Angular Acceleration Formula: Equation & Examples

Ans. Given: Initial angular speed (ω₀) = 1200 rpm

Final angular speed (ω) = 3120 rpm

Time (t) = 16 secs

We can use the formula

ω = ω₀ + αt

Where α is the angular acceleration

ω₀ = 1200 rpm

ω₀ = 2π × angular speed in rev/s

ω₀ = 2π × angular speed in rev/min60 s/min

ω₀ = 2π x 1200/60rad/s

ω₀ = 40π rad/s

Similarly,

ω₀ = 2π x 3120/60rad/s

ω₀ = 10 4π rad/s

Rearranging the formula ω = ω₀ + αt we get

α = ω - ω₀t

α =104π - 40π 16rad/s²

α = 4π rad/s²

∴ The angular acceleration of the motor wheel = 4π rad/s².

Ans. Given:

Mass of the flywheel (m) = 20 kg

Radius of the flywheel (r) = 20 cm = 0.2 m

Force acting on the flywheel (F) = 25 N

It is known that

I x α = τ…(1)

Where,

I is the moment of inertia of the flywheel about its axis

α is the angular acceleration

τ is the torque acting on the flywheel

The torque is given by the formula 

τ = F x r

τ = 25 N × 0.2 m

τ = 5.0 Nm...(2)

The moment of inertia is given by the formula

I = mr²2

I = 20 kg x 0.2 m x 0.2 m²

I = 0.4 kgm²...(3)

Substituting values obtained in (2) and (3) in equation (1)

0.4 kgm² x α = 5.0 Nm

α = 12.5 s^-2

∴ The angular acceleration of the motor wheel is 12.5 s^-2.

Ans. We are given,

Change in angular velocity of the wheel (dω) = 60 rad/s

Change in time (dt) = 10 secs

We know that 

α = dωdt

Substituting the given values in the above formula, we get:

α = 60/10rad/s² 

α = 6 rad/s²

∴ The angular acceleration of the rotating wheel is 6 rad/s².

Ans. We are given:

Angular acceleration of the blades of fan (α) = 1.2 rad/s²

The initial angular speed of the blades of the fan (ω₀) = 3.0 rad/s

Change in time (t) = 1.5 secs

We know that

ω = ω₀ + αt

Where ω is the final angular speed of the blades of a fan

Substituting given values in the above equation, we get:

ω = 3.0 rad/s + (1.2 rad/s² x 1.5 secs)

ω = 4.8 rad/s

∴ The final angular speed of the fan is 4.8 rad/s.

Ans. It is given:

The radius of the tire of the car (r) = 0.35 m

Linear acceleration of the car (a) = 2.8 m/s²

It is known that

α = a x r

Where α is the angular acceleration

Substituting given values in the above formula

α = 2.8 m/s² x 0.35 m

α = 8.0 rad/s²

∴ The angular acceleration of the tire of the car is 8.0 rad/s².

• This is the magnitude of the angular acceleration. To find the direction of angular acceleration, the right-hand thumb rule should be applied.
• By applying the right-hand thumb rule, it is found that the direction of angular acceleration when the car is moving in a particular direction lies towards the left-hand side of the car.
• Thus, the magnitude of angular acceleration is 8.0 rad/s² and the direction is towards the left.

Ans. It is given that

Initial angular velocity of the body in rotational motion (ω0) =π2 rad/s

Final angular velocity of the body in rotational motion (ω) = 3π4rad/s

Time (t) = 0.4 s

It is known that 

α = dωt

α = ω₀ - ωt

α = π2 rad/s - 3π4rad/s⁰.4

α = 5π8 rad/s².

∴ The angular acceleration of the body in rotational motion is 5π8 rad/s².

Ans. It is given:

Angle of rotation (θ) = 2πt₃ − πt₂ +3π − 6 rad

Time (t) = 2 secs

It is known as

ω = dθdt

Where ω is angular velocity.

ω = d(2πt₃ − πt₂ +3π − 6 )dt

ω = 6πt₂ − 2πt + 3π rad/s

So the angular velocity of the object is 6πt₂ − 2πt + 3π rad/s

It is also known that

α = dωdt

Where α is the angular acceleration

α = d(6πt₂ − 2πt + 3π)dt

α = 12πt −2π rad/s²

So the angular acceleration of the object is 12πt −2π rad/s²

αt at t = 2 secs is calculated as

αt = 12π(2) −2π

αt = 24π −2π

αt = 22π rad/s²

∴ The angular acceleration of the object in rotational motion at t = 2 secs is 22π rad/s².

Ans. It is given:

Angular acceleration of a merry-go-round (α) = 0.30 rad/s²

Time (dt) = 2.8 secs

Initial angular speed of the merry-go-round (ω₀) = 0 rad/s

It is known that

θ = ω₀t + 12αt₂

θ = [0 rad/s x 2.8 secs] + [12x 0.30 rad/s² x (2.8 secs)² ]

θ = 1.2 rad

∴ The angle at which the merry-go-round rotates is 1.2 rad.

Ans. Angular acceleration is defined as the rate of change of angular velocity per unit time. The Greek symbol α is used to denote angular acceleration. Angular acceleration is also called rotational acceleration. Angular acceleration is associated with rotational motion. It is analogous to the linear acceleration of an object having a linear motion. The SI unit of angular acceleration is radians per second squared i.e. rad/s².

Ans. There are two types of angular acceleration.

• Orbital angular acceleration
• Spin angular acceleration

Spin angular acceleration is associated with the change of spin angular velocity whereas orbital angular acceleration is associated with the change of orbital angular velocity.

Ans. dω = 100 and dt = 10

Using the formula: α = dω/dt
= angular velocity = 100/10
= angular velocity = 10 rad/s2