Moment of Inertia of Disc through diameter

In summary: But then I realized that √(a2-x2) in the equation must refer to the distance along the axis from the origin, and that didn't make any sense. So then I tried substituting x for h in the equation and got the same answer, which told me that x in the answer must be perpendicular to the axis against which the moment is being determined.
  • #1
unscientific
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Homework Statement



The problem is attached in the picture.


The Attempt at a Solution



I managed to solve it using a different method. I have no idea what the answer is talking about..

My method

Found dI of a strip = (1/3)*dm*h2

then i replace h by x, then integrate from -a to a to get the same answer.

But as for the answers, I have no clue.
 

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  • #2
As far as I can tell, x in the answer is perpendicular to the axis against which the moment is being determined. I think you can figure out the rest.
 
  • #3
voko said:
As far as I can tell, x in the answer is perpendicular to the axis against which the moment is being determined. I think you can figure out the rest.

The problem is, I can't fingure out what is the shape of their dM.

It looks as if they are taking a small square mass dM that is distance x away from axis (perpendicularly)

so I = x2 dM

but dM = 2σ √(a2-x2)

It looks as if √(a2-x2) is referring to the distance along the axis from the origin..But how does that make any sense? r in ∫ r2 dM refers to distance from the axis and not the origin.

Then are they using dM as a strip or a small square mass?
 
  • #4
They are taking the entire strip parallel to the axis at distance x from it.
 
  • #5
To understand the general statement, it may be easier to consider only a small mass element, dm, of the disc placed at coordinate x, y (with z being zero since its a plate). This one mass element contributes to the MOI around the x-axis as dIx = y2dm, to the MOI around the y-axis as dIy = x2dm, and to the MOI around the z axis as dIz = (x2+y2)dm = x2dm + y2dm, which, using the two previous equations, means that dIz = dIy + dIx (make a diagram if you have trouble seing this). Since this is valid for all the mass element of the planar disc it must also be valid for the sum (or rather the integral) of all the mass elements MOI, that is, Iz = Ix + Iy.
 
  • #6
voko said:
They are taking the entire strip parallel to the axis at distance x from it.

Yup, I finally figured that out. Initially I worked it out by using a strip perpendicular to the axis with an MOI of dI = (1/3)x2 dM
 

1. What is the moment of inertia of a disc through its diameter?

The moment of inertia of a disc through its diameter is a measure of its resistance to changes in rotational motion. It is a property of the disc's mass and distribution of mass around its axis of rotation.

2. How is the moment of inertia of a disc through its diameter calculated?

The moment of inertia of a disc through its diameter can be calculated using the formula I = (1/2) * m * r2, where m is the mass of the disc and r is the radius of the disc.

3. What is the difference between the moment of inertia through the diameter and through the center of mass?

The moment of inertia through the diameter is smaller than the moment of inertia through the center of mass. This is because the mass is more spread out when measured through the diameter, resulting in a smaller moment of inertia.

4. How does the moment of inertia of a disc change as its mass or radius is varied?

The moment of inertia of a disc increases as its mass or radius increases. This is because there is more mass distributed further from the axis of rotation, making it harder to change the disc's rotational motion.

5. What are some real-life applications of understanding the moment of inertia of a disc through its diameter?

Understanding the moment of inertia of a disc through its diameter is important in various fields, such as engineering and sports. In engineering, it is used in designing rotating machinery and calculating the stability of structures. In sports, it is used in analyzing the performance of discs in discus throwing events.

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