What does $\vec{\omega}\times\vec{r}$ equals to in circular motion?

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?

• A2A @ACuriousMind Feb 26 '20 at 14:33
• 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 . Feb 26 '20 at 14:38
• @trula Sorry, I don't understand. Feb 26 '20 at 14:40

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.

• Thank you for your detailed answer, sir. Does the magnitude of rotation vector $\vec{\omega}$ equals to angular velocity $\omega=\frac{d\theta}{dt}$? Feb 26 '20 at 15:08
• @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. Feb 26 '20 at 15:12
• Alright! Thanks :) Feb 26 '20 at 15:23
• Do you mean bivector in the geometric/clifford algebra sense? Feb 26 '20 at 18:27
• Oh, no. I didn't mean it. I just heard it first time. Thanks. Feb 26 '20 at 20:22

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 $$||\vec{n}||=1$$

• Thank you for your explanation. Feb 26 '20 at 20:22