Rotation of object on another object under rotation

First, I would like to know the rotational velocity of disk2 if disk1 turns at $\omega1$. The axis "x" is fixed to the ground and disk1 is allowed to turn around it. Axis "y" is fixed to disk1 and disk2 is allowed to turn around it. For me, disk2 turns around axis "x" and around axis "y" but not in the same manner. I can understand its rotational velocity is $\omega2$ and for me even disk2 turns around axis "x" and because disk1 is turning, this doesn't change its rotational velocity $\omega2$. I'm not sure. Can you explain please ? In fact, what's a rotational velocity in this case for disk2 ? Second, with $\omega1=-\omega2$. A motor is fixed between disk1 and disk2 (in axis "y"), stator on disk1 and rotor on disk2 (idea of Floris, thanks). If a torque from rotor increase $\omega2$ this would say another torque from stator increase $\omega1$. The motor need energy for give a torque to disk2, I'm agree with that. But I don't understand why the motor need energy for apply its torque from stator to disk1. For me, like stator and rotor turn at the same rotational velocity around axis "x" it's not difficult to add torque to disk2 even the stator+rotor turn around "x" in the same time. The motor turn around the axis "x", when stator turn $d\theta$ degree in one direction, rotor turn $d\theta$ degree in the same time. So, it's not "difficult" (need energy) to apply torque on disk1 for me. If rotor moves far away from stator, in this case I'm agree, I will be more difficult to accelerate it, but here stator and rotor never move from each other.

If I need to give forces that stator gives to disk1, I'm agree I need energy, but these forces are only a reaction from the stator because rotor apply a torque. I'm not sure I explain very good what I think, I hope someone can clear this idea. I would like to understand why the motor need energy for apply its torque to disk1.

I'm agree, if I'm on object that move back and I try to move an object in front of me, I need more energy. I drawn disks and 2 points A and B: Imagine, disk2 don't turn around its center of gravity (axis y). Motor and disk2 turn only around axis "x". You could look at distance AB, it is always the same even the stator rotates of 45°, because the rotor rotates of 45° too. It's not like stator move in one direction and rotor in other.

Now, the rotor turn around axis "y" but why it's not the same with a linear trajectory ? Now, distance AB change but only because stator "attack" rotor, in the same time stator rotates around axis "x" but rotor rotates around axis "x" too.

Maybe my confusion come from I dissociate 2 movements: first the rotation around axis "x" and second the rotation around axis "y". For me, inertia and centrifugal forces rotate motor and disk2 around axis "x" and torque from motor add rotational velocity of disk2 around axis "y", I don't see where the stator move "back" like your example, could you explain more please ? Maybe I don't have the right to think like that. Sorry if my level of physics is low.

I added forces, maybe like that it's easier to explain the problem: • Imagine mounting a laser pointer on disk 2. After how much time will it point in the same direction? $2\pi$ divided by that time is the angular velocity. – Floris May 30 '14 at 17:38
• In this case disk2 turn at w2, with or without w1. – user47586 May 30 '14 at 18:03
• That depends on whether $\omega_2$ is measured relative to disk 1, or not. – Floris May 30 '14 at 18:17
• Only from an external viewer, not relative rotational velocity please – user47586 May 30 '14 at 18:21
• That's my point. You need to decide how $\omega_2$ was defined. Is it defined in the frame of reference of disk 1, or the external frame? If it's in the external frame, then you have your answer - and disk 1 doesn't come into it. The entire center of mass of disk 2 is of course moving at $\omega_1$, so the total angular momentum is a more complex expression. But that's not what you asked. – Floris May 30 '14 at 18:27

To clear up your confusion about needing energy to add to the rotation of disk 1, let's take N easier linear problem for a second.

Imagine you are standing with your back against a solid wall, and you push for time $t$ with a force $F$ against an initially stationary mass $m$. You imparted momentum $Ft$ and the work you did was $(Ft)^2:(2m)$.

Imagine doing the same experiment in orbit. Now you would need to push against two masses - one in one direction and one in the opposite direction. In effect you are doing twice as much work - the two masses are separating twice as fast so you apply your force over twice as much distance and do twice the work. When you let go you have two masses with the same kinetic energy, so all is conserved.

The same is true in rotation - for force, read torque; for velocity read angular velocity, etc. it is harder to visualize but the underlying physics is the same: if just push off against something that is moving you have to do more work than of the something was fixed. The degree to which you do more work depends on the relative mass: the lighter your "thing you push against" the more work you have to do. A reason why the baseball pitcher firmly plants his feet before throwing ...?

• Ok, thanks Floris, I edited my question for add what I don't understand. I forgot this: if I apply a torque, its works is torque by angle, like moment of inertia of disk1 is different of moment of inertia of disk2, even torque are the same (in value), how the work can be the same ? – user47586 Jun 1 '14 at 20:35
• If moments of inertia are not the same then work done will not be the same - just the change in angular momentum. But energy is $L^2/(2I)$ so if delta L is the same it doesn't mean delta E is... – Floris Jun 1 '14 at 22:36

To define the (planar) motion of a rigid body you need the angular velocity $\omega$ as well as the point it is rotating about. You cannot deduce the center of rotation from $\omega$, but rather from the combination of linear and angular speed.

So if "y" axis has a linear speed of $v_1 = \omega_1 r$ as it orbits around "x", and disk2 has a rotational speed $\omega_2$ in an absolute reference frame (with relative rotation $\omega_2-\omega_1$) then the center of rotation is located at:

$$cor = r - \frac{\omega_1}{\omega_2} r$$

So if the two disks have no relative motion and $\omega_1=\omega_2$ then the center of rotation is at the origin $cor=0$ Duh!

But, like in a planetary gear system, you have have disk 2 rolling inside an outer ring of radius $R$, sp the center of rotation is $cor=R$ which happens when $$\omega_2 = -\frac{r}{R-r} \omega_1$$

Appendix

To find the center of rotation, add the linear velocity of disk1 as a function position along the radial direction with the linear velocity of disk2. Find at which position the combined speed is zero.

$$v = \omega_1 r + \omega_2 (x-r) = 0$$ $$cor = x = \frac{r (\omega_2-\omega_1)}{\omega_2}$$

• Maybe my question is not clear enough, axis of disk1 is fixed to the ground and the ground can't move (theoretical). – user47586 May 30 '14 at 19:43
• Yes I agree, but the axis of disk2 is orbiting about "x" and spinning about "y". The center of rotation is for disk2. – ja72 May 30 '14 at 20:04
• Ok, if w1=-w2 this would say cor = 2r ? – user47586 May 30 '14 at 20:14
• Yes. That would be epicyclic motion with ring radius of $2r$. – ja72 May 30 '14 at 20:31
• ok, in your last formula it's not w1-w2 ? you wrote w2-w2 – user47586 May 30 '14 at 21:02