Moment of inertia of an isosceles triangle

[Nexus99]

Homework Statement

Find the moment of inertia of an isosceles triangle of mass M = 1.0 kg, height h = 0.4 m and base angles equal to $\alpha = \frac{ \pi}{6}$, with respect to an axis passing through its vertex

Relevant Equations

moment of inertia

I did in this way:
$I = \int dm \rho^2$
Dividing the triangle in small rectangles with $dA = dy x(y)$ where $x(y) = 2 ctg( \alpha ) (h - y)$
we have : $dm = \sigma 2 ctg( \alpha ) (h - y)$
Now i have $\rho^2 = x^2 + (h-y)^2$
Now I don't know what I can do because it would be an integral in 2 variables that I don't know how to do

 

[PeroK]

Why not just set up an integral in $dxdy$?

 

[Nexus99]

Why not just set up an integral in $dxdy$?

I can't do for the moment integrals with two variables

 

[PeroK]

I can't do for the moment integrals with two variables

That might be a problem. I'm not sure how you could avoid a double integral.

 

[etotheipi]

As has been mentioned, it's easier to impose a Cartesian frame (e.g. origin at top of triangle, $x$ axis running along the height), then the region of integration is specified by $-\frac{x}{\tan{\alpha}} \leq y \leq \frac{x}{\tan{\alpha}}$ and $0 \leq x \leq h$, so you can write down the integral$$I_z = \int_S \rho r^2 dx dy = \int_0^h \int_{-\frac{x}{\tan{\alpha}}}^{\frac{x}{\tan{\alpha}}} \rho (x^2 + y^2) dy dx$$where $\rho$ is the area density. To evaluate this you can do the inner integral first, holding $x$ constant, and then feed the result into the outer integral.

 

[Nexus99]

As has been mentioned, it's easier to impose a Cartesian frame (e.g. origin at top of triangle, $x$ axis running along the height), then the region of integration is specified by $-\frac{x}{\tan{\alpha}} \leq y \leq \frac{x}{\tan{\alpha}}$ and $0 \leq x \leq h$, so you can write down the integral$$I_z = \int_S \rho r^2 dx dy = \int_0^h \int_{-\frac{x}{\tan{\alpha}}}^{\frac{x}{\tan{\alpha}}} \rho (x^2 + y^2) dy dx$$where $\rho$ is the area density. To evaluate this you can do the inner integral first, holding $x$ constant, and then feed the result into the outer integral.

I am sure that is easier, but i have never done a multivariable integral.
Is it possible tu use the fact that:
$I_z = I_x + I_y$?

 

[etotheipi]

I am sure that is easier, but i have never done a multivariable integral.
Is it possible tu use the fact that:
$I_z = I_x + I_y$?

That is indeed true for a lamina, it follows because of the linearity of the integral$$I_z = \int_S \rho r^2 dS = \int_S \rho (x^2 + y^2) dS = \int_S \rho x^2 dS + \int_S \rho y^2 dS = I_x + I_y$$but I don't think it will help you here, since the computations will be exactly the same, just in two parts (so in fact it is more long-winded).

The double integral is not too bad... I promise! For starters, can you evaluate$$\int_{-\frac{x}{\tan{\alpha}}}^{\frac{x}{\tan{\alpha}}} \rho (x^2 + y^2) dy$$whilst treating $x$ as a constant?

 

[jbriggs444]

Homework Statement:: Find the moment of inertia of an isosceles triangle of mass M = 1.0 kg, height h = 0.4 m and base angles equal to $\alpha = \frac{ \pi}{6}$, with respect to an axis passing through its vertex
Relevant Equations:: moment of inertia

View attachment 268183
I did in this way:
$I = \int dm \rho^2$
Dividing the triangle in small rectangles with $dA = dy x(y)$ where $x(y) = 2 ctg( \alpha ) (h - y)$
we have : $dm = \sigma 2 ctg( \alpha ) (h - y)$
Now i have $\rho^2 = x^2 + (h-y)^2$
Now I don't know what I can do because it would be an integral in 2 variables that I don't know how to do

I like to think of this sort of thing visually before starting down the road with formulas.

When you do a double integral, you are usually laying things out in rows and columns. You integrate down a row to get the moment of inertia of a single row. Then you add those moments up -- integrating up in the column direction.

The first challenge you always face is trying to figure out an orientation for your rows and columns. There are many possibilities here:

1. You could consider a bunch of parallel strips running perpendicular to the base, up in the general direction of the apex, find the moments of inertia of each of those and then integrate them up.

2. You could consider a bunch of narrow triangles running from apex down toward the base, find the moments of inertia of each of those and then integrate those up.

3. You could consider a bunch of strips parallel to the base, short rows near the apex and long rows near the base. This is where @etotheipi has taken you.

The moment of inertia of a thin strip about its midpoint is a well known formula. That plus the parallel axis theorem gets you to the moment of inertia of a thin strip about the apex here.

 

[Nexus99]

@jbriggs444 @etotheipi
thanks for explanation.
I watched some videos on youtube about double integrals and i got these results
$I_z = \frac{5 \rho h^4}{12 tan( \alpha) } = \frac{5 M h^2}{12}$
Hope it's correct

 

[Nexus99]

I noticed in this moment that is present the solution of this problem:

"For the calculation of the moment of inertia, can be considered the moment of inertia of the elementary rod with mass dm, thickness dy and length l = l (y) indicated in the figure, where y is the distance from the apex of the triangle . In this case, for Steiner's theorem the moment of inertia can be written as:"

Sorry if the solution is not written in english, but there was written nothing important, so I simply tried to explain what was written with arrows

 

[Lnewqban]

... with respect to an axis passing through its vertex
...

It seems that the axis of rotation is parallel to the plane of the triangle, according to angle θ shown in the schematic; could it be perpendicular instead?

https://en.m.wikipedia.org/wiki/Centroid#Of_a_triangle

"The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them."

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

"The moment of inertia about a parallel axis is the center of mass moment plus the moment of inertia of the entire object treated as a point mass at the center of mass."

 

[Nexus99]

Could it be perpendicular instead?

It isAnyway i don't understand what do you want to say with your previous message

 

[Lnewqban]

It isAnyway i didn't understand what do you want to say with your previous message

If I understand correctly, the axis of rotation is perpendicular to the plane of the triangle.
In other words, the triangle rotates about the dot, clockwise or counterclockwise on the plane over which it is represented.

Sorry about my previous post and links, perhaps I am over-simplifying the problem when I see it as estimating the moment of inertia (I) of a center of mass (m) located at the centroid of the triangle and rotating perpendicularly at a fixed distance (r) from an axis.

 

[Nexus99]

If I understand correctly, the axis of rotation is perpendicular to the plane of the triangle.
In other words, the triangle rotates about the dot, clockwise or counterclockwise on the plane over which it is represented.

Yes, it's a physical pendulum.
Sorry if it was not so clear but i had to summarise the problem