Find my Mistake? (Moment of inertia)

[Z90E532]

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.

 

[Charles Link]

I think your $R \cos(\theta)$ should be $R \sin(\theta)$. Would suggest you use spherical coordinates instead of integrating along z.

 

[SammyS]

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.

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 ?

 

[Z90E532]

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.

 

[SammyS]

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.

 

[Charles Link]

Suggestion: You can let the density $\delta$ of mass per unit area be $\delta=M/(4\pi R^2)$ .

 

[Z90E532]

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 $dz = R d\theta$.

 

[Charles Link]

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 $r$.)

 

[SammyS]

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.

 

[Charles Link]

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.