Moment of inertia (integration)

[xzibition8612]

Homework Statement

See attachment

Homework Equations

Integrating

The Attempt at a Solution

The answer is 16pi slug-ft^2.

Now I know your suppose to integrate it. Its in 2D, so a double integral at most, maybe a single integral? I'm not very good with integrals and their limits in finding moments of inertia, so any guidance would be appreciated.

 

[tiny-tim]

hi xzibition8612!

slice it into cylindrical shells of thickness dy

 

[SteamKing]

The ellipse is being rotated about the x-axis, according to the problem statement, which also asks you to calculate the MOI about the x-axis.

I think TT wanted to suggest trying to use cylindrical slices with thickness dx instead of dy.

 

[tiny-tim]

you can use cylinders with thickness dx,

or cylindrical shells with thickness dy ​

 

[xzibition8612]

let's keep it simple. So I slice that shape into thickness dx and then integrate it. That would only be 1 integral right? So its going to be integral of ... dx.
Now the problem is what do I integrate? What do you have to integrate to get moment of inertia? Area?

 

[tiny-tim]

hi xzibition8612!

So I slice that shape into thickness dx and then integrate it. That would only be 1 integral right? So its going to be integral of ... dx.
Now the problem is what do I integrate? What do you have to integrate to get moment of inertia? Area?

no, you use the ready-made formula for moment of inertia of a cylinder … 1/2 mr²

(if you used dy and cylindrical shells, you'd use the obvious and easy-to-remember formula for moment of inertia of a cylindrical shell … mr²)

 

[xzibition8612]

so m=15 slug, but what is r?
And I still don't understand how you can use 1/2mr^2 on the ellipsoid and not integrate. thanks.

 

[tiny-tim]

no, m is the mass of the cylindrical shell (or the cylinder), r is its radius, and yes do you have to integrate!

 

[xzibition8612]

I_xx = ∫(1/2)mr²dm

I think this equation is correct. Now how do you find the radius? And you can take the r² and (1/2) out of the integral correct?

 

[tiny-tim]

a cylinder is 1/2 mr²

a cylindrical shell is, obviously, mr²

"m" here is the mass of the cylinder or cylindrical shell, not the whole mass

r is the distance from the x-axis, ie y, and no you can't take it outside the ∫, it isn't constant

 

[xzibition8612]

I_xx=∫∫mr²dmdr

The limits of dm is 0 to 15 slugs. The limit of r is from -1 to 1 ft (for the range of dx). Is this correct?

 

[SteamKing]

You are not setting up the integral for the moment of inertia properly.

The density of the body is 15 slugs/cu.ft., not the total mass. The total mass of the body is its volume multiplied by the mass density.

One way to set up the integral for the elliptical body is to slice it up into a series of circular disks, each of thickness dx running along the x-axis. Each disk will have a mass of dm, which in turn, is a function of dx and the radius at the location of the disk.

 

[xzibition8612]

I_xx=ρ∫mr²dm

Ok, now dm is a function of dx and r, but I don't see/understand that and have no idea how to expand dm into dx and r...is it a double integral?

 

[SteamKing]

The mass of a disk dm = ρ dV. dV = area of the disk * thickness = A dx
The attachment shows the moment of inertia of the disk, but the mass of the disk is dm rather than M from the figure.

What you want to do is add up the inertia of all of the disks along the x axis.

 

[xzibition8612]

I_xx=∫(1/2)ρR²Adx

Limits of integration from -2 to 2 ft.

Now how do you find area A and radius R? Are they of each element of the circular disk? No idea how to do this thanks.

 

[tiny-tim]

you're slicing it into cylinders with thickness dx, so they have radius y

(if you sliced it into cylindrical shells with thickness dy, they would have radius y and height x(y) )

 

[xzibition8612]

ok so y would have limits from -1 to 1 ft. I understand that. Problem is my equation is in terms of dx...so how would I incorporate this? A double integral?

 

[tiny-tim]

ok so y would have limits from -1 to 1 ft. I understand that. Problem is my equation is in terms of dx...so how would I incorporate this? A double integral?

no, a single ∫

your only "d" is dx …

you've sliced it with slices perpendicular to the x axis, going along the x axis

(and your integrand is already an area thingy, so that only leave one dimension, doesn't it? )

 

[xzibition8612]

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