Welcome to Uniform Circular Motion!
Have you ever ridden a rollercoaster that loops upside down, spun a ball on a string, or felt pushed sideways as a car turns a tight corner? All of these everyday experiences are governed by the physics of circular motion.
In this chapter, we will explore what happens when an object moves in a circular path at a steady speed. Don't worry if this seems a little counter-intuitive at first — by breaking it down step-by-step with clear formulas and relatable examples, you will master this topic in no time!
1. Measuring Angles: The Radian
In everyday life, we measure angles in degrees (\(360^\circ\) in a full circle). However, in A Level Physics, it is much easier to use radians (\(\text{rad}\)).
What is a Radian?
One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.
The relationship between angle in radians (\(\theta\)), arc length (\(s\)), and radius (\(r\)) is given by:
\(\theta = \frac{s}{r}\)
Where:
• \(\theta\) = angular displacement in radians (\(\text{rad}\))
• \(s\) = arc length in metres (\(\text{m}\))
• \(r\) = radius of the circle in metres (\(\text{m}\))
Converting Between Degrees and Radians
Since the circumference of a full circle is \(s = 2\pi r\), the total angle in a full circle is:
\(\theta = \frac{2\pi r}{r} = 2\pi \text{ rad}\)
This gives us the fundamental conversion:
\(360^\circ = 2\pi \text{ rad}\) \(\implies\) \(180^\circ = \pi \text{ rad}\)
• To convert from degrees to radians: multiply by \(\frac{\pi}{180^\circ}\)
• To convert from radians to degrees: multiply by \(\frac{180^\circ}{\pi}\)
Example: Convert \(90^\circ\) to radians:
\(\theta = 90^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{2} \text{ rad} \approx 1.57 \text{ rad}\)
Key Takeaway
Always remember: Check that your calculator is in RAD mode whenever you are calculating circular motion problems!
2. Angular Displacement and Angular Velocity
Angular Velocity (\(\omega\))
Angular velocity (symbol: \(\omega\), Greek letter omega) is defined as the rate of change of angular displacement with respect to time.
\(\omega = \frac{\Delta \theta}{\Delta t}\)
Where:
• \(\omega\) = angular velocity in radians per second (\(\text{rad s}^{-1}\))
• \(\Delta \theta\) = angular displacement in radians (\(\text{rad}\))
• \(\Delta t\) = time taken in seconds (\(\text{s}\))
Connecting Angular Velocity to Period and Frequency
For an object completing one full revolution (\(\Delta \theta = 2\pi \text{ rad}\)):
• The time taken for one full revolution is the period (\(T\)) in seconds.
• The number of revolutions per second is the frequency (\(f\)) in Hertz (\(\text{Hz}\)), where \(f = \frac{1}{T}\).
Therefore, we can write:
\(\omega = \frac{2\pi}{T} = 2\pi f\)
Linear Speed (\(v\)) vs. Angular Velocity (\(\omega\))
Imagine two people riding a playground carousel: one sits near the centre, and one sits near the outer edge. Both complete one full turn in the exact same time, so they share the same angular velocity (\(\omega\)). However, the person on the outer edge travels a much greater distance in that time, meaning their linear speed (\(v\)) is larger!
The relationship connecting linear speed to angular velocity is:
\(v = r\omega\)
Where:
• \(v\) = linear (tangential) speed in metres per second (\(\text{m s}^{-1}\))
• \(r\) = radius of the circular path in metres (\(\text{m}\))
• \(\omega\) = angular velocity in radians per second (\(\text{rad s}^{-1}\))
Key Takeaway
Linear speed depends on how far you are from the centre (\(v = r\omega\)), but angular velocity (\(\omega\)) is the same everywhere on a rigid spinning body.
3. Centripetal Acceleration
The Big Question: Why is there acceleration at constant speed?
This is one of the most common stumbling blocks in physics! Remember that velocity is a vector quantity (it has both magnitude and direction).
• In uniform circular motion, the speed (magnitude) is constant.
• However, the direction of motion is constantly changing at every single instant.
• A change in direction means a change in velocity.
• By definition, a change in velocity over time means the object is accelerating!
Direction and Magnitude of Centripetal Acceleration
This acceleration is directed perpendicular to the velocity, towards the centre of the circle. It is called centripetal acceleration (\(a\)).
The equations for centripetal acceleration are:
\(a = \frac{v^2}{r}\)
Substituting \(v = r\omega\), we also get:
\(a = r\omega^2 = v\omega\)
Where:
• \(a\) = centripetal acceleration in metres per second squared (\(\text{m s}^{-2}\))
• \(v\) = linear speed in metres per second (\(\text{m s}^{-1}\))
• \(r\) = radius of the path in metres (\(\text{m}\))
• \(\omega\) = angular velocity in radians per second (\(\text{rad s}^{-1}\))
Key Takeaway
Even if an object moves at constant speed in a circle, it is always accelerating towards the centre because its direction is continuously changing.
