Angular Acceleration Definition
In Physics, we define angular acceleration as the time rate of change of angular velocity. This angular acceleration definition is plain because there are two types of angular velocity, spin angular velocity, and orbital angular velocity. This leads to two kinds of angular acceleration as well; spin angular acceleration and orbital angular acceleration.
Angular motion occurs when an object is moving in a rotational manner. The object’s velocity is always changing while it is moving in a circular path.
A good example to understand angular acceleration is a disk with an axle in its center.
The speed of the spinning disk is always changing. Therefore, the disk is said to have angular acceleration, since its velocity is constantly changing with time.
Velocity is a vector quantity that involves speed with velocity. When in a circular motion, the direction of every point on the object is changing constantly.
The unit of angular velocity can be expressed as radian/second. It explains the speed at which the object rotates and provides the direction of a rotating object.
There are two versions of angular acceleration, average angular acceleration and instantaneous angular acceleration. Average angular acceleration is an average calculation involving multiple values divided by the total number of values being considered.
The angular acceleration at a very specific moment of time is called instantaneous angular acceleration.
Types of Angular Acceleration
Spin angular acceleration is said to relate with the angular acceleration of the rigid body to its center of rotation, and orbital angular acceleration is defined as the angular acceleration of a point particle about a fixed origin. Angular acceleration is measured in units of angle per unit time squared (as per SI unit – radians per seconds squared) and represented by the 'α' symbol.
The angular acceleration formula is given by
α= ∆ω/∆t
where ∆ω is a change in angular velocity and ∆t is the time interval.
When uniform rotation is considered, both the average and instantaneous values opt to coincide. We will provide angular acceleration examples below –
Orbital Angular Acceleration of a Point Particle: Two-Dimensions
Orbital angular acceleration can be defined as the rate at which the two-dimensional orbital angular velocity of the particle changes from the origin. The instantaneous angular velocity ω at any point in the time is given by
ω=v˔/r
Where r is the distance from the origin and v˔ is the cross-radial component of the instantaneous velocity, which, by convention, is positive for counterclockwise motion and negative for a clockwise motion.
Therefore, the instantaneous angular acceleration α of the particle can be given by
α = d/dt (v˔/r)
expanding the right-hand side using the product rule from different calculus
α = 1/r d v˔/dt - v˔/r² dr/DT
In the special case where the particle undergoes circular motion about the origin, d v˔/dt becomes tangential acceleration a˔, and dr/dt goes out since the distance from the origin remains constant,
so the equation gets simplified as α = a˔/r
In two dimensions, angular acceleration is a number with a plus or minus sign indicating orientation but not pointing in a direction. If the angular speed increases counterclockwise or decreases clockwise, the sign is taken as positive. If the angular speed increases clockwise and decreases counterclockwise, then the sign is taken as negative.
Relation between Angular Acceleration and Linear Acceleration.
At = Δv/ Δt
However, in a circular motion, linear velocity v is a product of radius of curvature (r) and Angular velocity ω.
v = rω
Therefore linear acceleration At is given as,
A\[_{t}\] = Δ(rω)/ Δt
Now, The radius r is constant for circular motion.
So,
Δ(rω) = r(Δω).
Therefore,
A\[_{t}\] = rΔω/ Δt ….(i)
By definition, angular acceleration is given as
α = Δω/Δt
Therefore, the equation (i) changes to
A\[_{t}\] = rα
Or
α= \[\frac{A_{t}}{r}\]
Therefore, it can be proved that linear acceleration and angular acceleration are directly proportional.
Application of Angular Acceleration
The concept of angular acceleration is seen in a merry go round in a park. When a child sits on the horse and starts the merry go round from rest, there is an angular acceleration ω which constantly changes. The object, which is the merry go round, is spinning and its speed is also changing. Therefore, it can be said that the object is accelerating angularly.
The faster the machine moves, the greater is the angular acceleration. Care should be taken while designing the merry go round that the angular acceleration doesn’t exceed its limits and the rise is safe for children.
Angular acceleration can also be seen in a giant wheel. Once it starts operating it has to speed up and slow down. This gives it the angular velocity. If it speeds up at a constant rate then we can say that the angular acceleration is constant. Rarely does a giant wheel speed up at a constant rate. Hence we can say that the angular acceleration in a giant wheel is constantly changing.
Have you ever wondered why helicopter blades are so long and far away from the center point? Let’s explore why. The center point is called the pivot. This is the point from where the force is applied to move the blades in rotational motion. Now, the farther the force is applied from the pivot, the greater is the angular acceleration.
When the angular acceleration is greater, it then becomes easier for the helicopter to lift itself up in the air and fly at a safe distance above the earth, fighting the gravitational pull exerted by the earth.
This is the reason behind having large blades in helicopters. To provide greater angular acceleration and fly easily.
Test Angular Acceleration
There are two experiments illustrated here for students to understand angular acceleration.
Experiment 1:
Take a wire or a thin rope. Now hold near one of its ends and try rotating it. The wire or rope will not rotate freely and you will face difficulty in rotating it. Therefore, there will be little to no angular acceleration in this type of motion.
Experiment 2:
Take the same wire or rope. Now, instead of holding it near its end, hold it in the middle or closer to the other end. And rotate it. You will observe that the object (wire or thin rope) moves easily in rotational motion and can have greater angular acceleration.
Tips to study Angular Acceleration
Angular acceleration is one of the most exciting chapters in physics and can help you understand the mechanism behind the working of various objects in daily life. However, to understand the topic in-depth, here are a few steps you should take.
Read this article till you understand all the concepts stated here very well.
If students face difficulty in understanding something, you can ask us in the comments. You can also download Vedantu’s app and get a more clear understanding of topics.
