recall and use ω = 2π / T and v = rω

Kinematics of Uniform Circular Motion

Learning Objectives

• Define the radian and use it to express angular displacement.
• Recall and apply the fundamental relationships ω = 2π ⁄ T = 2π f and v = r ω.
• Convert confidently between revolutions per minute (rpm), hertz (Hz) and angular speed (rad s⁻¹).
• State the vector nature of angular speed and use the right‑hand rule.
• Derive and use the expressions for centripetal acceleration and force:
   a_c = r ω² = v²⁄r and F_c = m r ω² = m v²⁄r.

---

1. Radian, Angular Displacement & Angular Speed

• Radian (rad): the angle subtended at the centre of a circle by an arc whose length equals the radius. \[ \theta\;(\text{rad})=\frac{s}{r} \] where s is the arc length and r the radius.
• Angular displacement (Δθ) – measured in radians (or multiples of 2π for complete revolutions).
• Period (T) – time for one complete revolution (s).
• Frequency (f) – number of revolutions per second, \(f=\dfrac{1}{T}\) (Hz).
• Angular speed (ω) – rate of change of angular displacement, \[ \omega=\frac{\Delta\theta}{\Delta t}\;(\text{rad s}^{-1}) \] For uniform circular motion a full revolution gives Δθ = 2π rad, so \[ \boxed{\omega=\frac{2\pi}{T}=2\pi f} \]

---

2. Vector Nature of Angular Speed

Although many Cambridge 9702 problems treat ω as a scalar, it is fundamentally a vector directed along the axis of rotation. Its direction is given by the right‑hand rule:

**Key points**

• Magnitude of ω is the scalar used in the formulae \(v=r\omega\) and \(a_c=r\omega^2\).
• Direction follows the right‑hand rule: curl the fingers in the sense of rotation; the thumb points along ω.

---

3. Linear (Tangential) Speed from Angular Speed

The distance travelled along the circumference in one period is the circumference, \(2\pi r\). Hence

\[ v=\frac{2\pi r}{T} \]

Substituting the definition of ω gives the compact relationship

\[ \boxed{v=r\omega} \]

Thus, for a fixed radius, linear speed is directly proportional to angular speed.

---

4. Conversions – rpm ↔ rad s⁻¹ ↔ Hz

Quantity	Symbol	Conversion
Revolutions per minute	n (rpm)	\(\displaystyle \omega\;(\text{rad s}^{-1})=\frac{2\pi n}{60}\)
Angular speed	\(\omega\) (rad s⁻¹)	\(\displaystyle f\;(\text{Hz})=\frac{\omega}{2\pi}\)
Frequency	f (Hz)	\(\displaystyle n\;(\text{rpm})=60f\)

Numerical reminders:

• 1 rpm = \(2\pi/60\) rad s⁻¹ ≈ 0.1047 rad s⁻¹.
• 1 Hz = \(2\pi\) rad s⁻¹ ≈ 6.283 rad s⁻¹.

Worked conversion example

Problem: A motor runs at 1800 rpm. Find (a) the angular speed ω in rad s⁻¹ and (b) the linear speed of a point 0.05 m from the shaft.

Solution:

1. Convert rpm to ω: \[ \omega=\frac{2\pi\times1800}{60}=2\pi\times30=60\pi\;\text{rad s}^{-1}\approx 188.5\;\text{rad s}^{-1}. \]
2. Linear speed: \[ v=r\omega=0.05\times188.5\approx 9.4\;\text{m s}^{-1}. \]

---

5. Centripetal Acceleration and Force

Uniform circular motion requires a continuous change of direction, producing a centripetal (inward) acceleration.

\[ \boxed{a_c=r\omega^{2}=\frac{v^{2}}{r}} \]

The corresponding net inward force on a mass m is

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

These expressions are used in the next syllabus sub‑topic (12.2 – Forces in circular motion).

---

6. Worked Examples

1. Car on a circular track

   Radius \(r=50\;\text{m}\), period \(T=20\;\text{s}\).

