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I know that $\vec{v}=wr\hat{\theta}$ in uniform circular motion. This equation looks like a result of a cross product.

Yesterday, I started to learn Basic Dynamics of Rigid Bodies. My teacher wrote $\vec{v}=\vec{\omega}\times\vec{r}$ in the lecture. But I have never seen this. This equation also equals to $\vec{v}={\omega}{r}\sin(\vec{\omega},\vec{r})\hat{?}$ If $\sin(\vec{\omega},\vec{r})$ equals to $\sin(\pi/2)$ then the equation equals to $\vec{v}={\omega}{r}\hat{?}$

I wonder that how the unit vector $\hat{?}$ is equal to $\hat{\theta}$. When I try to figure it out, I can't. Can you explain it, please?

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  • $\begingroup$ A2A @ACuriousMind $\endgroup$ Feb 26 '20 at 14:33
  • $\begingroup$ what does 𝜃̂ mean in your first formula, since the formula of your teacher is the usual one. the next equations apply only to the absolut value of v not to the vector . $\endgroup$
    – trula
    Feb 26 '20 at 14:38
  • $\begingroup$ @trula Sorry, I don't understand. $\endgroup$ Feb 26 '20 at 14:40
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Your first equation works when you have already isolated the plane of rotation. It treats $\omega$ as a scalar. Sometimes we can't isolate it into a 2d case like this, such as if there are angular accelerations or other considerations. To handle the full 3d case, we define rotation using a vector, $\vec{\omega}$. This vector is defined to have a magnitude equal to the $\omega$ from the first equation, and a direction which is is at right angles to the rotation.

Now, if you're a purist, that cross product might bug you. In these cases, we're not actually using the cross product of two vectors, but the product of a bivector, which is easier to trace down to why it is the correct tool to use. It just so happens that in 3 dimensions, the math for cross products and bivectors is identical, and historically we've found teaching cross products easier than introducing bivectors.

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  • $\begingroup$ Thank you for your detailed answer, sir. Does the magnitude of rotation vector $\vec{\omega}$ equals to angular velocity $\omega=\frac{d\theta}{dt}$? $\endgroup$ Feb 26 '20 at 15:08
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    $\begingroup$ @ICCQBE Yes, it will. In fact, if you take a simple rotation in the x-y plane, where $\vec\omega$ is in the z direction, and crunch the numbers for the cross product, you'll end up with the 2d equations you are used to. It's a good simple test case to see how the cross product works its magic. $\endgroup$
    – Cort Ammon
    Feb 26 '20 at 15:12
  • $\begingroup$ Alright! Thanks :) $\endgroup$ Feb 26 '20 at 15:23
  • $\begingroup$ Do you mean bivector in the geometric/clifford algebra sense? $\endgroup$
    – roshoka
    Feb 26 '20 at 18:27
  • $\begingroup$ Oh, no. I didn't mean it. I just heard it first time. Thanks. $\endgroup$ Feb 26 '20 at 20:22
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you can write any vector with his magnitude and unity direction

$$\vec{v}=|\vec{v}|\,\vec{n}\tag 1$$

with

$|\vec{v}|=|\vec{\omega}|\,|\vec{r}|\,\sin(\theta) $

where $\theta$ is the angle between $\vec{\omega}$ and $\vec{r}$

thus equation (1)

$$\vec{v}=|\vec{\omega}|\,|\vec{r}|\,\sin(\theta)\,\vec{n}$$

where the vector n is perpendicular to the vector omega and r

$\vec{n}\perp\vec{\omega}\quad ,\vec{n}\perp\vec{r}$ with $||\vec{n}||=1$

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  • $\begingroup$ Thank you for your explanation. $\endgroup$ Feb 26 '20 at 20:22

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