What is the angular momentum of a spinning solid sphere

[joshuamr]

What is the angular momentum of a spinning solid sphere of mass 'm', equitorial velocity 'v', and radius 'r'?

(The mass of the sphere being uniformly distributed and the spin being along a single axis only)

I can't figure this out because the common textbook formula for angular momentum (mvr) only applies to cases where the mass is concentrated "at the end of the rope" so to speak. (like swinging around a bucket of water, or a planet going around in it's orbit)

If anyone out there can help me with this, I'd appreciate it.

 

[FulhamFan3]

first, your textbook is pure crap if it doesn't have something as basic as that or you're a high school student with high school text.

the formula for angular momentum is L=Iw
L-angular momentum
I-moment of inertia
w-angular velocity (r*v in your case)

moment of inertia for a uniform sphere is (2/5)m*r^2

I think you can do the rest

 

[FulhamFan3]

mistake

v=rw NOT v*r=w

 

[Sonty]

Originally posted by FulhamFan3
moment of inertia for a uniform sphere is (2/5)m*r^2

And how did you come to this? I get:
$$\int_{0}^{r} d\theta d\phi dr' sin\theta mr' = 2\pi mr^2$$

 

[Integral]

Originally posted by Sonty
And how did you come to this? I get:
$$\int_{0}^{r} d\theta d\phi dr' sin\theta mr' = 2\pi mr^2$$

The correct expression for the momement of inertia is

$$\int r^2 DM$$

Are you using the correct Volume element for the sphere?

The expression given by FulhamFan3 in correct, yours is not.

 

[Sonty]

Originally posted by Integral
The expression given by FulhamFan3 in correct, yours is not.

I thought it would be, that's why I asked.

Are you using the correct Volume element for the sphere?

Of course I'm not.
$$\int_{0}^{r} dr' d\theta d\phi mr'^4\sin\theta =\frac{4\pi}{5}mr^5$$

what am I screwing up now?

 

[Integral]

getting closer!

Perhaps you are using limits of 0-2π for both angular varibles one should only be 0-π

 

[Sonty]

Nope...

$$m\int_0^{2\pi}d\phi\int_0^\pi d\theta \sin\theta\int_0^rdr' r'^4=m*2\pi*2*\frac{r^5}{5}$$

i hate it when i get stuck in a mistake like this and can't detach.

 

[jamesrc]

Originally posted by Sonty
Nope...

$$m\int_0^{2\pi}d\phi\int_0^\pi d\theta \sin\theta\int_0^rdr' r'^4=m*2\pi*2*\frac{r^5}{5}$$

i hate it when i get stuck in a mistake like this and can't detach.

You've got two problems:

1. dm = ρdV as opposed to what you have which looks like dm = mdV. For the sphere, $\rho = \frac{m}{\frac{4}{3}\pi R^2}$.

2. You're mixing distances. The r in the formula for moment of inertia that Integral gave you is the distance from the axis of rotation. The r in you're formula for dV is the distance from the center of the sphere. If you keep your dummy variables straight, the integral should look more like this:
$I = \rho \int_0^{2\pi} \int_0^\pi \int_0^R(r\,\sin\theta)^2(r^2\,\sin\theta) dr d\theta d\phi$

(you can take the density outside of the integral since it is constant)

 

[Sonty]

Originally posted by jamesrc
The r in the formula for moment of inertia that Integral gave you is the distance from the axis of rotation. The r in you're formula for dV is the distance from the center of the sphere.

This was the problem. Thanks.

 

[Arcon]

Originally posted by joshuamr
What is the angular momentum of a spinning solid sphere of mass 'm', equitorial velocity 'v', and radius 'r'?

(The mass of the sphere being uniformly distributed and the spin being along a single axis only)

I can't figure this out because the common textbook formula for angular momentum (mvr) only applies to cases where the mass is concentrated "at the end of the rope" so to speak. (like swinging around a bucket of water, or a planet going around in it's orbit)

If anyone out there can help me with this, I'd appreciate it.

For more info on this please see
http://www.geocities.com/physics_world/mech/inertia_tensor.htm

It is incomplete but may be helpful.

I have not gotten to the part where I calculate the moment of inertia for a sphere of uniform mass density. I'm assuming that the mass density is uniform. I hope to get to that calculation soon. It's not that hard as I recall.

 

[Doc Al]

Originally posted by jamesrc
...the integral should look more like this:
$I = \rho \int_0^{2\pi} \int_0^\pi \int_0^R(r\,\sin\theta)^2(r^2\,\sin\theta) dr d\theta d\phi$

This is a perfectly good approach, but you can save yourself much work by dividing the sphere into cylindrical shells. If the shells have radius "x" (distance from the axis of rotation), then they have cross-sectional area of 2πx dx with a height of 2√(R²-x²). The integral then becomes:
$$I = \rho \int_0^{R} 4\pi x^3 \sqrt(R^2-x^2) dx$$
which is easy to evaluate with a simple change of variables.