Moment of inertia for a rhombus

In summary, 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, can be calculated using the integral I=(m/c^2)*2* int(from x=0 to x=c/sqrt2) int(from y=x to y=-x+c*sqrt2)x^2dydx. However, the correct origin for this calculation should be at (-c/sqrt2, -c/sqrt2) and the x limit should be from -c/sqrt2 to c/sqrt2. This will result in a double integral of 2 times the integral with x from 0
  • #1
teleport
240
0
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.
 
Physics news on Phys.org
  • #2
Hmm, I don't know why you have a two multiplying it.

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

I see the bounds to going from x=c to x=(x-c)c and y=0 to y=c/sqrt(2).
 
Last edited:
  • #3
No. I'm not sure I understand the bounds that u give for x.
 
  • #4
Oh sorry the bounds for x that i give in the expresion is wrong. The correct bounds for x are 0<=x<=c/sqrt2.
 
  • #5
Mindscrape said:
Hmm, I don't know why you have a two multiplying it.

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

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
 
  • #6
teleport said:
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?
 
  • #7
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.
 
  • #8
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?
 
  • #9
Oops, you're right. You don't need the angles of the rhombus if all the sides are equal, i.e. lozenge.
 
  • #10
Mindscrape said:
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.

teleport said:
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?
 
  • #11
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.
 
  • #12
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.
 
  • #13
I still don't understand ur question about the angle.
 
  • #14
ok great, minds, but ill wait for the other opinions too. I have a bad feeling about that integral.
 
  • #15
teleport said:
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.
 
Last edited:
  • #16
But then does it mean that I have to multiply by four the double integral?
 
  • #17
teleport said:
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.
 
  • #18
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?
 
  • #19
sorry my origin is only (0,-c/sqrt2) from yours. But what I last mentioned holds ground.
 
  • #20
teleport said:
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)
 
  • #21
teleport said:
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.
 
  • #22
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?
 
  • #23
could u try both integrals and compare? I think I might still be doing something stupid with the limits.
 
  • #24
teleport said:
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.
 
  • #25
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?
 
  • #26
teleport said:
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:

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

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

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

[tex] 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 [/tex]

[tex] 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 [/tex]

My way:

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

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

[tex] 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 [/tex]
 
  • #27
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.
 
  • #28
teleport said:
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?
 
  • #29
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?
 
  • #30
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.
 
  • #31
teleport said:
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?

A moment of inertia about a symmetry axis is just a sum over all the dm mass elements times their distances from the axis squared. There are many ways to do such sums. Your division of the plate into two parts and my division of it into four parts are examples. A flat rectangular plate rotating about an in-plane axis that divides it into equal rectangles can be thought of as a series of thin rods stacked side by side. The total moment of inertia is just the sum of the moments of inertia of all the individual rods.

In your problem, you could have divided the plate into strips of height dy and treated each strip as a rod. The problem is more complicated than an axis that divides the plate into equal rectangles because the strips have variable length and mass. But you could have used dI = dmL²/12 as the moment of inertia of a strip of height dy, with dm and L being functions of y and dy and integrated over y to get the sum. In fact, if you take your integral and do the x integral first and the y integral second, that is effecively what you would be doing. You don't get I = M(2c²)/12, which would be the result for a rod whose length is the diagonal of the square because most of the "rods" you are adding are shorter than the diagonal and have proportionately less mass. But for an axis that divides the plate into equal rectangles, all the little rods are identical so the moment of inertia looks just like the moment of inertia of a rod.
 
Last edited:
  • #32
Hmm, I wonder why there is no explanation of this in my book. Thanks again.
 

Related to Moment of inertia for a rhombus

1. What is moment of inertia for a rhombus?

The moment of inertia for a rhombus is a measure of its resistance to rotational motion. It is a property of the shape and mass distribution of the rhombus, and is calculated using the formula I = 1/12 * m * (a^2 + b^2), where m is the mass of the rhombus and a and b are the lengths of its diagonals.

2. How is moment of inertia different for a rhombus compared to other shapes?

The moment of inertia for a rhombus is unique to its shape and cannot be compared to other shapes. Each shape has its own moment of inertia formula and value, which depends on its mass distribution and geometry.

3. Does the orientation of a rhombus affect its moment of inertia?

Yes, the orientation of a rhombus does affect its moment of inertia. The moment of inertia is calculated with respect to a specific axis of rotation, so changing the orientation of the rhombus will change its moment of inertia value.

4. How is moment of inertia used in real-life applications?

Moment of inertia is used in various fields of science and engineering, such as in the design of structures and machines. It helps engineers and designers understand how a particular shape will behave when subjected to rotational motion and how to optimize its performance.

5. Can the moment of inertia for a rhombus be negative?

No, the moment of inertia for a rhombus cannot be negative. It is always a positive value, as it represents the resistance to rotational motion and cannot have a negative value.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
695
Replies
25
Views
560
  • Introductory Physics Homework Help
Replies
28
Views
585
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
324
  • Introductory Physics Homework Help
2
Replies
52
Views
2K
  • Introductory Physics Homework Help
2
Replies
40
Views
3K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
Back
Top