# Calculate the rotational inertia of a solid hexagonal

• Yalanhar
In summary: You can always find the perpendicular distance from the centre of mass to the axis. You don't need to integrate again.
Yalanhar
Homework Statement
Calculate the rotational inertia of a solid hexagonal (side = R) about an axis perpendicular through its center.
Relevant Equations
##I=\int r^2 dm##
What I did:

##\frac{dm}{dA} = \frac{M}{\frac{3\sqrt3 R^2}{2}}##

##dm = \frac{2M}{3\sqrt3 R^2} dA## (1)

##dA=3\sqrt3 rdr## (2)

(2) in (1)

##dm = \frac{2M}{3\sqrt3 R^2} 3\sqrt3 rdr##

Now in the integral

##I = \int \frac{r^2 2Mrdr}{R^2}##

How can I solve the integral interval? I think I need to change respect to ##\theta## and integrate over 2##\pi##. Or can I change r in the integral and integrate considering dA? Then the interval would be 0 to A (## A = \frac{3\sqrt3 R^2}{2}##.

I don't quite understand how i choose the interval. In a rod to it's center is -l/2 to +l/2, but in a ring it goes from 0 to R. Why not -R to +R?
Thanks

You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?

haruspex said:
You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?
Yes, i see now that if I use dr or d##\theta## it won't help me. I think I'm confusing r (distance of a infinitesimal part to the axis) to r (infinitesimal side of hexagon)

Yalanhar said:
Yes, i see now that if I use dr or d##\theta## it won't help me. I think I'm confusing r (distance of a infinitesimal part to the axis) to r (infinitesimal side of hexagon)
Well, you can use dθ and integrate along the strip, but there is an easier way using a couple of standard results, as I hinted.

haruspex said:
You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?
I would use ##dA = dxdy## The rectangles are for summing them from one side and then complete the hexagon?

Are you familiar with the parallel axis theorem ?

Yalanhar said:
I would use ##dA = dxdy## The rectangles are for summing them from one side and then complete the hexagon?
They are very thin rectangles, effectively rods.
Plus what BvU asked.

haruspex said:
They are very thin rectangles, effectively rods.
Plus what BvU asked.
I know the theorem. I saw someone doing this theorem for 6 triangles. He calculed the moment on the center for each triangle with the theorem and summed them up. But I didnt understand the rectangle, the rectangle itself can't fill the hexagon, for the x changes between the differents dy

I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon. Is that the case, or is the axis perpendicular to the hexagon? The problem as you have stated it in the original post leaves that question unanswered.

[QUOperpendicularmr post: 6260489, member: 639870"]
I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon. Is that the case, or is the axis perpendicular to the hexagon? The problem as you have stated it in the original post leaves that question unanswered.
[/QUOTE]
Its perpendicur. Sorry

Suppose you divide the hexagon into ##12## congruent triangles, each with one vertex at the center of the hexagon. Then you could integrate ##\rho r^2## over one triangle and multiply by ##12## to get the result.

Yalanhar said:
But I didnt understand the rectangle
You did not post a diagram, but I presumed from your algebra that your integration element consisted of a narrow trapezoidal strip parallel to one side of the hexagon (or maybe six such, forming a hexagon).
Concentrating one one such strip, being so narrow, it is effectively a rod. What is the moment of a rod about its centre?
tnich said:
I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon.
It's useful regardless.

## 1. What is rotational inertia and how is it calculated?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in its rotation. It is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

## 2. Can you explain the concept of rotational inertia in simpler terms?

Think of rotational inertia as the "rotational mass" of an object. Just as an object with more mass has more inertia and is harder to move, an object with a higher rotational inertia is harder to rotate.

## 3. How is rotational inertia different from linear inertia?

Rotational inertia is the tendency of an object to resist changes in its rotational motion, while linear inertia is the tendency of an object to resist changes in its linear motion. Rotational inertia is dependent on the distribution of mass in an object, while linear inertia is dependent on the object's mass and velocity.

## 4. How do you calculate the rotational inertia of a solid hexagonal?

The rotational inertia of a solid hexagonal can be calculated by using the formula I = (1/2)MR², where I is the rotational inertia, M is the mass of the hexagon, and R is the distance from the center of the hexagon to the axis of rotation. This formula assumes that the hexagon is a uniform, solid object.

## 5. What factors affect the rotational inertia of a solid hexagonal?

The rotational inertia of a solid hexagonal is affected by the mass and distribution of the mass within the hexagon. It is also affected by the distance from the axis of rotation, with a greater distance resulting in a higher rotational inertia. Additionally, any changes in the shape or size of the hexagon will also affect its rotational inertia.

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