Angular position graphs

θ–t slope = ω; ω–t slope = α; ω–t area = Δθ — rotation mirrors translation graph laws.

Angular position graphs0 min · Free lecture

In this lesson

  • θ–t slope = ω; ω–t slope = α; ω–t area = Δθ — rotation mirrors translation graph laws.

Every graph law from linear kinematics works here with θ, ω, α.

θ–t: slope ω. ω–t: slope α, area Δθ. α–t: area Δω. Read signs the same way.

A constant-slope θ–t line = uniform rotation; a horizontal ω–t = zero α.

ω=dθdt,α=dωdt\omega = \frac{d\theta}{dt},\quad \alpha = \frac{d\omega}{dt}

Angular slopes

Worked example

In uniform circular motion, the acceleration of the particle is directed:

  1. Speed is constant but velocity direction changes continuously.
  2. The change in velocity points toward the centre → centripetal acceleration a_c = v²/r toward the centre.

Answer: Toward the centre of the circle

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Angular position graphs — FemtoLearn.
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θ–t slope = ω; ω–t slope = α; ω–t area = Δθ — rotation mirrors translation graph laws.
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Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 1/5

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

  2. Q2 · Numerical · difficulty 1/5

    A point is at distance 1.7 m from the axis of a wheel rotating with angular velocity 11 rad/s. Find its linear speed.

  3. Q3 · Numerical · difficulty 2/5

    A particle moves on a circle of radius 1.5 m with constant speed 6 m/s. Find the magnitude of its centripetal acceleration.

  4. Q4 · MCQ · difficulty 1/5

    In uniform circular motion, the acceleration of the particle is directed:

    • Along the tangent to the circle
    • Toward the centre of the circle
    • Away from the centre of the circle
    • It is zero