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# Angular Acceleration

When an object is in motion then its velocity is changed continuously then it is called in acceleration state. We know that acceleration is the change in velocity of moving object with respect to time. When an object is moved on circular path then its velocity is the angular velocity. The angular velocity is in the perpendicular direction of the rotational plane and it is related to the change in the angular speed and measure with unit as radians per second or revolutions per second.

The angular acceleration shows the change in the angular velocity with time.
It is not required to be this rotational acceleration in the same direction of angular velocity; for example, if a car is rolling on highway with increasing its speed then the direction of angular acceleration is on the left side of the axis of the wheel and it becomes disappears when the car stops and maintained its constant velocity. When it slows down then the acceleration is in the reverse direction. It also helps to give relation between circular motion and curve motion.

Here, we are discussing about the angular acceleration and its mathematical formula, and relation with torque.

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## Angular Acceleration Definition

We know that the rate of change of displacement with respect to time is called velocity of the object. But if there is a change in the velocity while the body is going in uniform circular motion, what can we call it? how can we Calculate it?

The rate of change of angular Velocity with respect to time. It is a vector quantity.

It can be represented as:
$\alpha$ = $\frac{\mathrm{d}\omega}{\mathrm{d}t}$ = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$.

$\alpha$ = $\frac{a_{t}}{r}$
$a_t = \alpha r$
where,
$\omega$ is the angular Velocity
at is the linear tangential acceleration
r is the radius of Circular path

## Angular Acceleration Formula

If the angular velocity is constant, then algebraically it is defined as,
$\omega$ = $\frac{\theta}{t}$
where,
$\omega$ is angular velocity
$\theta$ is the angle rotated
‘t’ is the time taken

It may be realized when the angular velocity is constant, the angular acceleration is 0. If the velocity is not constant, then $\alpha$ is defined as

$\alpha$ = $\frac{\omega}{t}$ = $\frac{\theta}{t^{2}}$

If $\alpha$ is not constant and varies from time to time, then we can only refer to average angular acceleration and instantaneous acceleration.

In such a case formulas are,
$\alpha$av = $\frac{(\omega_{2} – \omega_{1})}{(t_{2} – t_{1})}$ and $\alpha$i = $\frac{\mathrm{d}\omega}{dt}$ = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$.These equations help us in finding angular acceleration.

## Angular Acceleration Units

As per the definition, an angular acceleration is the amount of angle covered per square of time. Hence dimensionally its unit must be the ratio of angle to the square of time. Then, in common terms, the angular acceleration unit is degrees/square of time like s2, min2, hr2. But in internationally accepted unit (SI unit), the unit of angle is radian (rad) and the unit of time is second (s). Hence the standard unit of Angular acceleration is rad/s2.
Hence it is more sensible to define the displacement in terms of the angle it has rotated. If the object completes one full rotation it comes back to the initial position on the circle, meaning the displacement is 0. But the angular displacement is 360o degrees or 2$\pi$ radians. Therefore, in Circular motions, the rate of change of angle is defined as angular velocity and the rate of change of angular velocity is defined as ‘Angular acceleration’.

## Average Angular Acceleration

Angular acceleration is the rate of change of angular velocity with respect to time. Suppose the body has moved some distance then if we consider initial and final point where the displacement gets an end, then its average angular acceleration is given by

Average angular acceleration,

$\alpha_{av} = \frac{\omega _{f} -\omega _{i}}{t_{f} - t_{i}}$
where,
ωf = final angular velocity
ωi = initial angular velocity
tf = final interval of time or the end of time
ti = initial interval of time
It has the same unit as the angular acceleration, i.e., rad/s2.

## Torque and Angular Acceleration

In case of linear motion, as per Newton’s second law a force is required to accelerate object. That force is defined as the product of the mass of the object and the acceleration created.

In case of circular motion the force that is required to impart angular acceleration is called ‘Torque’. In other words, torque is an angular force and it is denoted by the Greek letter $\tau$ (pronounced as ‘tau’).
Also in rotational motion the moment of inertia I of the object plays the role of mass.

The torque in a circular motion is defined as,
$\tau$ = I $\alpha$

## Constant Angular acceleration

If an object undergoes rotational motion about a fixed axis under a constant angular acceleration $\alpha$, its motion can be described with the following set of equations,

$\omega$ - $\omega_{0}$ = $\alpha$ t.
$\theta$ - $\theta_{0}$ = $\omega_{0}$ t + $\frac{1}{2}$ $\alpha$ t2
$\omega ^{2}$ = $\omega_{0}^{2}$ + 2 $\alpha$ ($\theta$ - $\theta_{0}$)

where $\omega_{0}$ = angular speed of the rigid body at time t=0
$\omega$ = angular speed of the rigid body at time t
$\alpha$ = angular acceleration.

