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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?

Irodov's answer implies that

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

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?

Meow
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2 Answers2

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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.

Ishan Jawale
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Brian Moths
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  • 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
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    @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. – John Alexiou May 19 '19 at 14:34
  • Hello sir, Sorry for pinging you 7 years after the answer but could you possibly tell me how the time derivative of $(\vec{\omega}_1-\vec{\omega}_2)$ is $(- \vec{\omega}_2 \times\vec{\omega}_1)$ – Linkin Jan 25 '21 at 15:36
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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.

John Alexiou
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