4. Centripetal Force
What is Centripetal Force?
According to Newton's Second Law of Motion (\(F = ma\)), an acceleration must be caused by a resultant force acting in the same direction.
The centripetal force (\(F\)) is the resultant force acting on an object moving in a circle, directed towards the centre of the circle.
Formulas for Centripetal Force
Using \(F = ma\) and our acceleration formulas:
\(F = \frac{mv^2}{r}\)
Or in terms of angular velocity:
\(F = mr\omega^2\)
Where:
• \(F\) = centripetal force in Newtons (\(\text{N}\))
• \(m\) = mass of the moving object in kilograms (\(\text{kg}\))
• \(v\) = linear speed in metres per second (\(\text{m s}^{-1}\))
• \(r\) = radius of the circle in metres (\(\text{m}\))
• \(\omega\) = angular velocity in radians per second (\(\text{rad s}^{-1}\))
Crucial Point: Centripetal force is NOT a "new" type of force!
Centripetal force is simply the role played by real physical forces that you already know. In any problem, ask yourself: What physical force is pulling or pushing towards the centre?
• Whirling a bung on a string: Provided by the tension in the string.
• A car cornering on a level road: Provided by the friction between tyres and the road surface.
• A satellite or planet orbiting: Provided by the gravitational force.
• An electron orbiting a nucleus: Provided by the electrostatic force.
Did You Know?
The term "centrifugal force" (a force pushing outward) is a fictitious force felt only from inside the rotating frame. In physics exam questions, always explain circular motion using the inward centripetal force!
5. Real-World Applications & Problem Solving
Scenario 1: Car Cornering on a Flat Road
When a car of mass \(m\) rounds a flat, horizontal bend of radius \(r\) at speed \(v\), the sideways friction (\(F_{\text{friction}}\)) between the tyres and the road must provide the necessary centripetal force:
\(F_{\text{friction}} = \frac{mv^2}{r}\)
• If the car goes too fast (\(v\) is too large) or the corner is too sharp (\(r\) is too small), the required force exceeds the maximum available friction, and the car will skid outwards in a straight line.
Scenario 2: Whirling an Object in a Vertical Circle
Motion in a vertical circle involves both the tension (\(T\)) in the string and the weight (\(mg\)) of the object. The resultant force towards the centre provides the centripetal force \(\frac{mv^2}{r}\).
1. At the Highest Point (Top):
Both tension and weight point downwards towards the centre:
\(T_{\text{top}} + mg = \frac{mv^2}{r}\)
\(\implies T_{\text{top}} = \frac{mv^2}{r} - mg\)
Condition to complete the loop: The string stays taut as long as \(T_{\text{top}} \ge 0\). The minimum speed occurs when \(T_{\text{top}} = 0\), so \(mg = \frac{mv^2}{r} \implies v_{\text{min}} = \sqrt{gr}\).
2. At the Lowest Point (Bottom):
Tension pulls upwards towards the centre, while weight pulls downwards away from the centre:
\(T_{\text{bottom}} - mg = \frac{mv^2}{r}\)
\(\implies T_{\text{bottom}} = \frac{mv^2}{r} + mg\)
Notice: Tension is always greatest at the bottom of the swing! This is why a string is most likely to snap at the very bottom of a vertical loop.
Key Takeaway
In a vertical circle, the tension varies continuously: it is at its minimum at the top (\(T = \frac{mv^2}{r} - mg\)) and at its maximum at the bottom (\(T = \frac{mv^2}{r} + mg\)).
6. Summary of Key Formulas & Common Mistakes
Formula Reference Table
• Angle in Radians: \(\theta = \frac{s}{r}\)
• Angular Velocity: \(\omega = \frac{\theta}{t} = \frac{2\pi}{T} = 2\pi f\)
• Linear Speed: \(v = r\omega\)
• Centripetal Acceleration: \(a = \frac{v^2}{r} = r\omega^2\)
• Centripetal Force: \(F = \frac{mv^2}{r} = mr\omega^2\)
Common Pitfalls to Avoid in Exams
1. Forgetting to convert units: Always ensure radius \(r\) is in metres (\(\text{m}\)), period \(T\) is in seconds (\(\text{s}\)), and mass \(m\) is in kilograms (\(\text{kg}\)).
2. Mixing up \(\omega\) and \(v\): Remember that \(\omega\) is measured in \(\text{rad s}^{-1}\), while \(v\) is measured in \(\text{m s}^{-1}\).
3. Incorrect Free-Body Diagrams: Never draw "centripetal force" as an extra independent force vector. Draw only the real forces (such as tension, friction, or gravity) and identify which combination points towards the centre.