Define centripetal acceleration and centripetal force
- Understand that circular motion requires a net inward force
- State the direction of centripetal acceleration
- Identify the force(s) that provide centripetal force in different situations
Solve problems using $a_c = v^2/r$ and $F_c = mv^2/r$
- Select the appropriate formula for the given quantities
- Rearrange equations to find unknown variables
- Work with angular velocity and relate it to linear speed
Relate angular velocity $\omega$ to linear speed $v$
- Use $v = \omega r$ to convert between angular and linear quantities
- Calculate period and frequency from angular velocity
- Connect rotational and translational descriptions of motion
A car travels around a roundabout at constant speed. Is it accelerating? Explain your reasoning.
Consider what you know about velocity as a vector quantity. Does constant speed mean constant velocity?
Uniform Circular Motion
Analyse objects moving in circular paths at constant speed. Understand centripetal acceleration, centripetal force, and their relationships to speed, radius, and angular velocity.
Describing Circular Motion
Period, frequency, angular velocity, and linear speed.
Work Circular Motion
When an object moves in a circle at constant speed, we describe its motion using several key quantities:
- Period ($T$): The time for one complete revolution. Measured in seconds (s).
- Frequency ($f$): The number of revolutions per second. $f = 1/T$, measured in hertz (Hz).
- Angular velocity ($\omega$): The rate at which the angle changes. $\omega = 2\pi/T = 2\pi f$, measured in rad/s.
- Linear (tangential) speed ($v$): The distance travelled per unit time along the circular path. $v = 2\pi r/T = \omega r$.
The relationship $v = \omega r$ is particularly important. It tells us that for a given angular velocity, points farther from the centre travel faster. This is why the outer edge of a spinning DVD moves faster than the inner part.
$\displaystyle f = \frac{1}{T}$ (Hz)
$\displaystyle \omega = \frac{2\pi}{T} = 2\pi f$ (rad/s)
$\displaystyle v = \frac{2\pi r}{T} = \omega r$ (m/s)
Centripetal Acceleration
Why constant speed does not mean zero acceleration.
In uniform circular motion, the speed is constant but the velocity is not. Velocity is a vector — it has both magnitude (speed) and direction. As the object moves around the circle, its direction continuously changes. A change in velocity means there is acceleration.
Consider the velocity vectors at two nearby points on the circle. The velocity is always tangent to the circle. When we subtract these vectors to find $\Delta \vec{v}$, the result points toward the centre of the circle. Taking the limit as the time interval approaches zero gives the instantaneous acceleration.
This acceleration is called centripetal acceleration ($a_c$), and it is always directed toward the centre of the circle.
$\displaystyle a_c = \frac{v^2}{r} = \omega^2 r$
Direction: always toward the centre of the circle.
Units: m s$^{-2}$ (same as any acceleration)
The two forms are equivalent since $v = \omega r$:
$$a_c = \frac{v^2}{r} = \frac{(\omega r)^2}{r} = \omega^2 r$$
Centripetal Force
The net force required to maintain circular motion.
By Newton's Second Law, any acceleration requires a net force. Since centripetal acceleration is directed toward the centre, there must be a net force directed toward the centre. This is called centripetal force ($F_c$).
$\displaystyle F_c = ma_c = \frac{mv^2}{r} = m\omega^2 r$
Direction: always toward the centre of the circle.
Units: N (newtons)
Critical understanding: Centripetal force is not a new type of force. It is simply the name we give to the net force directed toward the centre that causes circular motion. This net force can be provided by:
- Tension — a ball on a string being whirled in a horizontal circle
- Friction — a car turning on a flat road
- Gravity — a satellite in orbit, or the Moon around Earth
- Normal force — a car on a banked curve
- Any combination of these forces
If the net inward force is removed, the object travels in a straight line at constant velocity (Newton's First Law) — it does not fly radially outward.
Formula Summary: Uniform Circular Motion
Common Misconceptions
Right: Inertia wants you to travel in a straight line (Newton's First Law). Friction or tension provides an inward centripetal force that continuously changes your direction. There is no real outward force.
Right: Centripetal force is the name for the net inward force. It can be provided by tension, friction, gravity, normal force, or any combination of real forces. It is not a separate force that appears on a free-body diagram.
