← Mechanical Systems

3 Circular motion

Watch: Circular motion Opens a YouTube search: pick a video that clicks for you Angular velocity \omega=\frac{2\pi}{T}=2\pi f\qquad v=r\omega ω in rad s⁻¹; v is the tangential (linear) speed. Centripetal acceleration a_c=\frac{v^2}{r}=\omega^2 r Directed toward the centre. Centripetal force F_c=\frac{mv^2}{r}=m\omega^2 r The net inward force: provided by tension, gravity, friction, etc.

Key idea

  • Centripetal force is a net force, not a new kind of force: always identify what physically provides it.
  • Speed is constant in uniform circular motion, but velocity changes (direction), so there IS acceleration.
  • At the top of a vertical circle, minimum speed is when gravity alone supplies F_c: vmin = √(gr).
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Centripetal force is not an extra arrow on the diagram

Centripetal force is the REAL net force acting toward the centre — it is not a separate force alongside the ones already provided by tension, gravity, friction or the normal force. Drawing "Fc" as an additional arrow on a free-body diagram double-counts it. Identify which real force (or combination) IS providing the centripetal force, and use that. "Centrifugal force" is not a real force at all — it is the sensation of your own inertia in a rotating frame, and it should never appear on a diagram drawn from an outside (inertial) viewpoint.

Banked curves (no reliance on friction)

  • On a banked (tilted) curve, the NORMAL force itself has a horizontal component that can supply Fc — no friction needed at the "design speed".
  • Vertical: the normal force's vertical component balances gravity, N cos θ = mg.
  • Horizontal: the normal force's horizontal component supplies the centripetal force, N sin θ = mv²/r.
  • Dividing one equation by the other cancels N and m: tan θ = v²/(rg).
Worked example Banking angle A car track has a banked curve of radius 0.750 m, designed so no sideways friction is needed at 1.55 m s⁻¹. Find the banking angle. tan θ = v²/(rg). tan θ = 1.55² / (0.750 × 9.81) = 2.4025 / 7.3575 = 0.3266. θ = tan⁻¹(0.3266) = 18.1°. ✓ θ ≈ 18.1°.

Practise just this