understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration

Uniform Circular Motion – Centripetal Acceleration & Centripetal Force

Learning Objectives (AO1–AO3)

• Define the radian, angular displacement, angular speed (ω) and period (T). Recognise the radian as the derived SI unit for plane angles.
• Show that a force of constant magnitude that is always perpendicular to the instantaneous direction of motion produces a radial (centripetal) acceleration.
• Derive and use the three equivalent forms of the centripetal‑acceleration and centripetal‑force equations.
• Apply Newton’s II law to uniform circular motion and identify the source of the required centripetal force in a range of contexts (tension, gravity, normal reaction, magnetic force, etc.).
• Plan, carry out and interpret a simple experiment that demonstrates the relationship between force, speed and radius (AO3).

1. Kinematics of Uniform Circular Motion

• Radian (rad): the angle subtended at the centre of a circle by an arc whose length equals the radius. 1 rad ≈ 57.3°; it is the derived SI unit for plane angles.
• Angular displacement (Δθ): measured in radians.
• Angular speed (ω):

  \$\omega = \frac{\Delta\theta}{\Delta t}\qquad\text{[rad s}^{-1}\text{]}\$

• Period (T) – time for one complete revolution:

  \$T = \frac{2\pi}{\omega}\qquad\text{or}\qquad\omega = \frac{2\pi}{T}\$

• Linear speed (v) and its relation to ω:

  \$v = r\omega\qquad\text{or}\qquad\omega = \frac{v}{r}\$

2. Centripetal Acceleration

When an object moves round a circle of radius r at constant speed v, the magnitude of its acceleration is

\$a_c = \frac{v^{2}}{r}=v\omega=\omega^{2}r\qquad\text{[m s}^{-2}\text{]}\$

The direction of a_c is always towards the centre of the circle (radial).

Derivation (vector‑diagram method)

1. At time t the velocity vector v₁ is tangent to the circle.
2. After a short interval Δt the velocity has rotated through an angle Δθ, becoming v₂. Their magnitudes are equal: |v₁| = |v₂| = v.
3. The change in velocity is the vector difference

   \$\Delta\mathbf{v}= \mathbf{v}2-\mathbf{v}1\$

   For small Δθ,

   \$|\Delta\mathbf{v}| = 2v\sin\frac{\Delta\theta}{2}\approx v\Delta\theta.\$

4. Average acceleration during Δt:

   \$\mathbf{a}_{\text{avg}}=\frac{\Delta\mathbf{v}}{\Delta t}\approx\frac{v\Delta\theta}{\Delta t}=v\omega.\$

5. Taking the limit Δt → 0 gives the instantaneous centripetal acceleration

   \$a_c = v\omega = \frac{v^{2}}{r}= \omega^{2}r.\$

3. Centripetal Force

Newton’s II law, \(\mathbf{F}=m\mathbf{a}\), requires the net force to be directed radially inward because the acceleration is radial. Hence

\$Fc = m ac = m\frac{v^{2}}{r}=m\omega^{2}r\qquad\text{[N]}\$

Why must the force be perpendicular to the motion?

• If a component of the force were parallel (or anti‑parallel) to the velocity, it would do work \((\mathbf{F}\!\cdot\!\mathbf{v}\neq0)\) and change the kinetic energy – the speed would no longer be constant.
• A force that is always perpendicular satisfies \(\mathbf{F}\!\cdot\!\mathbf{v}=0\); therefore it can change only the direction of the velocity, not its magnitude.

4. Sources of the Required Centripetal Force

5. Worked Examples

Example 1 – String Tension (horizontal circle)

Problem: A 0.50 kg stone is whirled in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m s⁻¹. Find the tension in the string.

Solution:

1. Required centripetal force: \(F_c = \dfrac{mv^{2}}{r}\).
2. Substitute: \(F_c = \dfrac{(0.50)(4.0)^{2}}{0.80}=10\;\text{N}\).
3. Since tension is the only horizontal force, \(T = 10\;\text{N}\).

Example 2 – Banked Curve (no friction)

Problem: A car of mass 800 kg travels round a banked curve of radius 50 m that is banked at \(\theta = 15^{\circ}\). Find the speed for which the car can negotiate the curve without relying on friction.

Solution:

1. Vertical equilibrium: \(N\cos\theta = mg \;\Rightarrow\; N = \dfrac{mg}{\cos\theta}\).
2. Horizontal component provides centripetal force: \(N\sin\theta = \dfrac{mv^{2}}{r}\).
3. Combine:

   \$\$\frac{mg}{\cos\theta}\sin\theta = \frac{mv^{2}}{r}

   \;\Rightarrow\;

   v = \sqrt{rg\tan\theta}.\$\$

4. Insert numbers:

   \$\$v = \sqrt{(50)(9.81)\tan15^{\circ}}

   \approx 11.5\;\text{m s}^{-1}.\$\$

Example 3 – Charged Particle in a Magnetic Field

Problem: A proton (\(m_p = 1.67\times10^{-27}\) kg, \(q = +1.60\times10^{-19}\) C) enters a uniform magnetic field of magnitude \(B = 0.25\) T travelling perpendicular to the field with speed \(v = 2.0\times10^{6}\) m s⁻¹. Find the radius of its circular path.

