote . Please @downvoter review it again :( .I have edited. –  ABC May 16 '13 at 7:49
For an outline of a technique that's not too horribly messy, see ch. 15, problem 45 of this book: lightandmatter.com/mechanics –  Ben Crowell May 16 '13 at 22:00

It's actually not too hard to calculate the moment of inertia (MOI) of a right triangle. And you can make your triangle out of a large right triangle minus a smaller right triangle. So your MOI is just the MOI of the bigger triangle minus the MOI of the smaller one.

# Step 1:

Extend line $b$ (move vertex $C$) until you have a right triangle. We'll calculate the MOI relative to vertex $A$. I agree with @fibonatic that it's easiest to use polar coordinates. So we have $$I_{ABC'} = \int_0^\alpha \int_0^{R(\theta)} (\rho_A\, r^2) r\, dr\, d\theta~.$$ Here, $\alpha$ is the angle $\angle B A C$, which you can calculate using trig. Also, $R(\theta)$ is the length of a line $Aa$ that goes from vertex $A$ to line $BC=a$, where $\theta$ is the angle of that line above line $b$. A little simple trig says that $R(\theta) = b/\cos\theta$. \begin{align} I_{ABC'} &= \rho_A\, \int_0^\alpha \int_0^{b/\cos\theta} r^3\, dr\, d\theta \\ &= \rho_A\, \int_0^\alpha \left. \frac{r^4}{4} \right|_0^{b/ \cos\theta}\, d\theta \\ &= \frac{\rho_A\, b^4}{4}\, \int_0^\alpha \cos^{-4}\theta\, d\theta \\ &= \frac{\rho_A\, b^4}{24}\, \frac{3 \sin\alpha+\sin (3 \alpha )} {\cos^3\alpha} \end{align} [I'm terrible at integrals, so I just looked up the answer for $\cos^{-4}$. But I think it's easy enough to convince you that you can handle any reasonable MOI problem.]

# Step 2:

Use this formula and the parallel-axis theorem to get the MOI of the smaller right triangle that you need to remove from the bigger one to get your triangle.

# Step 3:

Subtract the result of step 2 from step 1 to get the MOI of your triangle about the vertex $A$.

# Step 4:

Use the parallel-axis theorem to get your MOI relative to whatever origin you want.

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Moment of Inertia is defined as: $$I={\sum}mr^2$$ which in this case can be rewritten into an integral: $$I=\rho\int_A{r^2dA}$$ Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. And I will use polar coordinates, in which case $dA=rd\theta dr$: $$I=\rho\left(\int_{\theta_1}^{\theta_2}\int_0^{r(\theta)}r^3drd\theta+\int_{\theta_2}^{\theta_3}\int_0^{r(\theta)}r^3drd\theta\right)$$ with $\theta_1=0$ radians, $\theta_2$ is equal to the angle between $AOB$ which is equals: $\theta_2=\pi-\alpha-\frac{\beta}{2}$, with $\alpha$ the angle between $BAC=\cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)$ and $\beta$ the angle between $ABC=\cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)$. And $\theta_3=\pi$ radians. For the first integral the range to which $r$ has to integrated equals: $r(\theta)=\frac{1/2b\sin\alpha}{\sin(\pi-\alpha-\theta)}$. For the second integral the range to which $r$ has to integrated equals: $r(\theta)=\frac{1/2b\sin\gamma}{\sin(\pi-\gamma-(\pi-\theta))}=\frac{1/2b\sin\gamma}{\sin(\theta-\gamma)}$, with $\gamma=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)$.

I hope this helped, and if you are still not able to solve this, I will try to add the solution to this equation.

Edit: here is the expression after solving the integral: $$I=\frac{\rho b^4}{192}\left(\sin\gamma\cos\gamma(2-\cos(2\gamma))-\frac{\sin^4\gamma\cos(\gamma-\theta_2)(2-\cos(2\gamma-2\theta_2))}{\sin^3(\gamma-\theta_2)}+\sin{\alpha}\cos{\alpha}(2-\cos(2\alpha))-\frac{\sin^4\alpha\cos(\alpha-\theta_2)(2-\cos(2\alpha-2\theta_2))}{\sin^3(\alpha-\theta_2)}\right)$$

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I don't know this complex integration. :( . Probably some easy method! –  ABC May 16 '13 at 14:30
It is not that complicated, the inner integral is simple: $\int r^3dr=1/4r^4$. But to solve the outer integral you use the interval for $r$ given as a function of $\theta$. The solution to this is quite difficult, I had to use wolframalpha to solve it. But it is solvable. –  fibonatic May 16 '13 at 14:40
Now after i see this, i'm afraid of MOI........:( –  ABC May 16 '13 at 15:01
For this you could use the parallel axis theorem as well, so you would have to subtract $md^2$ to get the moment of inertia at the center of mass. –  fibonatic May 16 '13 at 15:37

Now I have got a method to get it directly.And again it came out to be an easy problem. The answer comes out to be $$I= \dfrac{m}{12}(a^2+c^2)$$

See if we add another such plate along it's side $AC$ , then it comes out to be a parallelogram plate , whose MOI is known, same as rectangular plate

So, by symmetry arguments , both the triangular plates have the same MOI.

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