Uniform circular motion and gravitation

Study Notes: Uniform Circular Motion & Gravitation

Hey there! Welcome to one of the most exciting topics in Physics. Ever wondered how planets stay in orbit around the Sun, how a roller coaster completes a vertical loop, or why astronauts float in the space station? The answers lie in understanding Uniform Circular Motion and Gravitation. In these notes, we will break down these key HKDSE concepts into clear, structured sections. Let's get started!

1. Going in Circles: Uniform Circular Motion (UCM)

Imagine a car driving on a circular track at a steady 50 km/h. It is moving in a circle, and its speed remains constant. This is the essence of uniform circular motion.

What is UCM? The Basics

Uniform Circular Motion (UCM) is the motion of an object travelling at a constant speed along a circular path.

Remember that velocity is a vector with both magnitude (speed) and direction. In UCM:

• The speed is constant.
• The direction of motion is continuously changing (tangent to the circle at every point).

Because the direction changes continuously, the velocity is continuously changing, which means the object is continuously accelerating.

Analogy: Think about walking around a roundabout at a constant walking pace. Your speed stays the same, but you are constantly turning inward to follow the curve. Your velocity changes because your direction changes.

Angular Quantities: \(\omega\), \(T\), and \(f\)

To describe rotational motion, we define circular motion quantities:

• Angular displacement (\(\theta\)): The angle swept out by the radius vector (in radians, rad).
• Angular velocity (\(\omega\)): The rate of change of angular displacement: \(\omega = \frac{\Delta \theta}{\Delta t}\) (in rad s⁻¹).
• Period (\(T\)): The time taken to complete one full revolution (in s): \(T = \frac{2\pi}{\omega} = \frac{2\pi r}{v}\).
• Frequency (\(f\)): The number of revolutions per second (in Hz): \(f = \frac{1}{T} = \frac{\omega}{2\pi}\).

The relationship between linear speed \(v\) and angular velocity \(\omega\) is given by:

\(v = \omega r\)

The Direction Changer: Centripetal Acceleration

An object in UCM experiences an acceleration directed towards the center of the circle, known as centripetal acceleration (\(a\)).

• Direction: Always directed towards the center of the circular path (perpendicular to linear velocity).
• Purpose: It changes the direction of velocity without changing its magnitude (speed).

Centripetal acceleration can be expressed in linear or angular forms:

\(a = \frac{v^2}{r} = \omega^2 r = v\omega\)

The Center-Seeking Force: Centripetal Force

By Newton's Second Law (\(F = ma\)), the net resultant force required to keep an object in circular motion is the centripetal force (\(F\)):

\(F = \frac{mv^2}{r} = m\omega^2 r\)

Crucial concept: Centripetal force is not a new fundamental force. It is simply the net real force (or resultant force component) pointing towards the center of the circle.

Common HKDSE Applications of Centripetal Force:

• Horizontal circular motion (ball on a string): Tension provides the centripetal force: \(T = \frac{mv^2}{r}\).
• Car rounding a flat bend: Static friction provides the centripetal force: \(f_s = \frac{mv^2}{r}\). Maximum safe speed before slipping is \(v_{\max} = \sqrt{\mu g r}\).
• Banked roads (angle \(\theta\) to horizontal without friction): The horizontal component of the normal reaction force provides centripetal force: \(N \sin\theta = \frac{mv^2}{r}\) and \(N \cos\theta = mg\), giving \(\tan\theta = \frac{v^2}{rg}\).
• Conical pendulum: The horizontal component of tension provides centripetal force: \(T \sin\theta = m\omega^2 r\) and \(T \cos\theta = mg\).
• Vertical circular motion (non-uniform): At the top of a loop, \(T_{\text{top}} + mg = \frac{mv^2}{r}\); at the bottom, \(T_{\text{bottom}} - mg = \frac{mv^2}{r}\).

Common Mistake Alert: The Phantom "Centrifugal Force"

Centrifugal force is not a real force acting on the object in an inertial frame. What an occupant feels when turning is their own inertia—the natural tendency to continue in a straight line while the vehicle pushes them inward.

2. The Universal Pull: Gravitation

Isaac Newton established that gravity is a universal interaction that explains both falling bodies on Earth and planetary orbits.

Newton's Law of Universal Gravitation

Every point mass attracts every other point mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers:

\(F = \frac{GMm}{r^2}\)

Where:

• \(F\) is the gravitational force between the two masses (in N).
• \(G\) is the Universal Gravitational Constant (\(G \approx 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}\)).
• \(M\) and \(m\) are the masses of the interacting bodies (in kg).
• \(r\) is the center-to-center separation distance (in m).

Common Mistake Alert: 'r' is Center-to-Center!

For an object at an altitude \(h\) above a planet of radius \(R\), the total separation is \(r = R + h\). Never use just altitude \(h\) in Newton's formula!

Gravitational Field Strength (\(g\))

The gravitational field strength (\(g\)) at a point in space is defined as the gravitational force per unit mass experienced by a small test mass at that point:

\(g = \frac{F}{m} = \frac{GM}{r^2}\)

On Earth's surface (\(r = R\)), \(g \approx 9.81 \text{ N kg}^{-1}\) (or \(9.81 \text{ m s}^{-2}\)). Field strength decreases with the inverse square of distance as you move further from Earth's center.

3. Satellites in Orbit & Weightlessness

A satellite in a stable circular orbit around a planet has its required centripetal force supplied entirely by gravity.

Orbital Speed Formula

Setting centripetal force equal to gravitational force:

\(\frac{mv^2}{r} = \frac{GMm}{r^2}\)

Cancelling the satellite's mass \(m\) and solving for orbital speed \(v\):

\(v = \sqrt{\frac{GM}{r}}\)

Notice that orbital speed depends only on the central mass \(M\) and orbital radius \(r\), completely independent of the satellite's mass \(m\).

Orbital Period and Kepler's Third Law

Substitute \(v = \frac{2\pi r}{T}\) into the orbital speed equation:

\(\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \implies T^2 = \left(\frac{4\pi^2}{GM}\right) r^3\)

This proves Kepler's Third Law: the square of the orbital period is directly proportional to the cube of the orbital radius (\(T^2 \propto r^3\)).

Apparent Weightlessness

Why do astronauts inside an orbiting spacecraft float? They are not in zero gravity (at typical low Earth orbit altitudes, \(g\) is still about \(8.7 \text{ m s}^{-2}\)).

• Both the astronaut and the spacecraft experience the same gravitational acceleration towards Earth's center (\(a = g\)).
• They are in continuous free fall together.
• Because there is no relative acceleration between them, the normal contact force \(R\) exerted by the floor or scale on the astronaut is zero (\(R = 0\)). This condition is termed apparent weightlessness.