Right: Acceleration measures the rate of change of velocity, which is a vector. Even when speed is constant, the continuous change in direction means velocity is changing. Therefore $a_c = v^2/r \neq 0$.
Car on a Roundabout
A 1500 kg car travels at 35 km/h ($9.72\ \text{m/s}$) around a roundabout of radius 15 m.
$\displaystyle a_c = \frac{v^2}{r} = \frac{(9.72)^2}{15} = \frac{94.5}{15} = \mathbf{6.30\ \text{m/s}^2}$
$\displaystyle F_c = \frac{mv^2}{r} = \frac{1500 \times (9.72)^2}{15} = \frac{1500 \times 94.5}{15} = \mathbf{9450\ \text{N}}$
This 9450 N force is provided entirely by friction between the tyres and the road surface. On a dry road, static friction can easily provide this. But on a wet road, the maximum available friction is reduced. If friction is insufficient, the car slides outward — not because of a mysterious outward force, but because friction can no longer provide enough inward force to maintain the circular path. The car simply follows Newton's First Law and continues in a straight line.
This is why advisory speed limits on roundabouts exist: they are calculated from the coefficient of friction and the radius to ensure $F_c \leq F_\text{friction,max}$.
Work and Energy in Uniform Circular Motion
Syllabus 5.21 — Why the centripetal force does no work.
In uniform circular motion, speed is constant. Because $KE = \tfrac{1}{2}mv^2$, the kinetic energy never changes:
By the work–energy theorem, the net work done on the object equals its change in kinetic energy:
Why is the work zero? At every instant, the centripetal force points toward the centre, while the displacement is tangent to the circle. These two directions are perpendicular ($\theta = 90°$), so:
Work Done by Centripetal Force
Key insight: The centripetal force continuously redirects the velocity vector, but it never speeds up or slows down the object. No energy is transferred into or out of the system by the centripetal force. This is why satellites in circular orbits maintain constant speed without any engine thrust.
Right: Force only does work when it has a component parallel to displacement. In uniform circular motion, the net force is always perpendicular to motion, so $W = 0$ and total mechanical energy remains constant.
Drill Problems
Apply the formulas to three standard problems. Show all working.
Q1. An object moves in a circle of radius 20 m at a speed of 8 m/s. Calculate the centripetal acceleration.
Q2. A 0.5 kg mass is attached to a string of length 0.8 m and whirled in a horizontal circle at 2 revolutions per second. Calculate the tension in the string (which provides the centripetal force).
Q3. The Earth orbits the Sun at an average distance of $1.5 \times 10^{11}\ \text{m}$ with a period of 1 year ($3.16 \times 10^7\ \text{s}$). Calculate the orbital speed of the Earth.
Force Identification
For each scenario, identify what real force(s) provide the centripetal force.
In each scenario below, state which force or combination of forces provides the centripetal force required for circular motion.
(a) A satellite in orbit around Earth.
(b) A car travelling around a banked curve (no friction needed at design speed).
(c) An electron orbiting the nucleus in a simplified Bohr model.
(d) Wet clothes stuck to the inside wall of a spinning dryer drum.
Copy into Books
Key definitions and formulas to record in your workbook.
Period ($T$): Time for one complete revolution (s).
Frequency ($f$): Number of revolutions per second, $f = 1/T$ (Hz).
Angular velocity ($\omega$): Rate of change of angle, $\omega = 2\pi/T = 2\pi f$ (rad/s).
Linear speed ($v$): $v = 2\pi r/T = \omega r$ (m/s).
Centripetal acceleration: $a_c = v^2/r = \omega^2 r$, directed toward the centre.
Centripetal force: $F_c = mv^2/r = m\omega^2 r$, the net force toward the centre.
Key principle: Centripetal force is not a new force — it is the name for the net inward force provided by tension, friction, gravity, or normal force.
Revisit Think First
You predicted whether a car on a roundabout at constant speed is accelerating. Now, with your understanding of centripetal acceleration, revisit your answer:
- Is the car accelerating? What is the direction of acceleration?
- What force provides the centripetal force for the car?
- What happens if that force becomes insufficient?
