Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is Instantaneous Angular Acceleration Formula?
Definition: The rate of change of angular velocity of a particle with time is called the angular acceleration of that particle.
Let us consider a particle under rotational motion whose initial angular velocity is ω1 and final angular velocity after time t is ω2. So according to the definition
angular acceleration (α) \(=\frac{\text { change in angular velocity }}{\text { time }}\)
= \(\frac{\omega_2-\omega_1}{t}\)
Comparing with the definition of average acceleration in the chapter One-dimensional Motions this can be referred to as the average angular acceleration.
Instantaneous angular acceleration: The instantaneous angular acceleration of a particle at a given point is the limiting value of the rate of the change in velocity with respect to a time interval when the time interval tends to zero.
If the change in angular velocity of a particle in a small time interval Δt is Δω, then instantaneous angular acceleration,
α = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t}\) = \(\frac{d \omega}{d t}\).
Since, ω = \(\frac{d \theta}{d t}\), α = \(\frac{d}{d t}\left(\frac{d \theta}{d t}\right)\) = \(\frac{d^2 \theta}{d t^2}\)
If the angular velocity of a particle increases gradually, the particle is said to be moving with an angular acceleration. For example, when an electric fan is switched on, it under goes angular acceleration before attaining its maximum velocity. On the other hand, if the angular velocity of a body decreases gradually, it is said to be moving with an angular deceleration or angular retardation. For example, when an electric fan is switched off, it undergoes angular deceleration until it stops.
In the case of angular retardation, the initial angular velocity ω1 is greater than the final angular velocity ω2.
Hence, α = \(\frac{\omega_2-\omega_1}{t}\) = a negative quantity
So, angular retardation is nothing but a negative angular acceleration.
Angular acceleration is an axial vector: As \(\vec{\omega}_1\) and \(\vec{\omega}_2\) are axial vectors, \(\vec{\omega}_2\) – \(\vec{\omega}_1\) and, hence, \(\vec{\alpha}\) is also an axial vector. When \(\vec{\alpha}\) is positive, its direction is the same as that of \(\vec{\omega}\) and \(\vec{\alpha}\) when is negative, its direction is opposite to that of \(\vec{\omega}\).
Unit and dimension of angular acceleration:
Unit of angular acceleration
\(=\frac{\text { unit of angular velocity }}{\text { unit of time }}\)
= radian/second2 = (radᐧs-2)
Dimension of angular acceleration
\(=\frac{\text { dimension of angular velocity }}{\text { dimension of time }}\) = \(\frac{T^{-1}}{T}\) = T-2
Relation between Linear Acceleration and Angular Acceleration
Let us consider a particle moving along a circular path of radius r, whose linear velocity changes from \(\overrightarrow{v_1}\) to \(\overrightarrow{v_2}\) in time t. During this time-interval, its angular velocity changes from \(\vec{\omega}_1\) to \(\vec{\omega}_2\).
Thus, the magnitude of instantaneous linear acceleration = the magnitude of Instantaneous angular acceleration × radius of the circular path.
It should be remembered that in the case of pure rotation, linear acceleration changes its direction continuously, but the direction of angular acceleration remains unaltered. Whenever the axis of rotation remains fixed, the direction of angular acceleration does not change.
Geometrical representation: In Fig.(b), putting \(\vec{a}\) and \(\vec{\alpha}\) instead of \(\vec{v}\) and \(\vec{\omega}\), respectively, we will obtain the geometrical representation for the relation among \(\vec{a}\), \(\vec{a}\), and \(\vec{r}\).
Kinematical Equations of Rotational Motion
From the analogy of the three equations s = rθ, v = rω and a = rα, it can be inferred that in case of rotational motion, angular displacement θ, angular velocity ω and angular acceleration α, respectively, play the same role as that of displacement s, velocity v and acceleration a in the case of translational motion. For this reason θ, ω and α are called the rotational analogues of s, v and a respectively. (If a quantity has similarities with another quantity then that quantity is called the analogue of the other.)
