7.6 Angular velocity and its relation with linear velocity

In this section, we shall study what is angular velocity and its role in rotational motion. We have seen that every particle of a rotating body moves in a circle. The linear velocity of the particle is related to the angular velocity. The relation between these two quantities involves a vector product which we learnt about in the last section. We know in the rotational motion of a rigid body about a fixed axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. Fig. 7.16 shows a typical particle (at a point P) of the rigid body rotating about a fixed axis (taken as the z-axis). The particle describes a circle with a centre \(\mathrm{C}\) on the axis. The radius of the circle is \(r\), the perpendicular distance of the point P from the axis. We also show the linear velocity vector \(\mathbf{v}\) of the particle at \(P\). It is along the tangent at \(P\) to the circle.

Let \(\mathrm{P}^{\prime}\) be the position of the particle after an interval of time \(\Delta t\) (Fig. 7.16).
The angle PCP’ describes the angular displacement \(\Delta \theta\) of the particle in time \(\Delta t\).
The average angular velocity of the particle over the interval \(\Delta t\) is \(\Delta \theta / \Delta t\).
As \(\Delta t\) tends to zero (i.e. takes smaller and smaller values), the ratio \(\Delta \theta / \Delta t\) approaches a limit which is the instantaneous angular velocity \(\mathrm{d} \theta / \mathrm{d} t\) of the particle at the position P.
We denote the instantaneous angular velocity by \(\omega\) (the Greek letter omega). We know from our study of circular motion that the magnitude of linear velocity \(v\) of a particle moving in a circle is related to the angular velocity of the particle \(\omega\) by the simple relation
\(v=\omega r\), where \(r\) is the radius of the circle.

We observe that at any given instant the relation \(v=\omega r\) applies to all particles of the rigid body. Thus for a particle at a perpendicular distance \(r_i\) from the fixed axis, the linear velocity at a given instant \(v_i\) is given by
\(
v_i=\omega r_i \dots(7.19)
\)
The index \(i\) runs from 1 to \(n\), where \(n\) is the total number of particles of the body.

For particles on the axis, \(r=0\), and hence \(v=\omega r=0\). Thus, particles on the axis are stationary. This verifies that the axis is fixed.
Note that we use the same angular velocity \(\omega\) for all the particles. We, therefore, refer to \(\omega\) as the angular velocity of the whole body.

We have characterised pure translation of a body by all parts of the body having the same velocity at any instant of time. Similarly, we may characterise pure rotation by all parts of the body having the same angular velocity at any instant of time. 

For rotation about a fixed axis, the angular velocity vector lies along the axis of rotation and points out in the direction in which a right-handed screw would advance if the head of the screw is rotated with the body. (See Fig. 7.17a).
The magnitude of this vector is \(\omega=d \theta / d t\) is known as the angular velocity.

The rate of change of angular position is called angular velocity. Thus, the angular velocity is
\(
\omega=\lim _{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t}=\frac{d \theta}{d t} .
\)

We shall now look at what the vector product \(\boldsymbol{\omega} \times \mathbf{r}\) corresponds to. Refer to Flg. 7.17(b) which is a part of Fig. 7.16 reproduced to show the path of the particle \(P\). The figure shows the vector \(\boldsymbol{\omega}\) directed along the fixed \((z-)\) axis and also the position vector \(\mathbf{r}=\mathbf{O P}\) of the particle at \(\mathrm{P}\) of the rigid body with respect to the origin O. Note that the origin is chosen to be on the axis of rotation.

\(
\text { Now } \quad \boldsymbol{\omega} \times \mathbf{r}=\boldsymbol{\omega} \times \mathbf{O P}=\boldsymbol{\omega} \times(\mathrm{OC}+\mathrm{CP})
\)
\(
\text { But } \quad \boldsymbol{\omega} \times \mathbf{O C}=0 \text { as } \boldsymbol{\omega} \text { is along } \mathbf{O C}
\)
\(
\text { Hence } \quad \boldsymbol{\omega} \times \mathbf{r}=\boldsymbol{\omega} \times \mathbf{C P}
\)

The vector \(\boldsymbol{\omega} \times \mathbf{C P}\) is perpendicular to \(\boldsymbol{\omega}\), i.e. to the \(z\)-axis and also to \(\mathbf{C P}\), the radius of the circle described by the particle at P. It is, therefore, along the tangent to the circle at P. Also, the magnitude of \(\boldsymbol{\omega} \times \mathbf{C P}\) is \(\omega\) (CP) since \(\boldsymbol{\omega}\) and \(\mathbf{C P}\) are perpendicular to each other. We shall denote \(\mathbf{C P}\) by \(\mathbf{r}_{\perp}\) and not by \(\mathbf{r}\), as we did earlier.

