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Moment of inertia of a door about its hinge

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Homework Statement


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Homework Equations



[tex]I=p\int_{0}^{0.85}r^2dV[/tex] where p is density and r is perpendicular distance from the axis of rotation.

[tex]dV=dxdydz[/tex]

The Attempt at a Solution



I'm not sure where i'm going with this one as it seems I'll have to integrate dx*dy*dz which is kind of confusing to me.

Should I treat the door as a rod and just use dx?

Any help would be appreciated.
 

Answers and Replies

  • #2
cepheid
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The door is not a rod, it is a box! (Or, perhaps thin enough that you can consider all of the mass to be on a 2D rectangle).

In any solid, 3D distribution of mass, each infinitesimal mass element has moment of inertia:

[tex] dI = dm (r_\perp^2) [/tex]

where r_perpendicular is the perpendicular distance to the axis of rotation. This is just like the formula for the moment of inertia of a single point mass.

You can always write this as:

[tex] dI = \rho dV (r_\perp^2) [/tex]

where the mass element has been replaced by the density multiplied by an infinitesimal volume element. To get the total moment of inertia, you integrate this over the entire volume take up by the body:

[tex] I = \int_V dI = \int_V \rho dV (r_\perp^2) [/tex]

The derivation of the moment of inertia of any solid body about a given axis boils down to computing a volume integral of this type. In this case, since the mass density is constant (it is not a function of position within the body), it comes outside of the integral. The trick is finding a coordinate system that is convenient for this particular problem

For a box, I suppose Cartesian coordinates would work. However, for the 2D rectangle case, I can't help but wonder if cylindrical coordinates might not work better. The r coordinate in a cylindrical system would naturally be the perpendicular distance to the hinges (and the hinges would lie along the z-axis). Theta would be a constant, defining the plane of the door. I can't be 100% sure without trying it out myself.
 
  • #3
cepheid
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Hmm, sorry no. I think you should ignore what I said about approximating it as a 2D rectangle. I think you should also ignore what I said about cylindrical coords. They give you a 3D mass density, and the object is a rectangular box. So, I think that you should do the full 3D volume integral in Cartesian coordinates. The trick is figuring out what r_perpendicular is.

If orient your coordinate system so that the z-axis is the height direction (and the hinges lie along it), and the x-axis is the direction of the door's width (with the y-axis being the direction of its thickness), then you have simply that:

[tex] r_\perp = x [/tex]

and

[tex] I = \rho \int_0^{0.85} \int_0^{0.05} \int_0^{1.9} x^2 \, dx dy dz [/tex]
 
  • #4
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I'm pretty much aware of what you've said, although you explained it nicely however how am I supposed to calculate the dV needed to complete the calculation for this case?

EDIT: Posted this at the same time as your above post, going to read it through.

EDIT2: Ah yes this is exactly what I had in mind, however I've never come across one of these multi part integrals before and I'm not quite sure how to handle them, I'll post my final result in a minute and can you compare it to whatever you get?
 
  • #5
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Right I get a value of [tex]I=8.76375 kgm^2[/tex]
 
  • #6
cepheid
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EDIT2: Ah yes this is exactly what I had in mind, however I've never come across one of these multi part integrals before and I'm not quite sure how to handle them, I'll post my final result in a minute and can you compare it to whatever you get?
Aha! So you haven't done double or triple integrals yet. It's a case of the physics teacher asking you to apply math you haven't been taught yet (as happens far too often). This will be covered in a course on multi-variable calculus, but here is an overview:

You can think of a function that is a function of three independent variables, rather than just one: f(x,y,z). So, in other words, the value of the function is affected by changing x, or y, or z. So, the domain over which the function varies is a volume in 3D space, rather than just a set of real numbers (a 1D space). To integrate this function over a certain volume, you can set up a triple integral:

[tex] \int\int\int_V f(x,y,z) dV = \int_a^b \int_c^d \int_e^f f(x,y,z)\,dxdydz [/tex]

(Note: I was lazy about including all three integral signs to denote the volume integral some of my posts above, and it is a common shorthand to use only one until you introduce specific coordinates). The limits of integration here define the boundaries of the volume over which you are integrating (a box in this case).

The way to think about the right hand side above is as a set of NESTED 1D integrals:

[tex] \int_a^b \left( \int_c^d \left( \int_e^f f(x,y,z)\,dx\right)dy\right)dz [/tex]

So, FIRST you integrate the function over x, keeping y and z constant. Then you take the result of that, and integrate it over y. Then you take the result of that, and integrate it over z.

In your case, since the function you are integrating depends ONLY on x, the nested integrals become three separate 1D integrals that are just multiplied by each other (all of which are trivial)
 
  • #7
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Thanks alot, I used the method you provided (alot less daunting than it looks) and I got the answer above your post, does it check out?
 
  • #8
cepheid
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Right I get a value of [tex]I=8.76375 kgm^2[/tex]
I get an answer that matches yours...to 1 decimal place, anyway. (EDIT: meaning I get around 8.75 kg*m^2)

EDIT: To test your answer, before you plug the numbers in, does the formula you get for I match the formulas that are given in textbooks etc. for the moment of inertia of a rectangular box around an axis coinciding with one of its edges?

Typically, to check this, you have to re-write your formula in terms of the total mass M. So, whatever formula you get, factor out M = rho*l*w*h (length, width, and height), and see what remains multiplying M. It should match the formula given in the textbook.
 
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  • #9
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My textbook tells me that the formula is [tex]I=\frac{1}{3}Ma^2[/tex] Where a is width (in this case 0.85m).

Plugging in the numbers give me [tex]I=8.7513[/tex] which is pretty much the same.

Thanks alot for the help, keep up the good work! :D
 
  • #10
cepheid
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That's good, I'm glad to be of help. Again, remember I said you can check the formula by not quite plugging in numbers just yet?

So for a width of "a", a height of "b", and a thickness of "c", your integral gives you:

[tex]I = \rho \int_0^b dz \int_0^c dy \int_0^a x^2\,dx [/tex]

[tex] I = \frac{1}{3}\rho a^3 bc [/tex]

This is your result, the moment of inertia formula you had to derive. (You got your numerical result by plugging in a = 0.85 m, b = 1.9 m, c = 0.05 m).

The question is, does this result match the textbook formula? Well, as I said in my previous post, we have to relate it to M, the total mass of the door, which is equal to its density times its volume. The volume of the door is abc. So if we factor out the density*abc from the formula, we get:

[tex] I = \frac{1}{3} [\rho abc] a^2 [/tex]

[tex] I = \frac{1}{3} M a^2 [/tex]
 

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