Moment of inertia of rectangle without dd int

In summary, the student attempted to solve the double integration of a problem but messed up. He tried to use the double integrals and got an incorrect answer. He then tried to use the equation for the moment of inertia and got an incorrect answer. He then tried to use the average value of the y-values instead of the maximum value and got an incorrect answer.
  • #1
stateofdogma
30
0

Homework Statement


I found the moment of inertia of a rectangle, with the axis perpendicular and going through the center using double integration. But I also used what i thought to be an equivalent method, but it didn't work.

Homework Equations


[tex]dI=r^2dm[/tex]
[tex]\int dI [/tex]

The Attempt at a Solution


I was able to solve the double integration of the problem. But before that I tried using the double integrals, I used this method [tex]4((\int^\frac{a}{2}_0\frac{b^2}{4}+x^2)dx + (\int^\frac{b^2}{2}_0\frac{a^2}{4}+y^2))dy[/tex]

As you can imagine I toke the integral of the equation [tex] x^2+y^2[/tex] by one of the length like b/2 as constant and integrating over the width a, from a/2 to -a/2 and because its an even function it be taken from a/2 to 0 and you can take a 2 out. And lastly add another to for the symmetry, do the same for the same for the onther side.

You don't get the correct answer which is [tex]M\frac{a^2+b^2}{12}[/tex] and I don't understand why?
 
Last edited:
Physics news on Phys.org
  • #2
stateofdogma said:

Homework Statement


I found the moment of inertia of a rectangle, with the axis perpendicular and going through the center using double integration. But I also used what i thought to be an equivalent method, but it didn't work.

Homework Equations


[tex]dI=mr^2[/tex]
That should be ##dI = r^2\,dm##. You can't have a lone differential just sitting there.
[tex]\int dI [/tex]

The Attempt at a Solution


I was able to solve the double integration of the problem. But before that I tried using the double integrals, I used this method [tex]4((\int^\frac a2_0\frac{b^2}{4}+x^2)dx + (\int^\frac b2_0\frac{a^2}{4}+y^2))dy[/tex]

As you can imagine I toke the integral of the equation [tex] x^2+y^2[/tex] by one of the length like b/2 as constant and integrating over the width a, from a/2 to -a/2 and because its an even function it be taken from a/2 to 0 and you can take a 2 out. And lastly add another to for the symmetry, do the same for the same for the onther side.

You don't get the correct answer which is [tex]M\frac{a^2+b^2}{12}[/tex] and I don't understand why?
The simple answer is you did stuff that you're not justified doing. I'll assume you're using ##dm = dx\,dy##. Based on symmetry, you can say that
$$ \int (x^2+y^2)\,dm = 4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx,$$ but how can you say that
$$4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx = 4\left[ \int^\frac a2_0 \left(\frac{b^2}{4}+x^2\right)\,dx + \int^\frac b2_0 \left(\frac{a^2}{4}+y^2\right)\,dy\right]?$$How do you justify setting ##y=b/2## in one integral? Why couldn't it be ##y=b/4##? (I'm not saying that's right, but it would make more sense to use the average value of ##y## rather than the maximum.) How can you justify setting that variable to a constant value at all?

In math, you need to be able to justify each step. If you think you can and then get the wrong answer, only then you can ask why. At this point, to me, it just appears like you assumed certain things were true without having a good reason to. It's no surprise then that your calculation didn't work out.
 
  • #3
vela said:
The simple answer is you did stuff that you're not justified doing. I'll assume you're using ##dm = dx\,dy##. Based on symmetry, you can say that
$$ \int (x^2+y^2)\,dm = 4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx,$$ but how can you say that
$$4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx = 4\left[ \int^\frac a2_0 \left(\frac{b^2}{4}+x^2\right)\,dx + \int^\frac b2_0 \left(\frac{a^2}{4}+y^2\right)\,dy\right]?$$How do you justify setting ##y=b/2## in one integral? Why couldn't it be ##y=b/4##? (I'm not saying that's right, but it would make more sense to use the average value of ##y## rather than the maximum.) How can you justify setting that variable to a constant value at all?

In math, you need to be able to justify each step. If you think you can and then get the wrong answer, only then you can ask why. At this point, to me, it just appears like you assumed certain things were true without having a good reason to. It's no surprise then that your calculation didn't work out.

My calculation are base on the physical interpretation of the event, the length and width of the rectangle b and a, so by making the one of them constant, say b/2 and taking the integral from a/2 to zero of x^2+(b^2/4) and multiplying that by 4 I get the moment of inertia of 4 section of the rectangle and I do the same for the other 4. The place where I believe I went wrong was the density of the mass, I use M/ab, instead of the the two M/length. Then you would have r^2/r for the integral equation which turns into the equation sqrt(x^2 + (b^2/4)).
 
Last edited:
  • #4
stateofdogma said:
My calculation are base on the physical interpretation of the event, the length and width of the rectangle b and a, so by making the one of them constant, say b/2 and taking the integral from a/2 to zero of x^2+(b^2/4) and multiplying that by 4 I get the moment of inertia of 4 section of the rectangle.
What justification do you have to set y=b/2 in that integral? Can you articulate that? It's not enough to simply say you based it on the "physical interpretation of the event," whatever that means.

and I do the same for the other 4. The place where I believe I went wrong was the density of the mass, I use M/ab, instead of the the two M/length. Then you would have r^2/r for the integral equation which turns into the equation sqrt(x^2 + (b^2/4)).
 
  • #5
vela said:
What justification do you have to set y=b/2 in that integral? Can you articulate that? It's not enough to simply say you based it on the "physical interpretation of the event," whatever that means.

Now I get why you stress the b/2, by taking the b/2 as constant I am not considering the moment of inertia of the inner sections of the rectangle. Alright go it, thanks, I'll just use double integrals.
 

1. What is the equation for calculating the moment of inertia of a rectangle without any internal diagonal?

The equation for calculating the moment of inertia of a rectangle without any internal diagonal is I = (1/12) * m * (h^2 + w^2), where m is the mass of the rectangle, h is the height, and w is the width.

2. How do you determine the moment of inertia of a rectangle without any internal diagonal experimentally?

To determine the moment of inertia of a rectangle without any internal diagonal experimentally, you can use a torsion pendulum or a compound pendulum setup. By measuring the period of oscillation and the dimensions of the rectangle, you can calculate the moment of inertia using the equation I = (T^2 * m * g) / (4 * π^2 * d^2), where T is the period, m is the mass of the rectangle, g is the acceleration due to gravity, and d is the distance between the pivot point and the center of mass.

3. How does the moment of inertia of a rectangle without any internal diagonal affect its rotational motion?

The moment of inertia of a rectangle without any internal diagonal determines how difficult it is to rotate the object about its center of mass. A higher moment of inertia means the object will have a greater resistance to rotational motion, while a lower moment of inertia means it will be easier to rotate the object.

4. Can the moment of inertia of a rectangle without any internal diagonal be negative?

No, the moment of inertia of a rectangle without any internal diagonal cannot be negative. It is always a positive value, representing the distribution of mass around the object's axis of rotation.

5. How does the moment of inertia of a rectangle without any internal diagonal compare to other shapes?

The moment of inertia of a rectangle without any internal diagonal is generally lower than other shapes, such as a circle or a triangle, with the same mass and dimensions. This is because the rectangular shape has more of its mass concentrated closer to the axis of rotation, resulting in a lower moment of inertia.

Similar threads

Replies
25
Views
455
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
304
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
15
Views
652
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
Back
Top