Non-uniform circular motion: forces

Forces must provide BOTH components: ΣF_radial = mv²/r and ΣF_tangential = ma_t.

Non-uniform circular motion: forces0 min · Free lecture

In this lesson

  • Forces must provide BOTH components: ΣF_radial = mv²/r and ΣF_tangential = ma_t.

The tension in a string does double duty: it bends the path AND it can speed the mass up.

Resolve forces radially and tangentially. The radial equation always reads ΣF = mv²/r; the tangential equation reads ΣF = m dv/dt.

In vertical circles, gravity contributes to the radial part at every point — the classic source of the speed-dependent tension.

ΣFrad=mv2r,ΣFtan=mdvdt\Sigma F_{rad} = \frac{mv^2}{r},\quad \Sigma F_{tan} = m\frac{dv}{dt}

Force balance in non-uniform circular motion

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

Transcript (0 min)
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Non-uniform circular motion: forces — FemtoLearn.
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Forces must provide BOTH components: ΣF_radial = mv²/r and ΣF_tangential = ma_t.
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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