The unit of measure for angular acceleration in the International System is the radian per second squared. In this way, angular acceleration makes it possible to determine how angular velocity varies over time. Angular acceleration linked to uniformly accelerated circular motions is often studied.
On the Ferris wheel, angular acceleration is applied
Thus, in a uniformly accelerated circular motion, the angular acceleration value is constant. On the contrary, in a uniform circular motion, the angular acceleration value is zero. Angular acceleration is equivalent in circular motion to tangential or linear acceleration in straight motion.
In fact, its value is directly proportional to the value of the tangential acceleration. Thus, the greater the angular acceleration of a bicycle’s wheels, the greater the acceleration it experiences.
Therefore, angular acceleration is present in the wheels of a bicycle and in the wheels of any other vehicle, as long as there is a variation in the speed of rotation of the wheel.
Likewise, angular acceleration is also present in a Ferris wheel, as it experiences uniformly accelerated circular motion when it starts to move. Obviously, angular acceleration can also be found in a roundabout.
How to calculate angular acceleration?
In general, the instantaneous angular acceleration is defined from the following expression:
α = dω / dt
In this formula, ω is the angular velocity vector and t is the time.
The mean angular acceleration can also be calculated from the following expression:
α = ∆ω / ∆t
For the particular case of a plane motion, it turns out that the angular velocity and the angular acceleration are vectors with a direction perpendicular to the plane of motion.
On the other hand, the angular acceleration modulus can be calculated from the linear acceleration through the following expression:
α = a / R
In this formula, a is the tangential or linear acceleration; and R is the turning radius of the circular motion.
uniformly accelerated circular motion
As already mentioned above, angular acceleration is present in uniformly accelerated circular motion. For this reason, it is interesting to know the equations that govern this movement:
ω = ω + α ∙ t
θ = θ + ω ∙ t + 0.5 ∙ α ∙ t 2
ω 2 = ω 0 2 + 2 ∙ α ∙ (θ – θ )
In these expressions, θ is the angle traversed in the circular movement, θ is the starting angle, ω is the initial angular velocity and ω is the angular velocity.
Torque and Angular Acceleration
In the case of linear motion, according to Newton’s second law, a force is necessary for a body to acquire a certain acceleration. This force is the result of the multiplication of the body’s mass and the acceleration it has experienced.
However, in the case of a circular motion, the force required to transmit the angular acceleration is called torque. In summary, torque can be understood as an angular force. It is indicated by the Greek letter τ (pronounced “tau”).
Likewise, it must be taken into account that, in a rotational movement, the moment of inertia I of the body plays the role of the mass in the linear movement. In this way, the torque of a circular movement is calculated with the following expression:
τ = I α
In this expression, I am the moment of inertia of the body in relation to the axis of rotation.
Determine the instantaneous angular acceleration of a moving body subjected to a rotational motion, given its position in the rotation Θ (t) = 4 t 3 i. (The unit vector being in the x-axis direction).
Also, determine the value of the instantaneous angular acceleration when 10 seconds have elapsed since the start of motion.
The angular velocity expression can be obtained from the position expression:
ω (t) = d Θ / dt = 12 t 2 i (rad / s)
Once the instantaneous angular velocity has been calculated, the instantaneous angular acceleration can be calculated as a function of time.
α (t) = dω / dt = 24 ti (rad / s 2 )
To calculate the instantaneous angular acceleration value after 10 seconds, it is necessary to substitute the time value in the previous result.
α (10) = = 240 i (rad / s 2 )
Determine the mean angular acceleration of a body experiencing circular motion, knowing that its initial angular velocity was 40 rad / s and after 20 seconds it reached an angular velocity of 120 rad / s.
From the following expression, the average angular acceleration can be calculated:
α = ∆ω / ∆t
α = (ω f – ω) / (t f – t ) = (120 – 40) / 20 = 4 rad / s
What will be the angular acceleration of a giant wheel that starts moving with a uniformly accelerated circular motion until, after 10 seconds, it reaches an angular velocity of 3 revolutions per minute? What will be the tangential acceleration of the circular motion in that period of time? The radius of the Ferris wheel is 20 meters.
First, it is necessary to transform the angular velocity from revolutions per minute into radians per second. For this, the following transformation is performed:
ω f = 3 rpm = 3 ∙ (2 ∙ ∏) / 60 = ∏ / 10 rad / s
Once this transformation is performed, it is possible to calculate the angular acceleration, because:
ω = ω + α ∙ t
∏ / 10 = 0 + α ∙ 10
α = ∏ / 100 rad / s 2
And the tangential acceleration results from the operation of the following expression:
α = a / R
a = α ∙ R = 20 ∏ ∏ / 100 = ∏ / 5 m / s 2