Describe, qualitatively, motion in a circular path due to a force perpendicular to the motion as: (a) speed increases if force increases, with mass and radius constant (b) radius decreases if force increases, with mass and speed constant (c) an incre

1.5.1 Effects of Forces – Circular Motion

Learning objective

Explain, qualitatively, how an object moving in a circular path responds when a force acts perpendicular to its instantaneous direction of motion. In particular, describe the three situations required by the Cambridge IGCSE Physics (0625) syllabus:

1. How the speed changes when the centripetal force is increased while the mass and radius are kept constant.
2. How the radius changes when the centripetal force is increased while the mass and speed are kept constant.
3. Why a larger mass requires a larger centripetal force to maintain the same speed and radius.

Key concepts

• Centripetal force \(F_c\) – a real force directed toward the centre of the circular path that continuously changes the direction of the velocity vector.
• Centripetal acceleration \(a_c\) – the inward acceleration required for circular motion; magnitude

  \[

  a_c=\frac{v^{2}}{r}

  \]

  (direction: radially inward).

• From Newton’s 2nd law \(F=ma\) we obtain the fundamental relation

  \[

  Fc = m\,ac = \frac{m v^{2}}{r}.

  \]

• The force is always perpendicular to the instantaneous velocity, so it does not change the speed unless the magnitude of the force itself is altered.
• Terminology

  • Centripetal – “centre‑seeking”, a real inward force.
  • Centrifugal – an apparent outward force felt only in a rotating (non‑inertial) frame; it is not a real force on the object.

Derivation (reminder)

Uniform circular motion assumes constant speed \(v\) and a constant radius \(r\). The change in direction of the velocity vector produces an inward acceleration:

\(a_c = \dfrac{v^{2}}{r}\)

Applying Newton’s 2nd law (\(F = ma\)) gives the required centripetal force:

\(Fc = m ac = \dfrac{m v^{2}}{r}\)

This equation is the basis for the three qualitative cases below.

Qualitative cases (with numerical illustration)

(a) Increasing the force while keeping mass and radius constant

• Re‑arrange the formula for speed:

  \[

  v = \sqrt{\frac{F_c\,r}{m}}.

  \]

• Because \(r\) and \(m\) are fixed, a larger \(F_c\) makes the square‑root term larger → speed increases.
• Numerical example:

  \(m = 0.20\;\text{kg},\; r = 0.50\;\text{m}\).

  If \(F_c\) is raised from \(2.0\;\text{N}\) to \(4.0\;\text{N}\):

  \[

  v_1 = \sqrt{\frac{2.0\times0.5}{0.20}} = 2.2\;\text{m s}^{-1},\qquad

  v_2 = \sqrt{\frac{4.0\times0.5}{0.20}} = 3.2\;\text{m s}^{-1}.

  \]

  The speed rises by about 45 %.

• Interpretation: a stronger inward pull makes the object travel faster around the same circular track.

(b) Increasing the force while keeping mass and speed constant

• Solve the basic relation for radius:

  \[

  r = \frac{m v^{2}}{F_c}.

  \]

• With \(m\) and \(v\) fixed, a larger \(F_c\) gives a smaller \(r\) → the path becomes tighter.
• Numerical example:

  \(m = 0.20\;\text{kg},\; v = 3.0\;\text{m s}^{-1}\).

  Increase \(F_c\) from \(1.8\;\text{N}\) to \(3.6\;\text{N}\):

  \[

  r_1 = \frac{0.20\times3.0^{2}}{1.8}=1.0\;\text{m},\qquad

  r_2 = \frac{0.20\times3.0^{2}}{3.6}=0.5\;\text{m}.

  \]

  The radius halves.

• Interpretation: the same speed is retained, but the stronger inward pull forces the object onto a smaller circular track.

(c) Increasing the mass while keeping speed and radius constant

• From the basic equation:

  \[

  F_c = \frac{m v^{2}}{r}.

