## Angular Velocity

**Angular Velocity Definition:** The rate of change of the angular position of a particle with time is called the angular velocity of that particle.

The angle subtended by a particle, revolving in a circular path, at the centre of rotation in unit time is called the angular velocity of the particle.

So, angular velocity = \(\frac{\text { change of angular position }}{\text { time }}\)

= \(\frac{\text { angular displacement }}{\text { time }}\)

- Angular velocity is expressed by the symbol ω (Greek letter ‘omega’).
- While revolving in a circular path, if a particle subtends equal angles at the centre in equal intervals of time, the particle is said to be in uniform circular motion.
- During a uniform circular motion, if a particle subtends an angle θ at the centre in time t, then the angular velocity of the particle will be = \(\omega=\frac{\theta}{t}\)
- If the circular motion is not uniform, then the total angle subtended (θ) by the particle divided by the total time (t) is called the average angular velocity of the particle, i.e., the average angular velocity, \(\omega=\frac{\theta}{t}\)

**Read and Learn More: Class 11 Physics Notes**

**Instantaneous Angular Velocity:** The instantaneous angular velocity of a particle at a given point is the limiting value of the rate of the angular displacement, from that point with respect to a time interval when the time interval tends to zero.

In a time interval Δt, if the angular displacement of a particle is Δθ, then the instantaneous angular velocity,

= \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t}=\frac{d \theta}{d t}\)

Since angular displacement is an axial vector, angular velocity is also an axial vector. The direction of angular velocity is the same as that of angular displacement, i.e., along the axis of rotation.

**Unit And Dimension Of Angular Velocity:** Usually, the unit of angular displacement is radian; hence the unit of angular velocity

= \(\frac{\text { unit of angular displacement }}{\text { unit of time }}\)

= \(\text { radian } / \text { second }=\mathrm{rad} \cdot \mathrm{s}^{-1}\)

Since angular displacement is a dimensionless quantity, the dimension of angular velocity

= \(\frac{\text { dimension of angular displacement }}{\text { dimension of time }}=\frac{1}{\mathrm{~T}}=\mathrm{T}^{-1}\)

**Time-Period:** The time taken by a particle to make one complete revolution along a circular path is called the time period of that particle in rotational motion.

In one complete revolution of the particle along the circular path, the angular displacement of the particle, θ = 2π rad. So, if the time period is T, then the angular velocity of the particle is, \(\omega=\frac{2 \pi}{T} \quad \text { or, } \quad T=\frac{2 \pi}{\omega}\)

The unit of time period is second (s) and its dimension is T.

**Frequency: **The number of revolutions completed by a particle in unit time is called the frequency (n) of that particle. It represents the rotational speed of the particle.

For 1 complete revolution, angular displacement = 2π

So, for n complete revolutions, angular displacement =2πn

Since the time taken for these n revolutions is 1 second, the angular velocity,

⇒ \(\omega=\frac{2 \pi n}{1}=2 \pi n \quad \text { or, } \quad n=\frac{\omega}{2 \pi}\)

Hence, comparing with the time period we see that, n = 1/T

∴ The dimension of frequency = T^{-1}

The unit of frequency is s^{-1}, commonly known as Hertz (Hz). It is also called revolution per second or rps. For practical purposes, a widely used unit is revolution per minute or rpm. For very slow revolutions, revolution per hour or rph is also used.

It is clear that 1 rps = 60 rpm = 3600 rph

The frequencies of the second, minute and hour hands of a clock are 1 rpm, 1 rph and 1/12rph respectively.

Relation Between rpm And rad · s^{-1}:

1 rpm = \(\frac{1 \text { complete revolution }}{1 \text { minute }}=\frac{2 \pi}{60} \mathrm{rad} \cdot \mathrm{s}^{-1}=\frac{\pi}{30} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

**Relation Between Instantaneous Linear Velocity And Angular Velocity:** Let us consider that a particle moving with angular velocity ω along a circular path of radius r, covers a distance s along the circular path in time t. In that time, the angular displacement of the particle is θ.

According to the definition, \(\omega=\frac{\theta}{t}\).