## Angular Acceleration to Linear Acceleration

As mentioned earlier, the definition of average angular velocity is the change in angle $\theta$ with respect to time t.
$\omega$ = $\frac{\theta}{t}$
where,
$\theta$ is the angle rotated in the time t

Now we will bring in the linear velocity v and the radius of the circle r, in a circular motion.

Let $\iota$ be the actual distance moved by the object along the circumference. As per the geometry of circles,
$\theta$ = $\frac{\iota}{r}$,
Therefore, $\omega$ = $\frac{\theta}{t}$
= $\frac{\iota}{rt}$ = $\frac{\iota}{t}\frac{1}{r}$.

Now if we say, v = $\frac{\iota}{t}$
we can write
$\omega$ = $\frac{v}{r}$ and hence,
$\alpha$ = $\frac{\omega}{t}$
where v is the linear velocity of the object at any point and its direction is along the tangent at that point.

Now considering an infinitesimal study,
$\omega_{i}$ = $\frac{\mathrm{d}\theta}{dt}$ = $\frac{\mathrm{d} \iota}{\mathrm{d}t} \frac{1}{r}$ = $\frac{\mathrm{d}v}{\mathrm{r}}$
and hence,
$\alpha_{i}$ = $\frac{\mathrm{d}v}{\mathrm{d}t} \frac{1}{r}$
It may be noted that the angular acceleration is limited to only the change in angular velocity.

### Solved Examples

Question 1: If an object changes it angular velocity from 10 rad/s to 25 rad/s in 3 seconds. Calculate angular acceleration?
Solution:

if the angular velocity of an object is changed from 10 rad/s to 25 rad/s in 3 seconds, to calculate angular acceleration the formula for average angular acceleration must be used.
That is, $\alpha_{av}$ = $\frac{\omega^{2} – \omega^{1}}{(t^{2} – t^{1})}$
= $\frac{(25 - 10)}{(3 - 0)}$ rad/s$^{2}$
= $\frac{15}{3}$ rad/s$^{2}$
= 5 rad/s$^{2}$.

Question 2: An object is rotating according to the function θ(t) = t$^{3}$ + t, find the instantaneous acceleration when t = 2 seconds?
Solution:

$\alpha$i = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$
= 6t
Hence when t = 2 seconds, $\alpha_{i}$ = 12 rad/s$^{2}$.

## Finding Angular Acceleration

Using the following formulas, we can find the angular acceleration.
$\alpha_{av}$ = $\frac{(\omega_{2} – \omega_{1})}{(t_{2} – t_{1})}$
$\omega$ = $\frac{\theta}{t}$ = $\frac{v}{r}$

and hence $\alpha$ = $\frac{v}{r^2}$
$\omega_{i}$ = $\frac{\mathrm{d} \iota}{\mathrm{d}t} \frac{1}{r}$ = $\frac{v}{r}$
$\alpha_{i}$ = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$ = $\frac{\mathrm{d} \omega}{\mathrm{d}t}$ = $(\frac{1}{r})\frac{\mathrm{d}v}{\mathrm{dt}}$.

Where,
$\iota$ = length of the actual distance moved by the object
r = radius of Circular path
t = time taken to complete the path
v = linear velocity

Some problems are given below to find the angular acceleration
Example 1: If the angular velocity of an object is changed from 10 rad/s to 25 rad/s in 3 seconds, calculate the average angular acceleration.
The formula for average acceleration,
$\alpha_{av}$ = $\frac{(\omega_{2} – \omega_{1})}{(t_{2} – t_{1})}$
= $\frac{(25– 10)}{(3 – 0)}$
rad/s2 = $\frac{15}{3}$rad/s2 = 5 rad/s2

Example 2: If an object is rotating according to the function ?(t) = t3 + t, find the instantaneous acceleration when t = 2 seconds.
The angle of rotation varies with the time which means the angular acceleration is not uniform. Hence we can only find the instantaneous angular acceleration at a given time. The formula for instantaneous angular acceleration is,
$\alpha_{i}$ = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$
$\alpha_{i}$ = $\frac{d^2(t^3 + t)}{dt^2}$
= 6t and hence when t = 2 seconds, ai = 12 rad/s2

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