Angular velocity is $$\omega= \frac{dƟ}{dt},$$ here $\theta$ and $t$ are scalar quantities. But $\omega$ is a vector quantity. Why is it such?
So far I know the direction of $\omega$ is along the axis of rotation. If so, why?
Asked
Active
Viewed 4,588 times
0
-
2Possible duplicates: https://physics.stackexchange.com/q/69345/2451 , https://physics.stackexchange.com/q/286/2451 and links therein. – Qmechanic Mar 01 '18 at 09:49
-
Because angles can be expressed as vectors – Bardeen Mar 01 '18 at 10:00
-
you are saying ,angle can be expressed as vector . Then the direction of ω and angle are to be same. But they are not........... – Mohammad Mizanur Rahaman Mar 01 '18 at 10:11
-
The direction of your angle vector should be along the axis of rotation. – BRT Mar 01 '18 at 10:15
-
then the same question arise. Why should the direction of angle vector along the axis if rotation. The angle isn't changing along the axis.It's changing along a circle...... – Mohammad Mizanur Rahaman Mar 01 '18 at 10:20
-
Suppose we have particle in the $x$-$y$ plane. The idea is to represent its angle as $\theta \hat{z}$. – BRT Mar 01 '18 at 10:23
-
@MohammadMizanurRahaman when you write $\omega$ as a scalar you have already implictly assumed that the rotation is on a plane orthogonal to the rotation axis. Indeed, for a generic vector rotating about a generic axis, the contribution of the rotation is given by the vector product between the position and the rotation vector. Look for example here https://atmos.washington.edu/2005Q4/503/DDT.pdf – ndrearu Mar 01 '18 at 10:24
-
@BRT you are saying the same thing in different way.......... – Mohammad Mizanur Rahaman Mar 01 '18 at 10:27
-
4Possible duplicate of How is it that angular velocities are vectors, while rotations aren't? – SmarthBansal Mar 01 '18 at 11:48
-
@MohammadMizanurRahaman...You can also consider why $\tau$ is a vector ...Using the differential form... – Nehal Samee Mar 01 '18 at 16:53
3 Answers
1
Thus is actually a bit subtle: it’s not a vector quantity.
In two dimensions, a rotation has a sign (left or right) but the “direction” in otherwise constrained by the two-D plane: one component. But spatial vectors in 2D have two components.
In 3D, you need to specify three components to cover an arbitrary rotation, and displacement also need three.
In 4D, i.e. space-time, displacement takes 4 (x,y,z,t) but there are 6 rotations. What are those? In space-time, they’re the three rotations you know, plus the three Lorentz boosts that mix x&t, y&t and z&t.
Roughly, a rotation combines two axes. In 2D, there are only 2 to combine. In 3D, three pairs. The general formula is D(D-1)/2
Finally, even in 3D, those three components don’t quite have all the properties of a vector: seen in the mirror, the spin is reversed, so they have “opposite parity” of a regular displacement vector. (This is really just a consequence of the fact that 3 components in 3D is a coincidence). To distinguish that, when it matters, we remind ourselves by calling it an “axial vector”.
Bob Jacobsen
- 14,439
0
There are two axis where velocity and acceleration travel in a theta pulls a particle to the center of the curve just like the normal acceleration in a car pull you into the curve. Ar in the picture really changes the direction. The angular velocity should be on the axis of rotation because the axis are all arbitrary and you can make them where it is easiest to solve.
I hoped that helped
0
You can derive the angular rotation vector from trigonometry if you trying to track the changes of a vector that rides on a rotating body.
$$ \frac{{\rm d} \boldsymbol{x}}{{\rm d}t} \triangleq \boldsymbol{\omega} \times \boldsymbol{x} $$
It is a vector because it conveys magnitude and direction (for the rotation). Tt also obeys the regular rules of vector algebra. It can be multiplied with a scalar and it can added to other rotations vectors.
John Alexiou
- 38,341