Angular velocity

We can use the expression as defined above to calculate the magnitude of angular velocity. The evaluation of the first expression will require us to determine angle between two vectors. The second expression, which uses tangential velocity, gives easier option, when this component of velocity is known.

Direction of angular velocity

As the particle moves along a curved path, position of the particle changes with respect to point about which angular velocity is measured. In order to determine the direction of angular velocity, we need to know the plane of rotation at a particular position. We define plane of rotation as the plane, which contains the point (origin of the coordinate system in our case) and the tangential velocity. The direction of angular velocity is perpendicular to this plane.

Angular velocity

By convention, we consider anticlockwise rotation as positive. Looking at the figure, we can see that this direction is same that of the direction of vector product of "position vector" and "velocity".

Following the convention, we can write vector equation for angular velocity as :

It can be easily visualized that plane of rotation can change as particle moves in three dimensional space. Accordingly, direction of angular velocity can also change.

Example

Problem : The velocity of a particle confined to xy - plane is (8 i – 6 j ) m/s at an instant, when its position is (2 m, 4m). Find the angular velocity of the particle about the origin at that instant.

Solution : Angular velocity is related to linear velocity by the following relation,

$$\begin{array}{l}\mathbf{v}=\mathbf{\omega }\mathbf{x}\mathbf{r}\end{array}$$

From question, we can see that motion (including rotation) takes place in xy-plane. It, then, follows that angular velocity is perpendicular to xy-plane i.e. it is directed along z-axis. Let angular velocity be represented as :

$$\begin{array}{l}\mathbf{\omega }=a\mathbf{k}\end{array}$$

Putting the values, we have :

$$\begin{array}{l}8\mathbf{i}-6\mathbf{j}=a\mathbf{k}\mathbf{x}(2\mathbf{i}+4\mathbf{j})\end{array}$$

$$\begin{array}{l}\Rightarrow 8\mathbf{i}-6\mathbf{j}=2a\mathbf{j}-4a\mathbf{i})\end{array}$$

Comparing the coefficients of unit vectors on either side of the equation, we have :

$$\begin{array}{l}2a=-6\\ \Rightarrow a=-3\end{array}$$

and

$$\begin{array}{l}-4a=8\\ \Rightarrow a=-2\end{array}$$

We see here that we do not get unique value of angular velocity. As a matter of fact, we have included this example to highlight this aspect of the vector relation used here. It was pointed out that angular velocity is one of the operands of the vector product. From the relation, we can only conclude that angular velocity is perpendicular to linear velocity. We can see that this condition is fulfilled for any direction of velocity in the plane of motion. As such, we do not get the unique value of angular velocity as required.

Let us now apply the other vector relation that we developed to define angular velocity :

$$\begin{array}{l}\Rightarrow \mathbf{\omega }=\frac{\mathbf{r}\mathbf{x}\mathbf{v}}{r^2}\end{array}$$

Putting values, we have :

$$\begin{array}{l}\Rightarrow \mathbf{\omega }=\frac{(2\mathbf{i}+4\mathbf{j})\mathbf{x}(8\mathbf{i}-6\mathbf{j})}{{\surd (2^2+4^2)}^2}\end{array}$$

$$\begin{array}{l}\Rightarrow \mathbf{\omega }=\frac{(-12\mathbf{k}-32\mathbf{k})}{20}\end{array}$$

$$\begin{array}{l}\Rightarrow \mathbf{\omega }=-1.71\mathbf{k}\mathrm{rad}/s\end{array}$$

Instantaneous axis of rotation

The notion of rotation with the motion of a particle is visualized in terms of moving axis. We have discussed that the direction of angular velocity is perpendicular to the plane formed by tangential velocity and the point about which angular velocity is measured. We can infer this direction to be the axial direction at this point for the angular velocity at that instant.