MHB Finding the Rotation Vector $\omega$ of a Sphere

Click For Summary
The rotation vector $\omega$ of a sphere is equal to the angular velocity vector. In this case, the sphere rotates around the z-axis with an angular velocity of 4, indicating a counterclockwise direction when viewed from the positive z-axis. The relationship is defined mathematically by the equation $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{\vec{r}\cdot\vec{r}}$, where $\vec{v}$ represents the tangential velocity at a point on the sphere. Thus, the rotation vector directly corresponds to the angular velocity in this scenario. Understanding this relationship is crucial for analyzing rotational motion.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! (Wave)

A sphere with radius $10 cm$ and center $(0,0,0)$ turns around the $z$-axis with angular velocity $4$ and with such a direction that the rotation has counterclockwise direction, being seen my the positive semi-axis $z$.

I want to find the rotation-vector $\omega$.

Is this equal to the vector of the angular velocity? If so, why? (Thinking)
 
Mathematics news on Phys.org
evinda said:
Hello! (Wave)

A sphere with radius $10 cm$ and center $(0,0,0)$ turns around the $z$-axis with angular velocity $4$ and with such a direction that the rotation has counterclockwise direction, being seen my the positive semi-axis $z$.

I want to find the rotation-vector $\omega$.

Is this equal to the vector of the angular velocity? If so, why? (Thinking)

It is - by definition. The rotation vector $\vec{\omega}$ IS the angular velocity! It's defined by
$$\vec{\omega}=\frac{\vec{r}\times\vec{v}}{\vec{r}\cdot\vec{r}},$$
where $\vec{v}$ is the regular velocity vector in the tangential direction of a point on the circle or sphere.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
67
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
2K