Thus, \(\boldsymbol{\omega} \times \mathbf{r}\) is a vector of magnitude \(\omega r_{\perp}\) and is along the tangent to the circle described by the particle at \(P\). The linear velocity vector \(\mathbf{v}\) at \(P\) has the same magnitude and direction. Thus,
\(\mathbf{v}=\boldsymbol{\omega} \times \mathbf{r} \dots(7.20)\)
In fact, the relation, Eq. (7.20), holds good even for rotation of a rigid body with one point fixed, such as the rotation of the top [Fig. 7.6(a)]. In this case \(\mathbf{r}\) represents the position vector of the particle with respect to the fixed point taken as the origin.

We note that for rotation about a fixed axis, the direction of the vector \(\boldsymbol{\omega}\) does not change with time. Its magnitude may, however, change from instant to instant. For the more general rotation, both the magnitude and the direction of \(\boldsymbol{\omega}\) may change from instant to instant.

Angular acceleration

You may have noticed that we are developing the study of rotational motion along the lines of the study of translational motion with which we are already familiar. Analogous to the kinetic variables of linear displacement \((\mathbf{s})\) and velocity \((\mathbf{v})\) in translational motion, we have angular displacement \((\boldsymbol{\theta})\) and angular velocity \((\boldsymbol{\omega})\) in rotational motion. It is then natural to define in rotational motion the concept of angular acceleration in analogy with linear acceleration defined as the time rate of change of velocity in translational motion. We define angular acceleration \(\boldsymbol{\alpha}\) as the time rate of change of angular velocity; Thus,
\(
\boldsymbol{\alpha}=\frac{\mathrm{d} \boldsymbol{\omega}}{\mathrm{d} t} \dots(7.21)
\)
If the axis of rotation is fixed, the direction of \(\boldsymbol{\omega}\) and hence, that of \(\boldsymbol{\alpha}\) is fixed. In this case, the vector equation reduces to a scalar equation
\(
\alpha=\frac{\mathrm{d} \omega}{\mathrm{d} t} \dots(7.22)
\)

Example 7.8: A particle moves in a circle of radius \(20 \mathrm{~cm}\) with a linear speed of \(10 \mathrm{~m} / \mathrm{s}\). Find the angular velocity.

Solution:

The angular velocity is
\(
\omega=\frac{v}{r}=\frac{10 \mathrm{~m} / \mathrm{s}}{20 \mathrm{~cm}}=50 \mathrm{rad} / \mathrm{s} .
\)

Example 7.9: A particle travels in a circle of radius \(20 \mathrm{~cm}\) at a speed that uniformly increases. If the speed changes from \(5.0 \mathrm{~m} / \mathrm{s}\) to \(6.0 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s}\), find the angular acceleration.

Solution:

We know,
\(v=r \omega \dots(i)\)
where \(v\) is the linear speed of the particle. Differentiating equation (i) with respect to time, the rate of change of speed is
\(
a_t=\frac{d v}{d t}=r \frac{d \omega}{d t}
\)
\(
\text { or, } \quad a_t=r \alpha \text {. } \dots(ii)
\)
Remember that \(a_t=\frac{d v}{d t}\) is the rate of change of speed and is not the rate of the change of velocity. It is, therefore, not equal to the net acceleration.

We shall show that \(a_t\) is the component of acceleration along the tangent and hence we have used the suffix \(t\). It is called the tangential acceleration.

The tangential acceleration is given by
\(
\begin{aligned}
a_t & =\frac{d v}{d t}=\frac{v_2-v_1}{t_2-t_1} \\
& =\frac{6.0-5.0}{2.0} \mathrm{~m} / \mathrm{s}^2=0.5 \mathrm{~m} / \mathrm{s}^2 .
\end{aligned}
\)
The angular acceleration is \(\alpha=a_t / r\)
\(
=\frac{0.5 \mathrm{~m} / \mathrm{s}^2}{20 \mathrm{~cm}}=2.5 \mathrm{rad} / \mathrm{s}^2 .
\)