   • Angular speed: \(\displaystyle \omega=\frac{2\pi}{T}=0.314\;\text{rad s}^{-1}\).
   • Linear speed: \(\displaystyle v=r\omega=50\times0.314=15.7\;\text{m s}^{-1}\).
2. Stone on a string

   Mass \(m=0.20\;\text{kg}\), radius \(r=0.75\;\text{m}\), angular speed \(\omega=8.0\;\text{rad s}^{-1}\).

   • Centripetal acceleration: \(a_c=r\omega^{2}=0.75\times8^{2}=48\;\text{m s}^{-2}\).
   • Tension (centripetal force): \(F_c=m a_c=0.20\times48=9.6\;\text{N}\).
3. Turntable (conversion chain)

   Turntable rotates at 120 rpm; point of interest is 0.15 m from the centre.

   • ω = \(\dfrac{2\pi\times120}{60}=4\pi\;\text{rad s}^{-1}\approx12.57\;\text{rad s}^{-1}\).
   • Linear speed: \(v=r\omega=0.15\times12.57\approx1.89\;\text{m s}^{-1}\).

---

7. Common Mistakes to Avoid

• Confusing angular speed (rad s⁻¹) with frequency (Hz). Remember \(\omega=2\pi f\).
• Using the diameter instead of the radius in \(v=r\omega\) or \(a_c=r\omega^{2}\).
• Omitting the factor \(2\pi\) when converting between period and angular speed.
• For rpm problems, forgetting to divide by 60 to convert minutes to seconds.
• Thinking “centripetal force” is a new type of force; it is simply the net inward force required for circular motion.

---

8. Summary Table of Key Relationships

Quantity	Symbol	Formula	Units
Radian	θ	\(\displaystyle \theta=\frac{s}{r}\)	rad
Period	T	Time for one revolution	s
Frequency	f	\(\displaystyle f=\frac{1}{T}\)	Hz (s⁻¹)
Angular speed	ω	\(\displaystyle \omega=\frac{2\pi}{T}=2\pi f\)	rad s⁻¹
Linear (tangential) speed	v	\(\displaystyle v=r\omega=\frac{2\pi r}{T}\)	m s⁻¹
Centripetal acceleration	a_c	\(\displaystyle a_c=r\omega^{2}= \frac{v^{2}}{r}\)	m s⁻²
Centripetal force	F_c	\(\displaystyle F_c=m r\omega^{2}= \frac{m v^{2}}{r}\)	N

---

9. Diagram – Geometry of Circular Motion

Blue line = radius r; orange line = tangential speed v; green arc = angular displacement Δθ.

---

10. Quick Revision Questions

1. A satellite orbits Earth at a radius of \(7.0\times10^{6}\ \text{m}\) with an angular speed of \(1.1\times10^{-3}\ \text{rad s}^{-1}\).
   Calculate its orbital period T. Answer: \(T=\dfrac{2\pi}{\omega}= \dfrac{2\pi}{1.1\times10^{-3}}\approx5.7\times10^{3}\ \text{s}\) (≈ 1.6 h).
2. A turntable rotates at \(120\ \text{rpm}\).
   (a) Convert this to angular speed in rad s⁻¹.
   (b) Find the linear speed of a point \(0.15\ \text{m}\) from the centre. Solution: (a) \(\omega=\dfrac{2\pi\times120}{60}=4\pi\ \text{rad s}^{-1}\approx12.57\ \text{rad s}^{-1}\). (b) \(v=r\omega=0.15\times12.57\approx1.89\ \text{m s}^{-1}\).
3. A stone of mass \(0.25\ \text{kg}\) is whirled in a horizontal circle of radius \(0.60\ \text{m}\) at a speed of \(5.0\ \text{m s}^{-1}\).
   Determine the required centripetal force. Answer: \(F_c=\dfrac{m v^{2}}{r}= \dfrac{0.25\times5^{2}}{0.60}=10.4\ \text{N}\).