Kinematical equations for uniformly accelerated motion are:
- v = u + at
- s = ut + \(\frac{1}{2}\)at2
- v2 = u2 + 2as
For a uniformly accelerated rotational motion, the analogues of these equations are:
- ω2 = ω1 + αt
- θ = ω2t + \(\frac{1}{2} \alpha t^2\)
- \(\omega_2^2\) = \(\omega_1^2\) + 2αθ
Here, initial angular velocity = ω1, final angular velocity after time t = ω2, angular acceleration = α and angular displacement after time t = θ.
In the case of translational motion with uniform velocity, s = vt. in the case of uniform rotational motion, the analogue of this equation is θ = ωt.
Comparison of linear and angular motion with constant acceleration:
Straight line motion with constant linear acceleration:
For an object that starts moving along a straight line with initial velocity u and constant linear acceleration a, we have
- a = constant,
- u = u + at,
- s = ut + \(\frac{1}{2}\)at2,
- v2 = u2 + 2as
Fixed axis rotation with constant angular acceleration: For an object that starts revolving with initial
angular velocity ω0 and uniform angular acceleration α, we have
- α = constant,
- ω1 = ω0 + αt,
- θ = ω0t + \(\frac{1}{2} \alpha t^2\),
- ω2 = \(\omega_0^2\) + 2αθ
Numerical Examples
Example 1.
An electric fan is revolving with a velocity of 210 rpm. When its motion is increased with the help of a regulator, it attains a velocity of 630 rpm in 11 s. What is the angular acceleration of the fan? Also calculate the number of revolutions completed by the fan in that 11 s?
Solution:
Initial angular velocity,
ω1 = \(\frac{2 \pi \times 210}{60}\) = 7π rad ᐧ s-1
Final angular velocity, ω2 = \(\frac{2 \pi \times 630}{60}\) = 21π rad ᐧ s-1
Now, from the equation ω2 = ω1 + αt we get,
angular acceleration,
α = \(\frac{\omega_2-\omega_1}{t}=\) = \(\frac{21 \pi-7 \pi}{11}\) = \(\frac{14 \pi}{11}\) = \(\frac{14}{11} \times \frac{22}{7}\)
= 4 rad ᐧ s-1
Again, angular displacement,
θ = ω1t + \(\frac{1}{2} \alpha t^2\) = 7π × 11 + \(\frac{1}{2}\) × 4 × (11)2
= 11(7π + 22) = 484 rad
Since in a single revolution, the angular displacement is 2π, the number of revolutions
= \(\frac{\theta}{2 \pi}\) = \(\frac{484}{2 \pi}\) = 77
Example 2.
A wheel rolls on a horizontal path with uniform velocity. Prove that the velocity of any point on the
circumference of the wheel with respect to its centre is equal to the velocity of the wheel. What will be the instantaneous velocity of the point on the wheel which touches the ground?
Solution:
If the radius of the wheel is r then its circumference = 2πr, and velocity of the wheel = \(\frac{2 \pi r}{T}\) = \(\frac{2 \pi r}{\frac{2 \pi}{\omega}}\) = ωr; here, ω = angular velocity of the wheel.
Again, the linear velocity of any point on the circumference of the wheel with respect to its centre is, v = ωr. So, this velocity is the same as the velocity of the wheel.
At any moment, when a point on the wheel touches the ground, its linear velocity with respect to the centre of the wheel is v and its direction is just opposite to the direction of the linear velocity v of the wheel. Hence, the instantaneous resultant velocity of the point on the wheel which touches the ground = v – v = 0.
Example 3.
Starting from rest, a wheel, with uniform acceleration, attains an angular velocity of 60 rad ᐧ s-1 at the end of 30 complete revolutions. What is the angular acceleration of the wheel?
Solution:
Initial angular velocity, ω1 = 0.
Angular displacement at the end of 30 revolutions,
θ = 30 × 2π = 60π rad
Final angular velocity ω2 = 60 rad ᐧ s-1
Now, from the equation
\(\omega_2^2\) = \(\omega_1^2\) + 2αθ
(α = angular acceleration of the wheel)
we get, (60)2 = 0 + 2α × 60π
or, α = \(\frac{30}{\pi}\) = 9.55 rad ᐧ s-1