Practice problems. Do not directly jump to tough problems. Start with simple questions of angular acceleration and gradually move to difficult conceptual questions.
Create your own revision notes after understanding this topic well. It will help you remember and write well in exams.
Solved Questions Angular Acceleration Formula
1) If the Body's Angular Velocity in Rotational Motion Changes from \[\frac{\Pi }{2}\] rad/s to \[\frac{3\Pi }{4}\] in 0.4 sec. Find the Angular Acceleration.
Solution:
ω 1 = \[\frac{\Pi }{2}\] rad/s, ω2 = \[\frac{3\Pi }{4}\] rad/s, ∆t = 0.4
so by applying the formula for angular acceleration
α= ∆ω/∆t = ω1 – ω2/∆t = π/4-3π/4 / 0.4 = \[\frac{5\Pi }{8}\] rad/s²
2) The Angular Displacement of an Object in Rotational Motion is Usually Considered to be Depending on the Time t According to the Following Relation-
θ=2π t³ - -rt² + 3π - - 6, where θ is in rad and time in sec. Find angular acceleration at time t=2 sec.
Applying one of the formulas for angular acceleration angular velocity,
ω=dθ/dt = 6π t² - -2 π t + 3π rad
α=dω/dt = 12π t - -2π rad/s²
At t=2 sec, αt= 2s, 12π х 2 -2π = 22π rad/s²
The above examples are better to understand the angular acceleration. In two dimensions, angular acceleration is a pseudoscalar whose sign is taken to be positive if the angular speed increases counterclockwise or decreases clockwise. It is to be taken negatively if the angular speed increases clockwise and decreases counterclockwise.
In three dimensions, angular acceleration is a pseudovector.
In the case of rigid bodies, angular acceleration must be caused by a net external torque. This is not the same in the case of non-rigid bodies.
3. The radius of the tire of a car is about 0.35 meters. The car accelerates onto a straight path from rest at 2.8 m/s². Find the angular acceleration, both magnitude, and direction, of the front passenger-side tire?
Solution:
The angular acceleration is related to linear acceleration by the following equation.
α=a/r
In this case, a = 2.8 m/s²
and r = 0.35 meters.
Substituting these quantities into the equation, we get
α = a/r
= 2.8/0.35
= 8.0 rad/s²
By the use of the right-hand rule, we find that the direction of the angular acceleration is to the left side of the car when facing the direction in which the car is moving.
FAQs on Angular Acceleration Formula
1. What is the relation of Angular Acceleration with the Torque?
The net Torque that is present on a point particle is called the pseudovector
Torque= r х F
Where F is to be considered as the net force on the particle.
Torque is the rotational analogue of force. It induces a change in the rotational state of a system, just like force induces a change in the system's translational state. The net force on a particle may be connected to the acceleration of the particle by the equation "F = m a." Similarly, the net Torque on a particle is connected to the angular acceleration of the particle. Hence
Substituting F=ma in the above equation, we get
Ƭ = m (r х a) = mr² (r х a / r²)
From the previous section, it was derived that
α = r х a / r² – 2/r dr/dt ω
where α is the orbital angular acceleration of the particle and ω is the particle's orbital angular velocity. Therefore,
Ƭ = mr² (α + 2/r dr/DT ω)
= mr² α + 2mr dr/DT ω
In the special case where there the distance r of the particle from the origin does not change with time, the second term goes out, and the equation simplifies as
Ƭ = mr² α which is “rotational analogue” of F = ma.
2. What is Angular Acceleration in Three-dimensions? Is it Different from Two-dimensions?
The angular acceleration in three dimensions may not be associated with a change in angular speed. If the particle's position vector twists in space, then even if the angular speed is constant, there will still be a nonzero angular acceleration because there is a continuous change in the direction of the angular velocity vector with time. The same is not in two-dimensions because the position vector is restricted to a fixed plane so that the change occurs only through a change in its magnitude. Let us understand the example,
The instantaneous angular velocity vector ω at any point in time is given by
ω= r х v / r² where r is the particle’s position vector and v is the velocity vector
therefore, the orbital angular acceleration is the vector α is defined by
α= d/dt (r х v / r²)
Expansion done in this derivative by the usage of the product rule for cross-products and the ordinary quotient rule,
\[\alpha\] = \[\frac{1}{r^{2}}\] \[\left ( r\times \frac{dv}{dt}+\frac{dr}{dt}\times v \right )\] - \[\frac{2}{r^{3}}\] \[\frac{dr}{dt}\] \[\left ( r\times v \right )\]
= \[\frac{1}{r^{2}}\] (r х a + v х v) – \[\frac{2}{r^{3}}\] \[\frac{dr}{dt}\] (r х v)
= r х \[\frac{a}{r^{2}}\] – \[\frac{2}{r^{3}}\] \[\frac{dr}{dt}\] (r х v)
As ‘r х v’ is r²ω, the second term becomes – 2/r dr/dt ω. In the case where the distance r does not changes with time, the second term vanishes and the formula gets simplified as
α = r х a/r² and the cross-radial acceleration can be recovered as
a˔ = α х r.
3. What is the relationship between angular acceleration and linear acceleration?
Angular acceleration and Linear acceleration are directly proportional to each other. This implies that the greater the angular acceleration is, the bigger would be the linear (tangential) acceleration and the opposite. As an example we can conclude, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car.
4. What is the difference in the angular velocity of the second hand of radius 1cm on a wristwatch, when compared to the second hand of radius 5m on a large clock tower?
There is no difference in their angular velocities. The angular velocity should not change depending on the radius of the second hand. No matter what size the second hand, it will always cover one revolution every minute or 60s.
The linear velocity will be greater and the angular momentum will also be greater for the clocktower, however, its angular velocity will be the same. This can be understood by looking at the equation for angular velocity:
ω=θ/t