Find my Mistake? (Moment of inertia)

In summary, the moment of inertia of a spherical shell of mass M and radius R is given by:$$\int ^{R} _{-R} \rho\pi R^3 \cos ^3 \theta dz$$.
  • #1
13
0

Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations

The Attempt at a Solution


So I've been mulling over this for about two hours now and haven't figured out where I've made a mistake. Here's what I've done:

I know I can do the integral by doing ##\int ^{r} _{-r} \frac{8}{3}\pi \rho da##, with ##da## a surface element, but I wanted to do it sort of "from scratch" sort of.

I start with ##\int r^2 dm = \int r^2 A(r) dz## and then sub ##R \cos \theta = r## (the distance from ##dm## to ##z##) and ##A(r) = 2 \pi r \rho = 2 \rho\pi R \cos \theta## (circumference of the disk as a function of theta). We then have ##\int ^{R} _{-R} \rho\pi R^3 \cos ^3 \theta dz##. Next we sub in ##dz = \frac{r d \theta}{\cos \theta}## to get ##\int ^{\pi / 2} _{-\pi/2}2 \pi\rho R^4 \cos ^2 \theta d\theta ##. This doesn't come out to the correct expression, clearly.

I know there must be some dumb mistake in there, but I just can't seem to find it. Any help would be appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
I think your ## R \cos(\theta) ## should be ##R \sin(\theta) ##. Would suggest you use spherical coordinates instead of integrating along z.
 
Last edited:
  • #3
Z90E532 said:

Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations

The Attempt at a Solution


So I've been mulling over this for about two hours now and haven't figured out where I've made a mistake. Here's what I've done:

I know I can do the integral by doing ##\int ^{r} _{-r} \frac{8}{3}\pi \rho da##, with ##da## a surface element, but I wanted to do it sort of "from scratch" sort of.

I start with ##\int r^2 dm = \int r^2 A(r) dz## and then sub## R \cos \theta = r## (the distance from ##dm## to ##z##) and ##A(r) = 2 \pi r \rho = 2 \rho\pi R \cos \theta## (circumference of the disk as a function of theta). We then have ##\int ^{R} _{-R} \rho\pi R^3 \cos ^3 \theta dz##. Next we sub in ##dz = \frac{r d \theta}{\cos \theta}## to get ##\int ^{\pi / 2} _{-\pi/2}2 \pi\rho R^4 \cos ^2 \theta d\theta ##. This doesn't come out to the correct expression, clearly.

I know there must be some dumb mistake in there, but I just can't seem to find it. Any help would be appreciated.
Z90E532 said:

Homework Statement


Find the moment of inertia of a spherical shell

Homework Equations

The Attempt at a Solution


So I've been mulling over this for about two hours now and haven't figured out where I've made a mistake. Here's what I've done:

I know I can do the integral by doing ##\int ^{r} _{-r} \frac{8}{3}\pi \rho da##, with ##da## a surface element, but I wanted to do it sort of "from scratch" sort of.

I start with ##\int r^2 dm = \int r^2 A(r) dz## and then sub## R \cos \theta = r## (the distance from ##dm## to ##z##) and ##A(r) = 2 \pi r \rho = 2 \rho\pi R \cos \theta## (circumference of the disk as a function of theta). We then have ##\int ^{R} _{-R} \rho\pi R^3 \cos ^3 \theta dz##. Next we sub in ##dz = \frac{r d \theta}{\cos \theta}## to get ##\int ^{\pi / 2} _{-\pi/2}2 \pi\rho R^4 \cos ^2 \theta d\theta ##. This doesn't come out to the correct expression, clearly.

I know there must be some dumb mistake in there, but I just can't seem to find it. Any help would be appreciated.
How are you accounting for the thickness of the shell ?
 
  • #4
SammyS said:
How are you accounting for the thickness of the shell ?
I didn't but I shouldn't have to, right? It's a spherical shell of zero thickness.
 
  • #5
Z90E532 said:
I didn't but I shouldn't have to, right? It's a spherical shell of zero thickness.
In that case you will have a hard time if you use a finite density.
 
  • #6
Suggestion: You can let the density ## \delta ## of mass per unit area be ## \delta=M/(4\pi R^2) ## .
 
  • #7
SammyS said:
In that case you will have a hard time if you use a finite density.

Sure we can get it to work with just ##dm = \frac{M}{A} dA = \frac{M}{A}R d \theta 2 \pi r##, but this contradicts what I had found for it, which was (say ##\rho = 1##)##dm = 2 \pi r dz = 2 \pi (R \cos \theta )(\frac{R d \theta}{\cos\theta})## which leads to the integral after ##r^2## is introduced
$$\int ^{\pi / 2} _{-\pi/2}2 \pi R^4 \cos ^2 \theta d\theta$$. I guess my problem is seeing why ##dz = R d\theta##.
 
  • #8
The ## dA ## in spherical coordinates is ## dA=R^2 \sin(\theta) d\theta d\phi ##. The ## \phi ## integral is of course ## 2 \pi ##. The distance from the z-axis ## \rho=R \sin(\theta) ##. The integral is straightforward. (Your mistake is ## \cos(\theta) ## should be ## \sin(\theta) ##). Don't forget to include the density ## \delta=M/(4 \pi R^2) ##. I see you basically derived the ## dA ## for spherical coordinates yourself, but that needs to be ## \sin(\theta) ##. Incidentally, ## 2 \pi \rho dz ## is not equal to ## dA ##. (I'm using ## \rho ## in place of your ## r ##.)
 
Last edited:
  • #9
Charles Link said:
The ## dA ## in spherical coordinates is ## dA=R^2 \sin(\theta) d\theta d\phi ##. The ## \phi ## integral is of course ## 2 \pi ##. The distance from the z-axis ## \rho=R \sin(\theta) ##. The integral is straightforward. (Your mistake is ## \cos(\theta) ## should be ## \sin(\theta) ##). Don't forget to include the density ## \delta=M/(4 \pi R^2) ##. I see you basically derived the ## dA ## for spherical coordinates yourself, but that needs to be ## \sin(\theta) ##.
Don't give too much help . Let OP do some of this.
 
  • #10
One more input to the OP: (@SammyS I think the OP is trying, but his methodology is clumsy). Would recommend having the ## \theta ## polar coordinate integration go from ## 0 ## to ## \pi ##. That is normally how the spherical coordinates work.
 

1. What is the moment of inertia?

The moment of inertia is a measure of an object's resistance to rotational motion. It is a property that depends on the object's mass and how the mass is distributed around its axis of rotation.

2. How do I calculate the moment of inertia?

The moment of inertia can be calculated using the formula I = mr², where I is the moment of inertia, m is the mass of the object, and r is the distance between the object's axis of rotation and the mass element.

3. What are the units of moment of inertia?

The units of moment of inertia depend on the units used for mass and distance. In the SI system, the units are kg·m². In the British system, the units are slug·ft².

4. How does the moment of inertia affect an object's rotational motion?

The moment of inertia determines how easily an object can be rotated. Objects with a larger moment of inertia require more torque to make them rotate compared to objects with a smaller moment of inertia.

5. Can the moment of inertia change?

Yes, the moment of inertia can change depending on the distribution of mass and the axis of rotation of an object. For example, if an object's mass is moved further away from its axis of rotation, the moment of inertia will increase.

Suggested for: Find my Mistake? (Moment of inertia)

Back
Top