Uniform circular motion

Uniform circular motion occurs when an object moves in a circular path at a constant speed. Even though the speed stays the same, the object's velocity changes continuously because velocity is a vector quantity that includes both speed and direction. As the object follows the curve of the circle, its direction shifts constantly, which means it is always accelerating.

This acceleration is necessary to keep the object from moving in a straight line, as it would without any force acting to change its path. The term "uniform" refers specifically to the constant speed, not to constant velocity.

Centripetal acceleration

Centripetal acceleration is the acceleration that points toward the center of the circular path, causing the change in velocity direction required for circular motion. This acceleration is always perpendicular to the object's instantaneous direction of motion (its tangent to the circle) and acts inward, seeking the center.

The magnitude of centripetal acceleration depends on the object's speed and the radius of the path. Faster speeds or smaller radii result in greater acceleration because the direction changes more rapidly.

Formula for centripetal acceleration

$a=\frac{v^2}{r}$

Where:

• a = Centripetal acceleration (m/s²)
• v = Speed of the object (m/s)
• r = Radius of the circular path (m)

This formula shows that acceleration increases with the square of the speed but decreases as the radius gets larger.

Centripetal force

Centripetal force is the net force that provides the centripetal acceleration, pulling or pushing the object toward the center of the circle. It is not a separate type of force but rather the result of existing forces acting in a way that maintains the circular path.

According to Newton's second law, which states that force equals mass times acceleration (F = ma), the centripetal force can be calculated by combining this with the centripetal acceleration formula.

Formula for centripetal force

$F=m\frac{v^2}{r}$

Where:

• F = Centripetal force (N)
• m = Mass of the object (kg)
• v = Speed of the object (m/s)
• r = Radius of the circular path (m)

This equation allows you to find the force required to keep an object moving in a circle at a given speed and radius.

Sources of centripetal force

Centripetal force comes from everyday forces that act toward the center of the circular path. It is not a new fundamental force but simply the component of whatever force is causing the inward acceleration.

Common sources of centripetal force:

• String tension - In a situation like swinging a ball on a string, the tension in the string pulls the ball inward, providing the centripetal force.
• Friction - For a car turning a corner on a road, friction between the tires and the road surface acts toward the center, preventing the car from sliding outward.
• Gravitational attraction - In planetary orbits, the gravitational pull from a central body (like the Sun) provides the centripetal force that keeps planets moving in curved paths.

In each case, the force must be strong enough to match the required centripetal acceleration for the motion to remain circular.

Relating speed to radius and period

In uniform circular motion, the speed of the object is connected to the radius of its path and the time it takes to complete one full circle, known as the period (T). The period is measured in seconds and represents the duration of one revolution.

This relationship comes from the fact that the distance traveled in one period is the circumference of the circle (2πr), so speed is that distance divided by time.

Formula for speed in circular motion

$v=\frac{2\pi r}{T}$

Where:

• v = Speed of the object (m/s)
• r = Radius of the circular path (m)
• T = Period of one revolution (s)

This equation is useful for finding speed when you know the size of the path and how long a cycle takes, or for solving for other variables.

Worked example - Calculating centripetal acceleration

An object moves in a circle with a radius of 5 m at a constant speed of 10 m/s. Calculate the centripetal acceleration.

Step 1: Identify the values

• Speed (v) = 10 m/s
• Radius (r) = 5 m

Step 2: Formula

$a=\frac{v^2}{r}$

Step 3: Substitution and calculation

$a=\frac{{10}^2}{5}=\frac{100}{5}=20{\text{ m/s}}^2$

The centripetal acceleration is 20 m/s² toward the center.

Worked example - Calculating centripetal force

A 2 kg ball is swung in a horizontal circle on a string, with a radius of 1.5 m and a speed of 4 m/s. Calculate the centripetal force provided by the string tension.

Step 1: Identify the values

• Mass (m) = 2 kg
• Speed (v) = 4 m/s
• Radius (r) = 1.5 m

Step 2: Formula

$F=m\frac{v^2}{r}$

Step 3: Substitution and calculation

$F=2\times \frac{4^2}{1.5}=2\times \frac{16}{1.5}=2\times 10.67=21.33\text{ N}$

The centripetal force is approximately 21.3 N toward the center.

Worked example - Using period to find speed and force

A satellite orbits Earth with a period of 5400 s and an orbital radius of 8000 km (8 × 10⁶ m). The satellite has a mass of 500 kg. First, calculate its speed, then the centripetal force (assuming gravitational attraction provides it).

Step 1: Identify the values

• Radius (r) = 8 × 10⁶ m
• Period (T) = 5400 s
• Mass (m) = 500 kg

Step 2: Formula for speed

$v=\frac{2\pi r}{T}$

Step 3: Calculate speed

$v=\frac{2\times 3.14\times 8\times {10}^6}{5400}\approx \frac{5.024\times {10}^7}{5400}\approx 9304\text{ m/s}$

Step 4: Formula for centripetal force

$F=m\frac{v^2}{r}$

Step 5: Substitution and calculation for force

$F=500\times \frac{{9304}^2}{8\times {10}^6}\approx 500\times \frac{8.66\times {10}^7}{8\times {10}^6}\approx 500\times 10.825\approx 5413\text{ N}$

The speed is approximately 9300 m/s, and the centripetal force is approximately 5410 N toward the center.