Physics (9702) | Motion in a circle

Welcome to Motion in a Circle! (Topic 12)

Hello A-Level physicists! Get ready to explore a really satisfying topic: Motion in a Circle. This is where we combine kinematics (the study of motion) and dynamics (the study of forces) in a completely new way.

Why is this chapter crucial? Because circular motion is everywhere—from planets orbiting stars to the tiny rotors in a washing machine. The tricky part is realizing that even if an object moves at a constant speed in a circle (uniform circular motion), it is constantly accelerating. We will learn the physics behind this constant change in direction and calculate the force required to maintain it.

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12.1 Kinematics of Uniform Circular Motion

When an object is moving in a straight line, we use linear quantities like distance ($s$) and velocity ($v$). When dealing with circular motion, it is far more useful to describe the motion using angular quantities.

a) Defining the Radian and Angular Displacement

The standard SI unit for angles in circular motion is the radian (rad), not the degree.

Definition of a Radian:

A radian is the angle subtended at the centre of a circle when the length of the arc ($s$) along the circumference is equal to the radius ($r$) of the circle.

Key relationship: $$s = r\theta$$ (Where $\theta$ is the angular displacement, measured in radians.)

Quick Conversion Fact: $$2\pi \text{ radians} = 360^{\circ}$$ $$1 \text{ radian} \approx 57.3^{\circ}$$

b) Angular Speed (\(\omega\))

Angular speed (\(\omega\)) is the rate of change of angular displacement. It describes how quickly an object is rotating or revolving.

Definition: $$\omega = \frac{\text{change in angular displacement}}{\text{time taken}} = \frac{\Delta\theta}{\Delta t}$$

The unit for angular speed is radians per second ($\text{rad s}^{-1}$).

Relating Angular Speed to Period ($T$):

For one full revolution, the angular displacement is \(2\pi\) radians, and the time taken is the period ($T$).

Key Formula 1 (Syllabus Requirement): $$\omega = \frac{2\pi}{T}$$

Since the frequency ($f$) is $1/T$ (the number of rotations per second), we can also state: $$\omega = 2\pi f$$

c) Linking Linear Speed ($v$) and Angular Speed (\(\omega\))

Although every point on a rotating object has the same angular speed ($\omega$), points farther from the centre have to travel a greater distance in the same time, meaning they have a higher linear speed ($v$).

We use the radius ($r$) to connect these two types of speed.

Key Formula 2 (Syllabus Requirement): $$v = r\omega$$

Analogy: Think about a vinyl record or a spinning compact disc. Every point on the disc spins through the same angle at the same time ($\omega$ is constant), but the outside edge is moving much faster linearly ($v$) than a point near the spindle ($r$ is larger).

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12.2 Centripetal Acceleration and Force

a) The Centripetal Acceleration (\(a\))

Even if an object is moving in a circle at a constant speed (uniform circular motion), its velocity is constantly changing because velocity is a vector defined by both magnitude (speed) and direction. Since acceleration is the rate of change of velocity, there must be an acceleration!

Crucial Understanding (Syllabus Requirement 12.2/1):
The centripetal acceleration is caused by a force of constant magnitude that is always perpendicular to the direction of motion (the instantaneous velocity).

Direction: This acceleration is always directed inwards, towards the centre of the circle. This is why it is called centripetal (meaning "centre-seeking").

Centripetal Acceleration Formulas (Syllabus Requirement 12.2/3):

The magnitude of this acceleration is given by:

In terms of Linear Speed ($v$): $$a = \frac{v^2}{r}$$

In terms of Angular Speed ($\omega$): $$a = r\omega^2$$ (You can derive the second form by substituting $v = r\omega$ into the first equation.)

b) The Centripetal Force (\(F\))

According to Newton's Second Law ($F = ma$), if there is an acceleration ($a$), there must be a resultant force ($F$) causing it.

This resultant force required to keep an object moving in a circle is called the Centripetal Force.

• Direction: Always directed inwards, towards the centre of the circle.

Centripetal Force Formulas (Syllabus Requirement 12.2/4):

Using $F = ma$ and substituting the expressions for $a$:

In terms of Linear Speed ($v$): $$F = \frac{mv^2}{r}$$

In terms of Angular Speed ($\omega$): $$F = mr\omega^2$$

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12.3 Problem Solving: Identifying the Centripetal Force Provider

In any circular motion problem, the key to success is correctly identifying which actual force (or combination of forces) acts as the resultant centripetal force ($F_c$). Once you find $F_c$, you set it equal to $mv^2/r$ or $mr\omega^2$.

Example 1: The Car on a Flat Bend

When a car turns on a flat, horizontal road, it needs an inward force to change its direction.

• The force providing $F_c$ is the frictional force ($F_f$) between the tires and the road surface.
• If the speed ($v$) is too high, or the radius ($r$) is too small (a sharp turn), the required $F_c$ becomes very large. If this required force exceeds the maximum available friction, the car skids.

Equation: $F_f = \frac{mv^2}{r}$

(This is why engineers design banked tracks—to allow the horizontal component of the normal contact force to assist or replace friction in providing $F_c$.)

Example 2: Vertical Circular Motion (e.g., Swapping a bucket of water)

In a vertical circle, the weight ($W = mg$) of the object is constant but always points downwards, so its contribution to $F_c$ changes throughout the loop. Assume an object is tied to a string (tension $T$ provides force).

At the Bottom of the Loop (Max Speed)

• $T$ points upwards (towards the centre).
• $W$ points downwards (away from the centre).
• The resultant centripetal force is: $F_c = T - W$.

Equation: $$T - mg = \frac{mv^2}{r}$$

(Here, the tension $T$ must be greater than the weight $mg$.)

At the Top of the Loop (Min Speed)

• Both $T$ and $W$ point downwards (towards the centre).
• The resultant centripetal force is: $F_c = T + W$.

Equation: $$T + mg = \frac{mv^2}{r}$$

The Critical Speed (Minimum Speed)

For an object to just complete the loop (or for water in a bucket not to spill at the top), the tension ($T$) must be zero at the very top. In this critical case, the weight ($W$) is the *only* force providing the centripetal force:

When $T=0$: $$mg = \frac{mv^2}{r}$$ $$g = \frac{v^2}{r}$$ $$v_{\text{min}} = \sqrt{gr}$$

This minimum speed is independent of the mass of the object!