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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.

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 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 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.

1) If the Body's Angular Velocity in Rotational Motion Changes from π/2 rad/s to 3π/4 in 0.4 sec. Find the Angular Acceleration.

Solution:

ω 1 = π/2 rad/s, ω2 = 3π/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 = 5π/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.

FAQ (Frequently Asked Questions)

Q1: What is the Relation of Angular Acceleration with the Torque?

Ans: 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 analog 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.

Q2: What is Angular Acceleration in Three-dimensions? Is it Different from Two-dimensions?

Ans: 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,

α= 1/r² (r х dv/dt + dr/dt х v) – 2/r³ dr/dt (r х v)

= 1/r² (r х a + v х v) – 2/r³ dr/dt (r х v)

= r х a/r² – 2/r³ 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.