# How do u define

1. Aug 27, 2007

### pardesi

how do u define rotational kinetic energy when the body is rotting about more than one axis. as an example take a ball that ia rotaing about an axis passing through the centre of ineria whiloe the centre itself rotates in an axis perpendicular to this

2. Aug 28, 2007

### olgranpappy

the "centre itself" rotating about some other axis means that the center of mass is moving with some velocity which would not be put lumped into "rotational" kinetic energy, it's just translational kinetic energy of the whole sphere. Then the rotational kinetic energy is just given by the usual expression
$$1/2 I \omega^2$$
where \omega is the angular velocity about the "axis passing through the centre of [inertia]" and I is 2/5 MR^2.

3. Aug 28, 2007

### pardesi

no but the actual answer includes both the rotation anyway but as the centre of amss rotates sbouta point so do the rest

4. Aug 28, 2007

### rewebster

what do you mean?--like the earth, for example (or specifically)?

5. Aug 28, 2007

### pardesi

i just wanted to know the definition of rotational kinetic energy when the body is rotating about two axes

6. Aug 28, 2007

### rewebster

rotation and precession?

7. Aug 29, 2007

### olgranpappy

yeah but that because, if I choose to, I can write
$$m v^2/2 + I \omega^2/2 = m R^2\Omega^2/2 + I\omega^2/2$$
where $$\Omega$$ = v/R is the velocity (say, of the earth around the sun) divided by the earth-sun distance... whereas $$\omega$$ is 2\pi/24 hr^-1. The concept of "rotational" energy is most useful when dealing with extended bodies, and I can treat the first term as as translational energy of the center of mass if I like. Which I can't do for the second term

8. Aug 29, 2007

### Nesk

I assume that the movement of each partial mass relative to the total centre of inertia must be taken into account. If that is true, then the Earth-Sun system, in which the axes are roughly parallel, might not be a suitable model for a system with perpendicular axes as described by the OP. A deeper analysis of the situation might prove me wrong, though.

If not, another parameter one could look into is whether the orientation of the rotational axis (contrasted to the orbital axis) is constant or conically symmetric. (I realize that a few terms used in this paragraph are slightly arbitrary, but I hope y'all know what I mean).

9. Aug 29, 2007

### genneth

Just remember that rotational kinetic energy is simply kinetic energy, and that it's existence is a mathematical convenience. You can find the kinetic energy of a particle by the usual $$\frac{1}{2}mv^2$$. All other bodies are considered to be composed of particles, and the K.E. of the entire body is just the sum of the particles. For a rigid body rotating about one axis without translation, there is a simple (ahem) formula that relates the distribution of mass (encoded in the moment of inertia), rotational angular velocity $$\omega$$ and the total K.E. By a stroke of good luck, rotation about a single axis with translation just involves adding up the "rotating bit" and the "linear bit". You should decompose the body and try to see if your "rotation about 2 axis" falls apart into easy bits. Define your situation exactly, and try the calculation. I'm sure the denizens here will be happy to lend a hand if you get stuck.

10. Aug 30, 2007

### Europeman

The second rotation can never take place. (Unless you mean precession).

If it could, gyroscopes would not work! They do, believe me.