Angular acceleration is the time rate of change in the angular velocity. It is a pseudovector in 3 dimensions. In the SI unit, we measure it in radians/second squared (rad/\(s^{2}\)). Moreover, we denote it usually by the Greek letter alpha \(\alpha\). Learn the angularÂ acceleration formula here.
Types of Angular Acceleration

Spin Angular Acceleration

Orbital Angular Acceleration
These two represent the time rate of change of spin angular velocity and the orbital angular velocity respectively. Unlike linear acceleration, angular acceleration need not be caused by the next external torque. For example, a figure skater is able to speed up his/her rotation (thereby obtains an angular acceleration) very easily by contracting his/her arms inwards, and this does not involve any personal torque.
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Derivation of Angular Acceleration
While the body is performing a nonuniform circular motion, then its angular velocity changes. Hence, the body possesses an angular acceleration.
\(\alpha = \frac{d\omega }{dt}\)
We know that the acceleration is the rate of change in the velocity with respect to time.
i.e. a = \(\frac{dv}{dt}\)
âˆ´Â a = \(\frac{d}{dt}\left ( r\omegaÂ \right )\)Â âˆµ v = r\(\omega\)
Where \(\omega\) = angular velocity of the particle that performs UCM. Now radius of the circular path will be constant.
This is the relation between the angular acceleration and the linear acceleration in the UCM.
Where â€˜aâ€™ is the angular acceleration. Hence, linear acceleration = radius \(\times\) acceleration.
If the speed increases linear acceleration is in a similar direction as that of the linear velocity. If the speed decreases linear acceleration is in the direction that Is opposite to that of the linear velocity. We also call it as tangential acceleration.
For the UCM, \(\alpha\) = 0.
Solved ExamplesÂ onÂ Angular Acceleration Formula
Question:
When we switch our roomâ€™s fan from medium speed to high speed, the blades accelerate at 1.2 radians per second squared for 1.5 seconds. If the initial angular speed of the blades of fun is 3.0 radians per second, what is the final angular speed of the fan blades in radians per second?
Solution:
4.8 rad/s
Angular Acceleration is defined by:
\(\alpha\) = \(\frac{\Delta \omega }{\Delta t}\)
Where
\(\Delta\)\(\Omega\), is the angular speed â€“ initial angular speed, and
\(\Delta t\), is the time over which the angular speed starts changing. We are aware of the initial angular speed of the blades of the fan, so we can write:
\(\Delta \omega\) = \(\omega_{i}\) Where \(\omega_{i}\) = 3.0 radians per second.
Solve the equation for the acceleration for the final angular speed and plug it in the quantities that we know for getting the answer. The result is:
\(\alpha = \frac{\Delta \omega}{\Delta t}\)
\(\alpha \Delta t = \omega _{f} – \omega _{i}\)
\(\omega _{f} = \alpha \Delta t + \omega_{i}\)
1.2rad/\(s^{2}\) \(\times\) 1.5s + 3.0 rad/s
= 4.8 rad/s.
Question:
The radius of the tire of a car is about 0.35 meters. If the car is accelerating into a line that is straight from rest at 2.8 meters per second squared, what is the angular acceleration, both magnitude, and direction, of the front passengerside tire?
Solution:
8.0 rad/\(s^{2}\) to left
The angular acceleration has a relation the linear acceleration by
\(\alpha = \frac{a}{r}\)
In this case, (\alpha\) = 2.8 meters/second squared and r = 0.35 meters.
Plug these quantities into the equation:
\(\alpha = \frac{a}{r}\)
= \(\frac{2.8m/s^{2}}{0.35m}\)
= 8.0 rad/\(s^{2}\)
By the use of the righthand 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.
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