Moment of inertia | Integral form

In summary: A couple of things. For one, the axis goes through the center of the square. You should by integrating from -b to b. For another, it should be a double integral dx*dy. For a third thing, I seem to keep getting a different answer from the book answer. What do you get?
  • #1
heycoa
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Homework Statement


Starting from the sum: I=Ʃ mαα2 and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square with side length 2*b, lying in the x-y plane, rotated about the x-axis. Calculate its moment of inertia.

Homework Equations


The integral form turns to I=∫ρ(r)*r2dV

The Attempt at a Solution


I rotated it around the z-axis, giving me a cylinder with radius 2b and height 2b. I know the volume of this cylinder is 2*∏*r2*h. And that ρ=mass/volume. I need help in setting up and evaluating the integral because what I got seems wrong.

I used cylindrical coordinates, claiming that:
θ goes from 0 to 2∏
r goes from 0 to 2b
z goes from 0 to 2b

and my volume element dV=r*dr*dz*dθ. I end up with mass*b2, and I do not believe this is correct. Help please?
 
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  • #2
heycoa said:

Homework Statement


Starting from the sum: I=Ʃ mαα2 and replacing it by the appropriate integral, find the moment of inertia of a uniform thin square with side length 2*b, lying in the x-y plane, rotated about the x-axis. Calculate its moment of inertia.


Homework Equations


The integral form turns to I=∫ρ(r)*r2dV


The Attempt at a Solution


I rotated it around the z-axis, giving me a cylinder with radius 2b and height 2b. I know the volume of this cylinder is 2*∏*r2*h. And that ρ=mass/volume. I need help in setting up and evaluating the integral because what I got seems wrong.

I used cylindrical coordinates, claiming that:
θ goes from 0 to 2∏
r goes from 0 to 2b
z goes from 0 to 2b

and my volume element dV=r*dr*dz*dθ. I end up with mass*b2, and I do not believe this is correct. Help please?

You have the integral wrong because you have the concept wrong. You don't want to integrate over the cylinder, you want to integrate over the square in the x-y plane. Since they say the square is thin you can ignore the z coordinate and make it a 2d integral. The m_a should be replaced by the surface mass density of the square. The ρ should be replaced by the distance of a point in the square to the x-axis. And you don't want to do polar coordinates, cartesian will do fine. Try thinking about it that way for a bit.
 
  • #3
The problem actually has one more line that I neglected to add, it goes: rotated about the x-axis and passing through its center. I have no idea what they mean by "passing through its center".

I am very confused why this is a 2d problem, the problem tells me to rotate the square, so that makes me think cylinders and volumes. Is there a reason why this is 2d? Am I missing something?
 
  • #4
heycoa said:
The problem actually has one more line that I neglected to add, it goes: rotated about the x-axis and passing through its center. I have no idea what they mean by "passing through its center".

I am very confused why this is a 2d problem, the problem tells me to rotate the square, so that makes me think cylinders and volumes. Is there a reason why this is 2d? Am I missing something?

Moment of inertia is usually used for calculating rotational motion around an axis. The moment of inertia depends on the axis you rotate around. So the phrase "find the moment of inertia of the square rotated around the x-axis" doesn't mean "find the moment of inertia of the volume generated by rotating the square". It means find the moment of inertia of the square when it's rotated around the given axis.
 
  • #5
oh my god I am so stupid! I was stuck in calc 2 mode for some awful reason!

Ok I will give that a shot and hopefully get it, thank you very much Dick!
 
  • #6
So here is what I got: ∫(from 0 to 2b)M/A*x2dx

This turns out to be 2/3*M*b.

The answer in the back of the book says it should be 2/3*M*b2. What did I do wrong here?
 
  • #7
heycoa said:
So here is what I got: ∫(from 0 to 2b)M/A*x2dx

This turns out to be 2/3*M*b.

The answer in the back of the book says it should be 2/3*M*b2. What did I do wrong here?

A couple of things. For one, the axis goes through the center of the square. You should by integrating from -b to b. For another, it should be a double integral dx*dy. For a third thing, I seem to keep getting a different answer from the book answer. What do you get?
 
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FAQ: Moment of inertia | Integral form

1. What does the moment of inertia represent?

The moment of inertia represents an object's resistance to changes in its rotational motion. It is analogous to an object's mass in linear motion.

2. What is the formula for calculating moment of inertia in its integral form?

The integral form of the moment of inertia formula is I = ∫r²dm, where r is the distance from the axis of rotation and dm is the differential mass element.

3. How does the distribution of mass affect the moment of inertia?

The distribution of mass greatly affects the moment of inertia. Objects with more mass concentrated farther from the axis of rotation will have a higher moment of inertia compared to objects with the same mass but with the mass distributed closer to the axis.

4. What are some real-world applications of understanding moment of inertia?

Understanding moment of inertia is crucial in designing and analyzing the motion of rotating objects, such as wheels, gears, and flywheels. It also plays a role in sports, such as in figure skating and gymnastics.

5. Can the moment of inertia be negative?

No, the moment of inertia cannot be negative. It is always a positive quantity since it represents an object's resistance to rotational motion.

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