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Worked Examples · Example 6.10

Q.Obtain Eq. (6.36), ω=ω0+αt\omega = \omega_0 + \alpha t, from first principles.

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Starting from the definition of angular acceleration as the rate of change of angular velocity, we integrate with respect to time, assuming constant angular acceleration, to derive the first equation of rotational motion: ω=ω0+αt\omega = \omega_0 + \alpha t.

Why This Derivation Matters

The equation ω=ω0+αt\omega = \omega_0 + \alpha t is the rotational analogue of v=u+atv = u + at in linear motion. It connects angular velocity, initial angular velocity, angular acceleration, and time — but only when angular acceleration α\alpha is constant. Understanding where it comes from, rather than just memorising it, builds the foundation for all rotational kinematics.

The key idea: angular acceleration is the rate of change of angular velocity. That's a definition, not a derived result. If you know how fast ω\omega is changing (α\alpha) and for how long (tt), you can find the new ω\omega.


Step-by-Step Derivation

1. Start with the definition of angular acceleration

Angular acceleration α\alpha is defined as the instantaneous rate of change of angular velocity ω\omega with respect to time:

α=dωdt\alpha = \frac{d\omega}{dt}

This is the rotational equivalent of a=dv/dta = dv/dt. It tells us: "At any instant, how rapidly is the angular velocity changing?"

2. Rearrange to separate variables

We want to find ω\omega as a function of time. Multiply both sides by dtdt:

dω=α dtd\omega = \alpha \, dt

This is a differential equation. It says: a small change in angular velocity equals the angular acceleration multiplied by the small time interval during which it acts.

3. Integrate both sides

We integrate from the initial state (time t=0t=0, angular velocity ω0\omega_0) to the final state (time tt, angular velocity ω\omega):

∫ω0ωdω=∫0tα dt\int_{\omega_0}^{\omega} d\omega = \int_{0}^{t} \alpha \, dt

The left side is straightforward: the integral of dωd\omega is just ω\omega evaluated between the limits.

4. Handle the right side — the crucial assumption

Here's where the assumption of constant angular acceleration enters. If α\alpha is constant, it can be taken outside the integral:

∫0tα dt=α∫0tdt=α[t]0t=αt\int_{0}^{t} \alpha \, dt = \alpha \int_{0}^{t} dt = \alpha \left[ t \right]_{0}^{t} = \alpha t

Watch out

If α\alpha is not constant (e.g., it depends on time or angular position), you cannot pull it out of the integral. The equation ω=ω0+αt\omega = \omega_0 + \alpha t only holds for constant angular acceleration. In problems where α\alpha varies, you must integrate α(t)\alpha(t) directly.

5. Equate the two sides

Putting the left and right sides together:

ω−ω0=αt\omega - \omega_0 = \alpha t

6. Rearrange to the standard form …

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