Moment of inertia spinning disk integration

In summary, the homework statement is asking for a solution to integrating sin(x) over the range 0 to 2R, where R is the maximum radius of the disk. The student attempted to solve the problem by setting up an equation involving y=r sin(θ), and integrating along the x-axis from -R to R. However, they made a mistake in their original calculation when they replaced y with sqrt(R^2-x^2), which resulted in an incorrect answer. When they corrected the mistake, they were able to get the right answer.
  • #1
ShamelessGit
39
0

Homework Statement


I want to calculate the moment of inertia of a spinning disk via integration. I'm aware of the perpedicular axis theorem, but I want to integrate.


Homework Equations


I = ∫r^2dm


The Attempt at a Solution



if I set my coordinate axis op so that the origin of the xy plane lies in the center of my disk, and I decide to rotate the disk about the x axis, then I figure r in this case is equal to y and I integrate along the x-axis from -R to R.

So I set up ∫∫y^2dydx, which I can change to ∫∫r*y^2drdθ. y = rsin(θ), so

∫∫r^3*sin(θ)^2drdθ, where the limits of integration are 0< r < 2R (R = max radius), 0 < θ < 2PI

I look up the integral of sin(x) and I get x/2 - sin(2x), so I think my answer to this integral should be (PI/2)R^4. Of course the answer should be (PI/4)R^4. I don't understand what I did wrong. I think I know how to integrate so I assume something is wrong with my set-up.
 
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  • #2
Hi ShamelessGit! :smile:

(try using the X2 button just above the Reply box :wink:)
ShamelessGit said:
So I set up ∫∫y^2dydx, which I can change to ∫∫r*y^2drdθ. y = rsin(θ), so

∫∫r^3*sin(θ)^2drdθ, where the limits of integration are 0< r < 2R (R = max radius), 0 < θ < 2PI

no, 0 < r < R :wink:

I look up the integral of sin(x) and I get x/2 - sin(2x), so I think my answer to this integral should be (PI/2)R^4

not following this :confused:

can you show your working?​

(btw, learn your trigonometric identities … sin2x = (1 - cos2x)/2, which is easier! :smile:)
 
  • #3
I made a typo when I was giving the limits of integration. The way I set it up the limit of integration for r should be -r to r.

And also I figured out that when I integrate properly, I do get the right answer.

The mistake I made was that I changed y to sqrt(R^2-x^2) in my original calculation before realizing that that was unnecessary, and when I changed it back I got r*R^2 instead of r^3 because I got confused about which R was a constant and which was a variable. So I got R to the right power, but of course I got a factor of 1/2 instead of 1/4.
 
  • #4
ShamelessGit said:
I made a typo when I was giving the limits of integration. The way I set it up the limit of integration for r should be -r to r.

And also I figured out that when I integrate properly, I do get the right answer.

The mistake I made was that I changed y to sqrt(R^2-x^2) in my original calculation before realizing that that was unnecessary, and when I changed it back I got r*R^2 instead of r^3 because I got confused about which R was a constant and which was a variable. So I got R to the right power, but of course I got a factor of 1/2 instead of 1/4.

I made a type when trying to fix the previous typo. Yes the limit is only from 0 to r. Srry about that. I didn't make that mistake on the paper when I was actually trying to integrate
 
  • #5


Your approach to integrating for the moment of inertia of a spinning disk is correct. However, there are a few mistakes in your calculations.

Firstly, the limits of integration for r should be from 0 to R, not 2R. This is because the disk has a radius of R, so the maximum distance from the center to any point on the disk is R.

Secondly, the integral of sin(x)^2 is not x/2 - sin(2x), it is (x/2) - (sin(x)cos(x)/2). This can be found using the double angle formula for sin(2x).

Lastly, when substituting y = rsin(θ) into the integral, you should also take into account the differential element dy = rcos(θ)dθ. So the final integral should be ∫∫r^3sin(θ)^2rcos(θ)drdθ.

After correcting these mistakes, you should get the correct answer of (PI/4)R^4. It is important to double check your calculations and make sure all variables and limits are correct when integrating.
 

1. What is moment of inertia?

The moment of inertia of a spinning disk is a measure of its resistance to changes in its rotational motion. It is a property of the disk's mass distribution and can be thought of as the rotational equivalent of mass in linear motion.

2. How is moment of inertia calculated for a spinning disk?

The moment of inertia for a spinning disk can be calculated using the formula I = 1/2 * MR^2, where I is the moment of inertia, M is the mass of the disk, and R is the radius of the disk. This formula assumes that the mass of the disk is evenly distributed and that it is spinning around its central axis.

3. Why is moment of inertia important for spinning disks?

Moment of inertia is important for spinning disks because it affects their rotational motion. A larger moment of inertia means that more torque is needed to change the disk's rotational speed, while a smaller moment of inertia means that less torque is needed. This can have practical applications in fields such as engineering and physics.

4. How does the moment of inertia change if the mass distribution of the spinning disk is not uniform?

If the mass distribution of the spinning disk is not uniform, the moment of inertia will also not be uniform. This means that different parts of the disk will have different moments of inertia and will require different amounts of torque to change their rotational motion. In this case, the moment of inertia can be calculated using the parallel axis theorem.

5. Can the moment of inertia of a spinning disk be changed?

Yes, the moment of inertia of a spinning disk can be changed by altering its mass distribution. For example, if more mass is added to the outer edge of the disk, its moment of inertia will increase, making it harder to change its rotational motion. This can be seen in sports such as figure skating, where skaters increase their moment of inertia by extending their arms and legs, making it easier to perform spins.

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