# B Multiple Axes Of Rotation?

1. May 12, 2016

### Hoophy

So I am having trouble with this one, I was wondering if an object could have more than one axis of rotation. More than one axis of rotation goes against what I think is possible until I thought about a coin, at which point I was stumped, if the coin was rotating in the way a coin rotates when you spin it like a spinning top but was also spinning in a way in which it resembles a wheels rotation BOTH AT THE SAME TIME, would it have two axis of rotation? If so how is that possible and If not what would the single axis look like? Also for simplicity sake imagine that the coin is rotating in outer space. I am so utterly confused. I would really appreciate an explanation of why this is the way it is. Thank you for your time.

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2. May 12, 2016

### Staff: Mentor

Sure. Gyroscopes do it all the time...

EDIT -- But see post #6 below for how to add the various rotations vectorially.

Last edited: May 12, 2016
3. May 12, 2016

### wrobel

Theorem. Suppose that a rigid body has a fixed point and moves with nonzero angular velocity: $\boldsymbol \omega\ne 0$. Then at each moment of time there exists a unique line belonging to the rigid body such that the velocity of any point of this line is equal zero.

This line is referred to as axis of rotation

4. May 12, 2016

### olgerm

No.

It is analogues to how a car may move forward on road and same move to one side of road (change lane), but does not have 2 speed vectors.

Last edited: May 12, 2016
5. May 12, 2016

### Hoophy

I still do not understand. I don't know if the gyroscope applies to what I'm thinking, I meant a more "solid" object. How would an object rotate on 2 axis using the analogy of a coin? Imagine you are trying to explain this to a dummy. (That's me)

6. May 12, 2016

### A.T.

7. May 12, 2016

### anorlunda

How would you describe this?

8. May 12, 2016

### olgerm

1. Take a coin and spin it the way you described.
2. Now take a ball and spin it the same way (same direction) as coin.
3. Notice 2 points on the ball, which are not moving(because of spinning).
4. Imagine line, which includes both of these points. This line is rotation axis.
5. Consider, that rotation axis of the coin is parallel with rotation axis of the ball.

9. May 13, 2016

### wrobel

That's a good training question by the way. The coin loses its height because of the energy expires for friction of table. Consider an ideal model. The edge of coin does not slip on table's surface. It is not hard to write formulas and describe motion when the centre of coin runs by the horizontal circle and the absolute value of the center's velocity remains constant. For example, precise formulation may be as follows. Assume that we know absolute value of center's velocity and we know the angle between the table and the plane of coin. Find the radius of the circle which the center of coin describes. The radius of coin is also known.

Last edited: May 13, 2016
10. May 13, 2016

### Hoophy

So then would the two axis of any object have to be perpendicular to each other? And is there a limit to the number of axis an object can have at the same time?

11. May 13, 2016

### A.T.

You can decompose the total angular velocity vector into arbitrary many vectors.

12. May 13, 2016

### Hoophy

13. May 13, 2016

### A.T.

14. May 13, 2016

### anorlunda

I hoped that the OP would respond to that video. I wanted to find out whether precession and rotation mean the same thing in his vocabulary.

15. May 13, 2016

### wrobel

Standard fact.

Let $\Sigma$ stand for a rigid body that moves in space. Assume that its angular velocity does not vanish: $\boldsymbol\omega\ne 0.$

Theorem. There exists a unique line $\ell=\ell(t)\subset \Sigma$ such that for any point $B\in\ell$ it follows that the velocity $\boldsymbol v_B$ (of the point $B$) is parallel to the line $\ell$.
This line is parallel to the vector $\boldsymbol\omega$ and velocities of each point of this line are the same.
Moreover, if we know velocity of some point $O\in\Sigma$ then we can find a point $A\in\ell$ by the formula $\boldsymbol {OA}=\frac{1}{|\boldsymbol\omega|^2}\boldsymbol\omega\times\boldsymbol v_O.$

16. May 13, 2016

### Hoophy

I think perhaps my question is way to broad and not very clear. I don't think that this can be explained to me because I need a more solid understanding of physics. Thanks for trying, I will do some more research on my own. Thanks again everybody.

Last edited: May 13, 2016