27  Circular Motion

27.1 Syllabus inquiry question

  • How is the force on an object related to its motion in a circular path?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 21:

In circular motion at constant speed, the velocity is always changing—not in magnitude, but in direction. This change requires a force, and that force must point toward the center.

27.2 Learning Objectives

  • Explain centripetal acceleration and centripetal force.
  • Apply circular motion equations to various situations.
  • Analyse forces on objects in circular motion.
  • Investigate the relationship between energy, work, and circular motion.
  • Apply torque concepts to rotational systems.

27.3 Content

27.3.1 Uniform Circular Motion

In uniform circular motion, an object moves in a circle at constant speed. Although the speed is constant, the velocity is changing because its direction changes continuously.

Key Principle

Any change in velocity requires acceleration. In circular motion, this acceleration points toward the center and is called centripetal acceleration.

27.3.2 Interactive: Circular Motion Visualisation

27.3.3 Centripetal Acceleration

The centripetal acceleration always points toward the center of the circle:

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

where: - \(a_c\) = centripetal acceleration (m/s²) - \(v\) = speed (m/s) - \(r\) = radius of circle (m)

Alternative Form

Using \(v = 2\pi r/T\), centripetal acceleration can also be written as: \[a_c = \frac{4\pi^2 r}{T^2}\]

27.3.4 Speed and Period

For an object completing one revolution:

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

where \(T\) is the period (time for one complete revolution).

Frequency (\(f\)) is the number of revolutions per second: \[f = \frac{1}{T}\]

27.3.5 Centripetal Force

By Newton’s second law, centripetal acceleration requires a centripetal force:

\[F_c = ma_c = \frac{mv^2}{r}\]

Critical Understanding

Centripetal force is NOT a new type of force. It is the name for the net inward force that causes circular motion. It could be: - Tension (mass on a string) - Friction (car on a bend) - Gravity (satellites) - Normal force (banked tracks)

27.3.6 Interactive: Forces in Circular Motion

27.3.7 Angular Velocity

Angular velocity (\(\omega\)) measures how fast an angle changes:

\[\omega = \frac{\Delta\theta}{t}\]

Relationship to linear speed: \[v = \omega r\]

Units

Angular velocity is measured in radians per second (rad/s). To convert from degrees: multiply by \(\pi/180\).

27.3.8 Applications: Cars on Bends

Flat bend: Friction provides the centripetal force.

\[\mu_s mg \geq \frac{mv^2}{r}\]

Maximum safe speed: \(v_{max} = \sqrt{\mu_s gr}\)

Banked bend: Normal force component provides centripetal force.

For ideal banking (no friction needed): \[\tan(\theta) = \frac{v^2}{gr}\]

27.3.9 Interactive: Car on Banked Track

27.3.10 Energy and Work in Circular Motion

In uniform circular motion: - Centripetal force is perpendicular to velocity - No work is done by the centripetal force - Kinetic energy remains constant

\[W = F \cdot d \cdot \cos(90°) = 0\]

27.3.11 Torque

Torque measures the effectiveness of a force in causing rotation:

\[\tau = r_\perp F = rF\sin(\theta)\]

where: - \(\tau\) = torque (N·m) - \(r\) = distance from pivot to force - \(F\) = applied force - \(\theta\) = angle between \(r\) and \(F\)

Maximum Torque

Torque is maximum when force is perpendicular to the lever arm (\(\theta = 90°\)).

27.4 Worked Examples

27.4.1 Example 1: Centripetal acceleration

A car travels at \(15\ \text{m/s}\) around a bend of radius \(30\ \text{m}\). Find the centripetal acceleration.

Solution:

  1. Use \(a_c = v^2/r\)

  2. \(a_c = 15^2/30 = 225/30\)

  3. \(a_c = 7.5\ \text{m/s}^2\) toward the center

27.4.2 Example 2: Centripetal force

A \(0.80\ \text{kg}\) mass moves in a circle of radius \(2.0\ \text{m}\) at \(4.0\ \text{m/s}\). Find the centripetal force.

Solution:

  1. Use \(F_c = mv^2/r\)

  2. \(F_c = 0.80 \times 4.0^2 / 2.0 = 0.80 \times 16 / 2.0\)

  3. \(F_c = 6.4\ \text{N}\) toward the center

27.4.3 Example 3: Speed from period

A satellite orbits at radius \(7.0 \times 10^6\ \text{m}\) with period \(6000\ \text{s}\). Find its speed.

