Moment of inertia for a rhombus

[teleport]

Hi, this is the question:

Give the moment of inertia for a rhombus with sides of length c and mass m, about an axis that is parallel to the plane of the rhombus and goes from one corner to the opposite corner.

I have set up the integral:

I=(m/c^2)*2* int(from x=0 to x=c) int(from y=x to y=-x+c*sqrt2)x^2dydx.

where I have used the y-axis as the axis of rotation.

I don't know if this is the correct integral. In particular I suspect of the 2 multiplying the first integral. The thing is that this 2 would be correct if I were just finding the area of the rhombus, but since i have introduced the
x^2 that doesn't have anything to do with the area, then I don't know if it works here.

 

[Mindscrape]

Hmm, I don't know why you have a two multiplying it.

Did you use this as your moment?
$$I = \iint(x^2 + y^2) \rho (x,y) dA$$

I see the bounds to going from x=c to x=(x-c)c and y=0 to y=c/sqrt(2).

 

[teleport]

No. I'm not sure I understand the bounds that u give for x.

 

[teleport]

Oh sorry the bounds for x that i give in the expresion is wrong. The correct bounds for x are 0<=x<=c/sqrt2.

 

[OlderDan]

Hmm, I don't know why you have a two multiplying it.

Did you use this as your moment?
$$I = \iint(x^2 + y^2) \rho (x,y) dA$$

I see the bounds to going from x=c to x=(x-c)c and y=0 to y=csqrt(2).

I think you have the wrong axis. The axis is in the plane of the rhombus

 

[OlderDan]

Hi, this is the question:

Give the moment of inertia for a rhombus with sides of length c and mass m, about an axis that is parallel to the plane of the rhombus and goes from one corner to the opposite corner.

I have set up the integral:

I=(m/c^2)*2* int(from x=0 to x=c) int(from y=x to y=-x+c*sqrt2)x^2dydx.

where I have used the y-axis as the axis of rotation.

I don't know if this is the correct integral. In particular I suspect of the 2 multiplying the first integral. The thing is that this 2 would be correct if I were just finding the area of the rhombus, but since i have introduced the
x^2 that doesn't have anything to do with the area, then I don't know if it works here.

Don't you need to know the angle of the rhombus?

 

[teleport]

I put the 2 in there because I gave those boundaries for x. The actual area of the rhombus is the double of what those boundaries in the integrals represent. But again, I'm not sure that is right. Help please.

 

[teleport]

Sure. Once the rhombus is cut in half by the axis, there are four 45 degree angles and two 90. But why do u ask?

 

[Mindscrape]

Oops, you're right. You don't need the angles of the rhombus if all the sides are equal, i.e. lozenge.

 

[OlderDan]

Oops, you're right. You don't need the angles of the rhombus if all the sides are equal, i.e. lozenge.

You do need the angle. A rhombus has 4 equal sides and can have any smaller angle. I take it from the other post that this is actually a square.

Sure. Once the rhombus is cut in half by the axis, there are four 45 degree angles and two 90. But why do u ask?

 

[teleport]

yes that's right but since i put the y-axis as the axis of rotation then the square won't 'seat' on a base but on a corner by convential methods anyways.

 

[Mindscrape]

I think the original integral is correct, but OlderDan is right and you do technically need an angle because assuming the small angle is 45 is a little sketchy.

 

[teleport]

I still don't understand ur question about the angle.

 

[teleport]

ok great, minds, but ill wait for the other opinions too. I have a bad feeling about that integral.

 

[OlderDan]

yes that's right but since i put the y-axis as the axis of rotation then the square won't 'seat' on a base but on a corner by convential methods anyways.

OK.. I have the picture now. There are actually 4 equal contributions from the 4 triangles in the 4 quadrants. The lower right quadrant boundary is y = x - c/sqrt(2). The upper right is bounded by y = -x + c/sqrt(2). I think your sqrt(2) is in the wrong place.

Added: and your x is from 0 to c/sqrt(2) for one of the 4 triangles.

 

[teleport]

But then does it mean that I have to multiply by four the double integral?

 

[OlderDan]

But then does it mean that I have to multiply by four the double integral?