Solution:

1. Magnetic Lorentz force: \(F_B = qvB\) acts as the centripetal force.
2. Set \(qvB = \dfrac{mv^{2}}{r}\) → \(r = \dfrac{mv}{qB}\).
3. Calculate:

   \$\$r = \frac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.60\times10^{-19})(0.25)}

   \approx 8.3\times10^{-2}\;\text{m}.\$\$

Example 4 – Satellite in Circular Orbit

Problem: A satellite of mass 500 kg orbits the Earth at an altitude where the distance from the Earth’s centre is \(r = 7.0\times10^{6}\) m. Find the orbital speed.

Solution:

1. Gravitational force provides the centripetal force: \(\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}\).
2. Cancel m and solve for v:

   \$v = \sqrt{\frac{GM}{r}}.\$

   Using \(GM = 3.986\times10^{14}\;\text{m}^{3}\text{s}^{-2}\):

   \$\$v = \sqrt{\frac{3.986\times10^{14}}{7.0\times10^{6}}}

   \approx 7.5\times10^{3}\;\text{m s}^{-1}.\$\$

Example 5 – Vertical Circular Motion (pendulum at the bottom)

Problem: A 0.20 kg ball is attached to a light string of length 0.50 m and swings in a vertical circle. At the lowest point its speed is 6.0 m s⁻¹. Find the tension in the string at that instant.

Solution:

1. Required centripetal force: \(\dfrac{mv^{2}}{r} = \dfrac{(0.20)(6.0)^{2}}{0.50}=14.4\;\text{N}\).
2. Weight acts downward, so the tension must supply both the weight and the centripetal force:

   \$\$T = mg + \frac{mv^{2}}{r}

   = (0.20)(9.81) + 14.4 \approx 16.4\;\text{N}.\$\$

6. Practical Investigation – Demonstrating the Relationship \(F_c \propto v^{2}/r\)

Planning Checklist (AO3)

• Aim: To verify experimentally that the required centripetal force is directly proportional to the square of the speed and inversely proportional to the radius.
• Variables

  • Independent: speed (by varying the pulling force) and radius (by using different circular tracks).
  • Dependent: tension in the string (measured as the centripetal force).
  • Controlled: mass of the object, air currents, plane of motion (keep the motion as horizontal as possible), length of the string.

• Safety: Ensure the swinging mass is well‑secured, keep observers clear of the path, use a safety net or clear area.
• Apparatus: Light string, small mass (≈ 50 g), interchangeable circular tracks (different radii), spring balance (or force sensor), stopwatch, metre ruler, digital balance.

Method (outline)

1. Measure the mass \(m\) of the object with the digital balance.
2. Choose a track, measure its radius \(r\) from the centre to the point where the string is attached.
3. Attach the mass to the string, swing it in a horizontal circle, and use the spring balance to record the tension \(T\) (which equals the required centripetal force).
4. Time 20 complete revolutions with the stopwatch; compute the period \(T{\text{period}} = \dfrac{\text{time}}{20}\) and the linear speed \(v = \dfrac{2\pi r}{T{\text{period}}}\).
5. Repeat steps 2–4 for at least three different radii and for three different speeds (by changing the pulling force). Record all data in a table.
6. Analyse:

   • For each radius, plot \(T\) (or \(F_c\)) against \(v^{2}\); the slope should be \(m/r\).
   • For a fixed speed, plot \(T\) against \(1/r\); the slope should be \(mv^{2}\).

7. Discuss sources of error (air resistance, slight vertical motion, reaction time of the stopwatch, spring‑balance calibration) and suggest improvements.

7. Formula Summary

8. Conceptual Questions (AO1)

• Explain why a force that is always perpendicular to the velocity does no work on the object.
• How does the required centripetal force change if the speed is doubled while the radius remains the same?
• For a satellite in a circular orbit, why can we replace the gravitational force by the expression \(F_c = mv^{2}/r\)?
• Describe how the normal reaction on a banked curve provides the centripetal force when the coefficient of friction is zero.
• In vertical circular motion, why is the tension greatest at the bottom of the path and smallest (or zero) at the top?
• What would happen to the speed of a mass on a string if the tension were suddenly reduced while the mass was at the bottom of a vertical circle?

9. Suggested Diagram

10. Summary

A constant‑magnitude force that remains perpendicular to the instantaneous direction of motion produces a radial (centripetal) acceleration given by

\$a_c = \frac{v^{2}}{r}=v\omega=\omega^{2}r.\$

Newton’s II law then gives the corresponding centripetal force

\$Fc = m ac = \frac{mv^{2}}{r}=m\omega^{2}r.\$

The source of this force varies with the situation – tension, gravity, normal reaction, magnetic Lorentz force, or a combination of forces – but the underlying relationship between force, speed and radius is universal. Mastery of these concepts enables you to analyse any uniform circular‑motion problem on the Cambridge IGCSE/A‑Level physics syllabus, from simple string‑whirling experiments to satellites in orbit.