Now, if the linear speed of the particle is v, then,

v = \(\frac{s}{t}=\frac{r \theta}{t} \quad \text { or, } \quad v=\omega r\)

i.e., speed = angular velocity x radius of the circular path

- In this case, the magnitude of instantaneous linear velocity is equal to the magnitude of instantaneous speed; hence, it can be inferred that at any moment, magnitude of instantaneous linear velocity = magnitude of instantaneous angular velocity x radius of the circular path
- If an extended body undergoes perfect revolution (i.e., different points on the body revolve in different circular paths around a definite axis), then the speed of different particles situated at different distances from the axis of rotation will be different, though the magnitude of the angular velocity of every particle remains the same.

Greater the distance of a particle from the axis, greater is its linear velocity.

**Proof Of v = ωr With The Help Of Calculus:**

We know that, s = rθ

Differentiating with respect to time we get, \(\frac{d s}{d t}=r \frac{d \theta}{d t}\) (because r is constant)

or, v = rω (instantaneous linear velocity, v = \(\frac{d s}{d t}\); instantaneous angular velocity, = \(\frac{d \theta}{d t}\)

**Vector Notation:** The vector notation of the relation between linear velocity and angular velocity is \(\vec{v}=\vec{\omega} \times \vec{r}\)

The vectors \(\vec{v}\), \(\vec{w}\) and \(\vec{r}\) are shown.

## Angular Velocity Numerical Examples

**Example 1. Determine the angular velocities of the second, minute and hour hands of a clock.**

**Solution:**

Since the second hand of a clock completes a circular revolution in 60 seconds, the angular velocity of the second hand is, \(\omega=\frac{2 \pi}{60}=\frac{\pi}{30}=0.105 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Since the minute hand of a clock completes a circular revolution in 60 minutes, the angular velocity of the minute hand is, \(\omega=\frac{2 \pi}{60 \times 60}=\frac{\pi}{1800}=1.75 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1} .\)

Since the hour hand of a clock completes a circular revolution in 12 hours, the angular velocity of the \(\omega=\frac{2 \pi}{60 \times 60}=\frac{\pi}{1800}=1.75 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1} .\)

**Example 2. A car is moving along a circular path of radius 20 m with a speed of 40 km · h ^{-1}. Find its angular velocity.**

**Solution:**

Linear velocity, \(v=\frac{40 \times 1000}{60 \times 60}=\frac{100}{9} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Angular velocity, \(\omega=\frac{v}{r}=\frac{100}{9 \times 20}=0.56 \mathrm{rad} \cdot \mathrm{s}^{-1}\).

**Example 3. Calculate the angular velocity of the earth about its own axis.**

**Solution:**

Earth completes one rotation about its axis in 24 h i.e., it covers an angle 2π in 24 x 60 x 60 s.

∴ Angular velocity of the earth = \(\frac{2 \pi}{24 \times 60 \times 60}\)

= \(\frac{\pi}{43200} \mathrm{rad} / \mathrm{s}\)

**Example 4. Find the time interval between two successive overlaps of the hour hand and the minute hand of a clock.**

**Solution:**

Let the time interval after which the two hands meet each other be t hours.

In the case of the hour hand, the angular velocity \(\omega=\frac{2 \pi}{12}=\frac{\pi}{6} \mathrm{rad} \cdot \mathrm{h}^{-1}\)

Hence, in t hours, the angular displacement of the hour hand, \(\theta=\omega t=\frac{\pi}{6} t \mathrm{rad}\)

In the case of the minute hand, the angular velocity \(\omega^{\prime}=\frac{2 \pi}{1}=2 \pi \mathrm{rad} \cdot \mathrm{h}^{-1} .\)

Hence, the angular displacement of the minute hand, \(\theta^{\prime}=\omega^{\prime} t=2 \pi t \mathrm{rad} .\)

After meeting each other, when the hands coincide again after t hours, the minute hand completes one revolution more than the hour hand, i.e., the angular displacement of the minute hand is 2π rad more than that of the hour hand.

So, \(\theta^{\prime}-\theta=2 \pi\) or, \(2 \pi t-\frac{\pi t}{6}=2 \pi\)

or, \(t \times \frac{11}{6}=2 or, \quad t=\frac{12}{11} \mathrm{~h}=1 \mathrm{~h} 5 \min 27 \mathrm{~s}(approx.).\)