You can do it that way.. 4 times the integral over 1 of the 4 triangles. See my previous note for the added comment about the x interval.

 

[teleport]

If so the result should be the same from the original integral (with a corrected x boundary that I mentioned) the only difference is that your origin is at the center of the rhombus but in mine the origin is at
(-c/sqrt2, -c/sqrt2) from ur origin. But if u actually do both integrals in this way, (with the 2 or the 4) u get different results. Why?

 

[teleport]

sorry my origin is only (0,-c/sqrt2) from yours. But what I last mentioned holds ground.

 

[OlderDan]

If so the result should be the same from the original integral (with a corrected x boundary that I mentioned) the only difference is that your origin is at the center of the rhombus but in mine the origin is at
(-c/sqrt2, -c/sqrt2) from ur origin. But if u actually do both integrals in this way, (with the 2 or the 4) u get different results. Why?

Ah. You were close in the first place. Sorry

If you are integrating over x², the origin has to be on the y axis. If you put the corner at the origin, your limits should be x from - c/sqrt(2) to + c/sqrt(2) and y from x to -x + c*sqrt(2), so it was just your x limit that was off. Then you can do 2 times the integral with x from 0 to c/sqrt(2) as you intended, or 4 times the integral using those x limits and y from x to c/sqrt(2) (the lower right of the 4 triangles)

 

[OlderDan]

sorry my origin is only (0,-c/sqrt2) from yours. But what I last mentioned holds ground.

I realized that. I'm sure the two ways are equivalent if the limits are stated correctly.

 

[teleport]

Hey I just did both integrals (mine and yours) and I get different answers. I have checked them and the same. Mine gives I = (8-3sqrt2)mc^2/(12sqrt2)
Yours is giving me a negative number which seems illogical. What is happening?

 

[teleport]

could u try both integrals and compare? I think I might still be doing something stupid with the limits.

 

[OlderDan]

could u try both integrals and compare? I think I might still be doing something stupid with the limits.

OK I'll be back in a bit.

 

[teleport]

I tried ur origin cutting in four the rhombus and it gives me that negative number. However, when I do it with only two parts, with ur origin also, (each part separated by the axis of rotation/y-axis ) it gives me the same number I got with my origin. Why is it that it doesn't work dividing by four the region?

 

[OlderDan]

Hey I just did both integrals (mine and yours) and I get different answers. I have checked them and the same. Mine gives I = (8-3sqrt2)mc^2/(12sqrt2)
Yours is giving me a negative number which seems illogical. What is happening?

They are indeed the same.

Your way:

$$I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c\sqrt 2 - x} {dydx} }$$

$$I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 dx} \left( {c\sqrt 2 - 2x} \right)$$

$$I = 2\sigma \int_0^{c/\sqrt 2 } {\left( {c\sqrt 2 x^2 - 2x^3 } \right)dx}$$

$$I = 2\sigma \left( {c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2}} \right) = 2\sigma \left( {\frac{2}{3} - \frac{1}{2}} \right)\left( {c/\sqrt 2 } \right)^4$$

$$I = 2\sigma \left( { - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2} + c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3}} \right) = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \frac{M}{{12}}c^2$$

My way:

$$I = 4\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c/\sqrt 2 } {dydx} }$$

$$I = 4\sigma \int_0^{c/\sqrt 2 } {\left( {c/\sqrt 2 x^2 - x^3 } \right)dx}$$

$$I = 4\sigma \left( {\frac{{\left( {c/\sqrt 2 } \right)^4 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{4}} \right) = 4\sigma \left( {\frac{1}{3} - \frac{1}{4}} \right)\left( {c/\sqrt 2 } \right)^4 = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \left( {\frac{M}{{12}}} \right)c^2$$

 

[teleport]

Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.

 

[OlderDan]

Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.

You're welcome. Anything about that result look familiar?

 

[teleport]

It is identical to the case of a rod with the axis of rotation through the center. Other rotational lines for the same square also give similar rotational inertias to other cases of axis for a rod. I can speculate why this happens but I'm not sure. Why is it that this happens?

 

[teleport]

sorry the comparison should have used at least a rectangle (more general), not a square. However, the results for the rectangle are identical to those of a rod.