Circular Motion: Angular Kinematics

Angular displacement, angular velocity and angular acceleration describe rotation exactly like their linear cousins — and v = ωr, a_c = ω²r connect the two worlds. Learn the framework JEE uses for every rotation problem.

Circular Motion1 min · Free lecture

In this lesson

  • Define angular displacement, velocity and acceleration and their units
  • Apply v = ωr and split acceleration into radial and tangential parts
  • Convert between rpm, rad/s and period/frequency fluently

For a particle moving in a circle of radius rr, the angle swept from a reference line is the angular displacement θ\theta (radians). Its rate of change is angular velocity ω=dθ/dt\omega = d\theta/dt, and the rate of change of that is angular acceleration α=dω/dt\alpha = d\omega/dt. Every 1D kinematics equation transfers directly with the replacements s→θs \to \theta, u→ω0u \to \omega_0, v→ωv \to \omega, a→αa \to \alpha.

ω=dθdt,α=dωdt,v=ωr,at=αr\omega = \frac{d\theta}{dt}, \qquad \alpha = \frac{d\omega}{dt}, \qquad v = \omega r, \qquad a_t = \alpha r

Angular quantities and their connection to linear motion

  • \theta — angular displacement (rad)
  • \omega — angular velocity (rad/s)
  • \alpha — angular acceleration (rad/s^2)
  • r — radius of the circle (m)
ac=v2r=ω2r,a=at2+ac2a_c = \frac{v^2}{r} = \omega^2 r, \qquad a = \sqrt{a_t^2 + a_c^2}

Centripetal (radial) acceleration and the total acceleration

  • a_c — centripetal acceleration (towards centre) (m/s^2)
  • a_t — tangential acceleration (along motion) (m/s^2)
In uniform circular motion, speed is constant but acceleration is NOT zero — the direction of velocity keeps changing. Centripetal acceleration always points toward the centre, never along the tangent.

Frequency and period give the fastest route in most numericals: one full revolution is 2π2\pi radians, so ω=2πf=2π/T\omega = 2\pi f = 2\pi / T. Rotational speed is often quoted in rpm (revolutions per minute): ω=2π×rpm/60\omega = 2\pi \times \text{rpm} / 60.

ω=2πf=2πT,ω=2π⋅rpm60\omega = 2\pi f = \frac{2\pi}{T}, \qquad \omega = \frac{2\pi \cdot \text{rpm}}{60}

Angular velocity from frequency, period or rpm

  • f — frequency (Hz)
  • T — time period (s)
  • \text{rpm} — revolutions per minute

Worked example

A wheel rotates at 300 rpm. Find (a) its angular velocity in rad/s, (b) the linear speed and centripetal acceleration of a point 0.20 m from the axis.

  1. (a) ω = 2π × 300 / 60 = 10π ≈ 31.4 rad/s
  2. (b) v = ωr = 31.4 × 0.20 ≈ 6.28 m/s
  3. a_c = ω²r = (31.4)² × 0.20 ≈ 197 m/s²

Answer: ω ≈ 31.4 rad/s; v ≈ 6.28 m/s; a_c ≈ 197 m/s²

Transcript (1 min)
WEBVTT

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Circular Motion: Angular Kinematics
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ω = dθ/dt · α = dω/dt
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v = ωr · a_t = αr · a_c = ω²r
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Uniform circular motion: a_c ≠ 0, always toward the centre
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ω = 2π × rpm / 60
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Example: 300 rpm → ω ≈ 31.4 rad/s, v ≈ 6.28 m/s, a_c ≈ 197 m/s²
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Recap: θ–ω–α mirror s–v–a · v = ωr · a_c = ω²r
Printable notes (free)

Practice

3 free · 1 premium
  1. Q1 · Numerical · difficulty 1/5

    A wheel rotates at 270 rpm. What is its angular velocity in rad/s?

    Hint available · solution with Premium.

  2. Q2 · Numerical · difficulty 1/5

    A point is at distance 0.8 m from the axis of a wheel rotating with angular velocity 7 rad/s. Find its linear speed.

    Hint available · solution with Premium.

  3. Q3 · Numerical · difficulty 2/5

    A particle moves on a circle of radius 4.5 m with constant speed 26 m/s. Find the magnitude of its centripetal acceleration.

    Hint available · solution with Premium.

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