1. Centripetal acceleration is always:
2. A car rounds a curve at constant speed. The centripetal force is provided by:
3. What happens to $F_c$ if speed doubles and radius stays constant?
4. The angular velocity $\omega$ for 3 revolutions per second is:
5. "Centrifugal force" is:
1. A 0.3 kg ball on a 0.6 m string makes 2.5 revolutions per second. Calculate the tension in the string.
2. A 1000 kg car rounds an unbanked curve of radius 25 m. The coefficient of static friction between the tyres and the road is 0.65. Calculate the maximum safe speed of the car.
3. Evaluate the statement: "An object in uniform circular motion is in equilibrium because its speed is constant." Use Newton's laws and the concept of acceleration to justify your answer.
Answers
▾Activity 1: Drill Problems
Q1: $a_c = \dfrac{v^2}{r} = \dfrac{8^2}{20} = \dfrac{64}{20} = \mathbf{3.2\ \text{m/s}^2}$
Q2: $f = 2\ \text{Hz}$, so $\omega = 2\pi f = 2\pi \times 2 = 4\pi = 12.57\ \text{rad/s}$. Then $F_c = m\omega^2 r = 0.5 \times (12.57)^2 \times 0.8 = 0.5 \times 157.9 \times 0.8 = \mathbf{63.2\ \text{N}}$ (or use $F_c = mv^2/r$ with $v = \omega r = 10.05\ \text{m/s}$).
Q3: $v = \dfrac{2\pi r}{T} = \dfrac{2\pi \times 1.5 \times 10^{11}}{3.16 \times 10^7} = \dfrac{9.42 \times 10^{11}}{3.16 \times 10^7} = \mathbf{2.98 \times 10^4\ \text{m/s}}$ (approximately 30 km/s).
Activity 2: Force Identification
(a) Gravitational force (weight) — the gravitational attraction between Earth and the satellite provides the centripetal force for orbit.
(b) Component of the normal force — on a banked curve, the normal force has a horizontal component directed toward the centre of the turn.
(c) Electrostatic force (Coulomb force) — the attraction between the positively charged nucleus and negatively charged electron provides the centripetal force.
(d) Normal force from the drum wall — the wall pushes inward on the clothes, providing the centripetal force. Water escapes through the holes (not pushed out by a force).
MCQ Answers
1. B (Toward the centre)
2. C (Friction between tyres and road)
3. D (Quadruples, since $F_c \propto v^2$)
4. B ($6\pi$ rad/s, since $\omega = 2\pi f$)
5. C (A fictitious force in rotating frames only)
SA1: Tension in String (2 marks)
$f = 2.5\ \text{Hz}$, so $\omega = 2\pi f = 2\pi \times 2.5 = 5\pi = 15.7\ \text{rad/s}$ (1 mark for $\omega$)
$F_c = m\omega^2 r = 0.3 \times (15.7)^2 \times 0.6 = 0.3 \times 246.5 \times 0.6 = \mathbf{44.4\ \text{N}}$ (1 mark)
Since tension provides the centripetal force, $T = 44.4\ \text{N}$.
SA2: Maximum Safe Speed (3 marks)
At maximum speed, friction provides the centripetal force: $F_c = F_\text{friction,max}$
$\dfrac{mv^2}{r} = \mu_s N = \mu_s mg$ (1 mark for equating $F_c$ and max friction)
$v^2 = \mu_s g r = 0.65 \times 9.8 \times 25 = 159.25$ (1 mark)
$v = \sqrt{159.25} = \mathbf{12.6\ \text{m/s}}$ or $\mathbf{45.4\ \text{km/h}}$ (1 mark)
SA3: Evaluate Equilibrium Statement (4 marks)
The statement is false. (1 mark)
An object in uniform circular motion is not in equilibrium because its velocity vector is continuously changing direction. By Newton's First Law, an object in equilibrium has constant velocity (both constant speed and constant direction) or is at rest. (1 mark)
The continuous change in direction of velocity means there is a centripetal acceleration $a_c = v^2/r$ directed toward the centre. By Newton's Second Law ($F_\text{net} = ma$), a nonzero acceleration requires a nonzero net force. (1 mark)
For equilibrium, $F_\text{net} = 0$, which is not satisfied in circular motion. The centripetal force $F_c = mv^2/r$ is the required net inward force. Therefore, the object is accelerating (changing direction) and cannot be in equilibrium, even though its speed is constant. (1 mark)
Circular Motion Mastery Challenge
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