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Circular Motion

Source lecture(s): SC133 Lec 5

Intuition

A ball on a string moves in a circle at constant speed — yet it is accelerating the whole time, because its velocity's direction keeps changing. The acceleration points toward the center (centripetal, "center-seeking"): the string constantly pulls the velocity vector around without ever changing its length. No inward force, no circle — let go of the string and the ball flies off along the tangent, not outward.

The formulas

For speed \(v\) on a circle of radius \(r\):

\[\boxed{\,a_c = \frac{v^2}{r} = \omega^2 r\,} \qquad \text{directed toward the center}\]

with angular speed \(\omega = v/r\) (rad/s), period \(T = 2\pi r/v = 2\pi/\omega\).

The inward force required, by Newton's second law:

\[F_c = \frac{m v^2}{r}\]

Centripetal force is not a new force — it is a role played by whatever real force points inward: string tension, gravity (orbits), friction (cornering car), normal force (banked turn, loop-the-loop).

Derivation of \(a_c = v^2/r\)

In time \(\Delta t\) the position vector rotates by \(\Delta\theta = \omega\Delta t\); so does the velocity vector. For small angles the change in velocity has magnitude \(|\Delta \vec v| \approx v\,\Delta\theta\), pointing toward the center. Hence

\[a = \frac{|\Delta\vec v|}{\Delta t} = v\frac{\Delta\theta}{\Delta t} = v\omega = \frac{v^2}{r}.\]

The geometry of the velocity triangle mirrors the position triangle — that similarity is the entire proof.

Non-uniform circular motion

If the speed changes too, add a tangential component \(a_t = dv/dt\) along the motion; total acceleration \(a = \sqrt{a_c^2 + a_t^2}\). The centripetal part handles turning, the tangential part handles speeding up.

Worked example: how fast can a car corner?

Static friction supplies the centripetal force: \(\mu_s m g \geq mv^2/r\), so

\[v_\text{max} = \sqrt{\mu_s\, g\, r}\]

For \(\mu_s = 0.8\), \(r = 50\,\text{m}\): \(v_\text{max} = \sqrt{0.8\times9.8\times50} \approx 19.8\,\text{m/s} \approx 71\,\text{km/h}\). Independent of the car's mass — heavier cars need more force but also grip harder.

Common mistakes

  • Inventing "centrifugal force". In an inertial frame there is no outward force on the ball; the outward feeling is your body trying to go straight while the car turns under you.
  • Thinking constant speed means zero acceleration. Acceleration is change of velocity — direction counts.
  • Drawing \(F_c\) as an extra arrow on a free-body diagram. Label the actual forces (tension, gravity, friction, normal); their inward resultant is the centripetal force.

Knowledge graph position

Prerequisites: Kinematics, Vectors. Leads to: Rotation, Gravitation & orbits.

Quiz

Q1 (computational). A satellite orbits at \(r = 7000\,\text{km}\) with \(v = 7.5\,\text{km/s}\). Its centripetal acceleration?

Answer

\(a_c = v^2/r = (7500)^2 / 7\times10^6 \approx 8.0\,\text{m/s}^2\) — nearly \(g\)! Orbiting is falling; the satellite just keeps missing the ground.

Q2 (conceptual). A ball on a string is swung in a vertical circle. Where is the string tension largest?

Answer

At the bottom: tension must supply the centripetal force and fight gravity, \(T = mv^2/r + mg\) (and \(v\) is also largest there by energy conservation).

Q3 (multiple choice). The string breaks at the top of a horizontal circle. The ball initially flies: (a) radially outward (b) along the tangent (c) spirally

Answer

(b). Remove the force and Newton's first law takes over: straight-line motion along the instantaneous velocity — the tangent.