# Relative angular velocity and acceleration

Background: (Irodov 1.55) Two bodies rotate around intersecting perpendicular axes with angular velocities $\hat\omega_1,\hat\omega_2$. Relative to one body, what is the angular-velocity and -acceleration of the other?

$$\hat \omega=\hat\omega_1-\hat \omega_2$$ $$\hat \alpha=\hat\omega_1\times\hat \omega_2.$$

I have a hard time grokking why the above are true above (for the first) vague analogies with linear velocity. Does anyone, willing to share, have an intuitive grasp on the above equations?

• Do you have a link or other reference. – ja72 Mar 25 '14 at 19:15
• @ja72 Yes, here, although I'm less sure that my interpretation of $\alpha$ as a cross product is correct. – Meow Mar 25 '14 at 19:28

ja72's answer is probably right, but I was confused by his notation, so I will give my own answer. Suppose we have two object rotating with angular velocity $$\vec{\omega}_1$$ and $$\vec{\omega}_2$$. Then the velocity of a point $$\vec{r}$$ of object $$1$$ in the lab frame is $$\vec{v}_{1,lab}=\vec{\omega}_1 \times \vec{r}$$. Similarly, the velocity of a point $$\vec{r}$$ of object $$2$$ in the lab frame is $$\vec{v}_{2,lab}=\vec{\omega}_2 \times \vec{r}$$.

Now to someone in the second object, a point $$\vec{r}$$ that is stationary in the lab frame will have an apparent velocity $$-\vec{v}_{2,lab}=-\vec{\omega}_2 \times \vec{r}$$. From this, you can see that a point $$\vec{r}$$ in the first object will appear to have a velocity $$\vec{v}_{1,lab}-\vec{v}_{2,lab} = \vec{\omega}_1 \times \vec{r}-\vec{\omega}_2 \times \vec{r} = (\vec{\omega}_1-\vec{\omega}_2)\times \vec{r}$$. Thus the first object appears to have angular velocity $$\vec{\omega}_1-\vec{\omega}_2$$ to an observer in the second object.

Now since $$\vec{\omega}_1$$ is stationary in the lab frame, it's time derivative is $$- \vec{\omega}_2 \times\vec{\omega}_1 =\vec{\omega}_1\times \vec{\omega}_2$$ in the frame of the second object. Thus object one appears to have an angular acceleration of $$\vec{\omega}_1\times \vec{\omega}_2$$ in the frame of the second object.

• In "will have an apparent velocity $-\vec{v_{2,lab}}=-\vec{\omega_2} \times \vec{r}$", isn't $r$ the position vector of the observer? Then the 2 $r$s in the relative velocity expression $\vec{v}_{1,lab}-\vec{v}_{2,lab} = \vec{\omega}_1 \times \vec{r}-\vec{\omega}_2 \times \vec{r} = (\vec{\omega}_1-\vec{\omega}_2)\times \vec{r}$ are different. – Aritra Das Nov 13 '15 at 16:46
• @Aritra Das No, an object that is stationary in the lab from appears to be rotating with angular velocity $-\vec{\omega}_2$ in the frame of the second object. So if it has a position $\vec{r}$, it would appear to have a velocity $-\vec{\omega}_2 \times \vec{r}$ in the frame of the second object, regardless of where it is being observed from. – Brian Moths Nov 13 '15 at 17:01
• I agree this answer is better than mine since it speaks to why and I answered the how. – ja72 May 19 at 14:34

Lets pick a coordinate system where the second body rotates about the local $Z$ axis.

The rotational kinematics if the second body are defined as

$$E_2 = E_1 {\rm Rot}(\hat{z}, \theta)$$

where $E_i$ are the 3×3 rotation matrices, and $\hat{z}=(0,0,1)$ . If the angular velocity of the first body is $\hat{\omega}_1$ then differentiating the above expression yields the angular velocity of the second body

$$\hat{\omega}_2 \times E_2 = \hat{\omega}_1 \times E_1 {\rm Rot}(\hat{z}, \theta) + (E_1 \hat{z} \dot\theta) \times (E_1 {\rm Rot}(\hat{z}, \theta))$$ $$\hat{\omega}_2 \times (E_1 {\rm Rot}(\hat{z}, \theta)) = \left( \hat{\omega}_1 + E_1 \hat{z} \dot\theta \right) \times (E_1 {\rm Rot}(\hat{z}, \theta))$$ $$\hat{\omega}_2 = \hat{\omega}_1 + E_1 \hat{z} \dot\theta$$

Further differentiation yields the angular acceleration kinematics

$$\hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times ( E_1 \hat{z} \dot\theta )$$ $$\hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times ( \hat{\omega}_2 - \hat{\omega}_1)$$ $$\hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times \hat{\omega}_2$$

So the relative velocity and acceleration are

$$\hat{\omega} = \hat{\omega}_2 -\hat{\omega}_1 = E_1 \hat{z} \dot\theta$$ $$\hat{\alpha} = \hat{\alpha}_2 -\hat{\alpha}_1 = E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times \hat{\omega}_2$$

If there is no relative joint acceleration ($\ddot\theta =0$) then you get the expression stated in the question.

• I just noticed that I flipped the [1] and [2] indices. You can manage I hope. – ja72 Mar 25 '14 at 19:39
• PS. Time derivative of a rotation matrix is $$\dot E = \hat\omega \times E$$. See derivatives in rotating frame. – ja72 Mar 25 '14 at 19:41