  \]

• Because \(F_c\) is directly proportional to \(m\), a heavier object needs a larger centripetal force to stay in the same circular motion.
• Numerical example:

  \(v = 2.5\;\text{m s}^{-1},\; r = 0.40\;\text{m}\).

  For \(m = 0.10\;\text{kg}\): \(F_c = 0.10\times2.5^{2}/0.40 = 1.56\;\text{N}\).

  For \(m = 0.20\;\text{kg}\): \(F_c = 3.13\;\text{N}\). The required force doubles.

• Interpretation: more inertia (greater mass) must be continually redirected, so a larger inward pull is necessary.

Momentum and energy links

• Momentum: Momentum \(\mathbf{p}=m\mathbf{v}\) is a vector. Even when the speed is constant, the direction of \(\mathbf{v}\) (and therefore \(\mathbf{p}\)) changes continuously, which means a net force must act (Newton’s 1st law). The inward force provides the required change in momentum.
• Work‑energy relation (relevant to case a): when the speed increases, the kinetic energy \(KE=\tfrac12 m v^{2}\) also increases. The extra centripetal force does work over the arc length \(s = v\,t\):

  \[

  W = F_c \, s = \Delta KE.

  \]

  This explains why a larger force can increase the speed – it supplies the additional energy.

Summary table

Exam‑style practice questions

1. Short answer: Explain why a car tyre that is pulled tighter (smaller radius) must experience a larger tension force if it continues to travel at the same speed.
2. Multiple‑choice: In a horizontal circular motion experiment the mass of the object is kept constant while the speed is doubled. Which of the following statements is true?
   

   A. The required centripetal force is unchanged.
   

   B. The required centripetal force is doubled.
   

   C. The required centripetal force is quadrupled.
   

   D. The required centripetal force is halved.
   

   Answer: C

3. Structured response: A 0.5 kg stone is whirled in a horizontal circle of radius 0.8 m at a constant speed of 4 m s⁻¹.
   

   (i) Calculate the required centripetal force.
   

   (ii) If the stone’s speed is increased to 6 m s⁻¹ while the radius is unchanged, state how the required force changes and give the new value.

Practical demonstration (classroom activity)

1. Apparatus: Light wooden block (≈ 200 g), string (≈ 0.60 m), smooth tabletop, stopwatch, ruler or measuring tape.
2. Procedure

   • Secure one end of the string and swing the block in a horizontal circle at a comfortable speed. Measure the radius (length of the string) and record the time for 10 revolutions; compute the speed \(v = 2\pi r / T\).
   • Increase the speed by swinging faster. Observe that the tension in the string (the centripetal force) increases. If you shorten the string, the radius decreases while the speed is kept roughly the same – illustrating case (b).
   • Replace the block with a heavier one (add a small weight) and repeat the experiment at the same speed and radius. Feel the greater pull required on the string – illustrating case (c).

3. Safety considerations

   • Clear the area of people and fragile objects.
   • Never swing the block near faces or loose clothing.
   • Keep the string taut at all times to avoid sudden jerks.

Real‑world examples

• Planetary orbits: Gravity supplies the centripetal force. A more massive planet needs a stronger gravitational pull to stay in the same orbit.
• Roller‑coaster loops: The track provides the normal force. Tighter loops (smaller radius) require larger normal forces to keep the car on the track.
• Satellite manoeuvres: To raise a satellite’s speed at a fixed altitude, the thrusters must produce a larger inward thrust.

Connections to other syllabus sections

• Newton’s 2nd law: The centripetal‑force equation is a direct application of \(F = ma\) with \(a = v^{2}/r\).
• Momentum: Continuous change of the direction of the momentum vector explains why a net force is required even when speed is constant.
• Kinetic energy: When speed changes (case a) the kinetic energy changes, linking to the work‑energy theorem.
• Forces and fields: In planetary motion the centripetal force is provided by a gravitational field; in a string‑whirl experiment it is the tension force.