Solution:

  1. Use \(v = 2\pi r/T\)

  2. \(v = 2\pi \times 7.0 \times 10^6 / 6000\)

  3. \(v = 7.3 \times 10^3\ \text{m/s}\) (7.3 km/s)

27.4.4 Example 4: Torque calculation

A force of \(40\ \text{N}\) is applied at \(0.25\ \text{m}\) from a pivot at an angle of \(90°\). Find the torque.

Solution:

  1. Use \(\tau = rF\sin(\theta)\)

  2. \(\tau = 0.25 \times 40 \times \sin(90°) = 0.25 \times 40 \times 1\)

  3. \(\tau = 10\ \text{N·m}\)

27.4.5 Example 5: Maximum speed on a flat curve

A car can safely navigate a curve of radius \(50\ \text{m}\) if \(\mu_s = 0.60\). Find the maximum speed.

Solution:

  1. At maximum speed, friction equals required centripetal force: \(\mu_s mg = mv^2/r\)

  2. \(v_{max} = \sqrt{\mu_s gr} = \sqrt{0.60 \times 9.8 \times 50}\)

  3. \(v_{max} = \sqrt{294} = 17.1\ \text{m/s}\) (about 62 km/h)

27.5 Common Misconceptions

Common Misconceptions
  • Misconception: Centripetal force is a separate “extra” force. Correction: It’s the name for whatever net force points inward—tension, friction, gravity, etc.

  • Misconception: There’s an outward “centrifugal” force. Correction: In an inertial reference frame, there is no outward force. The sensation of being “pushed outward” is due to inertia.

  • Misconception: Acceleration is tangential in uniform circular motion. Correction: In uniform (constant speed) circular motion, acceleration is purely centripetal (toward center).

  • Misconception: Centripetal force does work. Correction: The force is perpendicular to motion, so it does no work. Kinetic energy stays constant.

  • Misconception: Torque is largest when force is parallel to the lever arm. Correction: Torque is maximum when force is perpendicular to the lever arm.

27.6 Practice Questions

27.6.1 Easy (2 marks)

An object moves in a circle of radius \(3.0\ \text{m}\) at \(6.0\ \text{m/s}\). Find the centripetal acceleration.

  • Use \(a_c = v^2/r\) correctly (1)
  • Correct answer: \(a_c = 36/3.0 = 12\ \text{m/s}^2\) toward center (1)

Answer: 12 m/s² toward the center

27.6.2 Medium (4 marks)

A \(500\ \text{g}\) mass on a string moves in a horizontal circle of radius \(0.80\ \text{m}\) with a period of \(0.50\ \text{s}\). Find the speed and centripetal force.

  • Speed: \(v = 2\pi r/T = 2\pi \times 0.80 / 0.50 = 10.1\ \text{m/s}\) (2)
  • Force: \(F_c = mv^2/r = 0.50 \times 10.1^2 / 0.80 = 63.7\ \text{N}\) (2)

Answer: \(v = 10\ \text{m/s}\), \(F_c = 64\ \text{N}\)

27.6.3 Hard (5 marks)

A car travels at \(25\ \text{m/s}\) around a banked curve of radius \(100\ \text{m}\). Find the ideal banking angle where no friction is needed.

  • Identify that \(\tan(\theta) = v^2/(gr)\) for ideal banking (2)
  • Substitute: \(\tan(\theta) = 25^2/(9.8 \times 100) = 625/980 = 0.638\) (2)
  • \(\theta = \tan^{-1}(0.638) = 32.5°\) (1)

Answer: \(\theta = 33°\)

27.7 Multiple Choice Questions

Test your understanding with these interactive questions:

27.8 Quick Quiz: Circular Motion

27.9 Extended Response Practice

27.10 Summary

Key Takeaways
  • Centripetal acceleration: \(a_c = v^2/r = 4\pi^2 r/T^2\)
  • Centripetal force: \(F_c = mv^2/r\) (net inward force, not a new force type)
  • Speed from period: \(v = 2\pi r/T\)
  • Angular velocity: \(\omega = \Delta\theta/t\), and \(v = \omega r\)
  • No work done by centripetal force (perpendicular to motion)
  • Torque: \(\tau = rF\sin(\theta)\)

27.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to: