Moment of inertia of a thin, square plate

  • #1
simphys
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Homework Statement
the moment of inertia about an axis through the center of and perpendicular to a uniform, thin square plate. mass M and dimension L x L.
Relevant Equations
d
I don't really understand what the 2 integrals (dx and dxdy) for I_x represent. Could I get some explanation here please? Thanks in advance.
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  • #2
The second line in the derivation of ##I_x## has one ##dx## too many. Is that what bothers you?
 
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  • #3
Also, the second line is missing the ##r^2## and the limits of integration for ##x## are not correct. Keep in mind that this integral represents the moment of inertia about the x-axis, and the x-axis lies in the plane of the thin plate. Think about how to express ##r^2## in terms of ##x## and/or ##y##.
 
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  • #4
If you put in the correct $$r^2=x^2+ y^2$$ in the second line then you see that the x and y integrals give you the "perpendicular axis theorem" without issue. Please redo this.
 
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  • #5
TSny said:
Also, the second line is missing the ##r^2## and the limits of integration for ##x## are not correct. Keep in mind that this integral represents the moment of inertia about the x-axis, and the x-axis lies in the plane of the thin plate. Think about how to express ##r^2## in terms of ##x## and/or ##y##.
Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.
 
  • #6
hutchphd said:
If you put in the correct $$r^2=x^2+ y^2$$ in the second line then you see that the x and y integrals give you the "perpendicular axis theorem" without issue. Please redo this.
this is the solution. not mine :)
 
  • #7
So where is your attempt? How are we to help?
 
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  • #8
hutchphd said:
So where is your attempt? How are we to help?
well.. that is exactly what I don't understand..
I didn't understand why the solution has used two integrals in such a way.. I haven't used a volume integral or smtn. I did differently by using a mass element and summing over that with one definite integral.
 
  • #9
One must sum each mass element (mass density times volume element) over the entire volume of the object and scaled by the square of the distance to the chosen axis . The world is three dimensional and so is the integral. Show us your work.
 

1. What is the moment of inertia of a thin, square plate?

The moment of inertia of a thin, square plate is a measure of its resistance to rotational motion. It is similar to the concept of mass in linear motion and is calculated by multiplying the mass of the plate by the square of its distance from the axis of rotation.

2. How is the moment of inertia of a thin, square plate different from other shapes?

The moment of inertia of a thin, square plate is different from other shapes because it depends on the distribution of mass within the plate. For example, a square plate with its mass concentrated at the edges will have a different moment of inertia than a square plate with its mass evenly distributed.

3. What factors affect the moment of inertia of a thin, square plate?

The moment of inertia of a thin, square plate is affected by its mass, the distance of the mass from the axis of rotation, and the distribution of mass within the plate. It is also affected by the shape and size of the plate, as well as the material it is made of.

4. How is the moment of inertia of a thin, square plate calculated?

The moment of inertia of a thin, square plate can be calculated using the formula I = (1/12) * m * a^2, where I is the moment of inertia, m is the mass of the plate, and a is the length of one side of the square plate.

5. Why is the moment of inertia of a thin, square plate important?

The moment of inertia of a thin, square plate is important because it helps us understand how the plate will behave when subjected to rotational forces. It is also a crucial factor in designing structures and machines that involve rotational motion, such as wheels, gears, and flywheels.

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