# Angular Acceleration of a Disc

1. Jun 7, 2004

### e(ho0n3

I think I'm starting to abuse this forum. Anyways, it seems that I haven't grasped the concept of angular acceleration yet since I'm having trouble solving this problem:

Suppose I have a disc laying on the x-y plane with center at the origin. Suppose the disc is rotating about the z-axis with angular velocity $\vec{w}$ and about the y-axis with angular velocity $\vec{u}$. What is the angular acceleration of the disc?

The resultant angular velocity of this disc is $\vec{w} + \vec{u}$ right? I guess since this resultant angular velocity is changing direction as the disc spins, it gives rise to an angular acceleration. I can't really picture the angular acceleration though (i.e. where is it at?).

2. Jun 7, 2004

### e(ho0n3

What the heck am I saying!? The resultant angular velocity vector doesn't change direction. Then where is this angular acceleration I'm asked for comming from? This is weird.

3. Jun 8, 2004

### Gza

Remember from $$\vec{\tau} = I \vec{\alpha}$$ that a torque is required to cause an angular acceleration. This body can't somehow cause a force(or torque for that matter) on itself by newton's third law, since all forces(or torques) are interactions involving more than one body. Therefore the author of your problem must be high. :tongue:

4. Jun 8, 2004

### arildno

Where is it stated that the resulting angular velocity $$u\vec{j}+w\vec{k}$$ is a constant in time??
The angular acceleration is simply $$\vec{\alpha}=\dot{u}\vec{j}+\dot{w}\vec{k}$$

5. Jun 8, 2004

### e(ho0n3

I'm stating it now then: the angular velocity vectors are constant (i.e. do not change with time). If I applied your formula, then I would get an angular acceleration of zero. I'm starting to loose confidence with this problem.

6. Jun 8, 2004

### arildno

Well, then the angular acceleration is zero.
Let $$\vec{\omega}$$ be the resultant angular velocity vector.
We may write:
$$\vec{\omega}=\omega\vec{i}_{\omega}$$

where:
$$\vec{i}_{\omega}=\frac{u}{\sqrt{u^{2}+w^{2}}}\vec{j}+\frac{w}{\sqrt{u^{2}+w^{2}}}\vec{k}$$
$$\omega=\sqrt{u^{2}+w^{2}}$$

In your case, $$\vec{i}_{\omega}$$ is a constant vector defining the rotation plane (for which it is the unit normal)
$$\omega$$ is the scalar angular velocity.

Last edited: Jun 8, 2004
7. Jun 8, 2004

### TALewis

Maybe the correct answer is zero?

8. Jun 8, 2004

### arildno

Sorry about the Latex mess; it should be fixed by now.

9. Jun 8, 2004

### e(ho0n3

Funny, the book has a different answer. I didn't give a verbatim copy of the problem since it's lengthy and has a picture. Maybe my summary of the problem I gave here is not accurate. No matter, since my doubts have been cleared.

10. Jun 8, 2004

### arildno

Could you post the answer from the book?

11. Jun 8, 2004

### e(ho0n3

The answer is numerical, so I don't think that will help unless I quote the problem exactly. Don't worry about it. Thanks again.