Introduction to Circular Motion

• conduct investigations to explain and evaluate, for objects executing uniform circular motion, the relationships that exist between:

Circular Motion: Centripetal Force and Acceleration

Introduction to Circular Motion

Circular motion is a fundamental concept in physics, often encountered in various real-world situations like the orbits of planets, the motion of a roller coaster, or the spinning of a Ferris wheel. To understand circular motion, we delve into its key elements: centripetal acceleration and force, period, frequency, and angular speed.

Uniform circular motion is defined as when an object travels in a circular motion at constant speed. 

Centripetal Acceleration

 

When an object moves in a circle, it constantly changes direction, implying that it accelerates even if it moves at a constant speed. This acceleration, directed towards the centre of the circle, is known as centripetal (centre-seeking) acceleration.

Mathematically, it is given by

$$a_c = \frac{v^2}{r}$$​

where:

• `v` is the linear speed of the object
• $`r`$is the radius of the circle

Centripetal Force

To cause this centripetal acceleration, a force must be applied. This force, known as centripetal force, keeps the object moving in a circular path.

 

 

According to Newton's second law `F = ma`, the centripetal force is given by:

$$m \times a_c = m \times \frac{v^2}{r}$$

$$F_c = \frac{mv^2}{r}$$

where: 

• `m` is the mass of the object undergoing circular motion in kg
• `v` is the linear or tangential velocity of the object in metres per second (m/s)
• `r` (m) is the radius of this circular motion in metres (m)

 

The direction centripetal force is orthogonal (right-angled) to the direction of the object’s velocity. In other words, it is always towards the centre of the circular motion. 

Centripetal force is a non-real net force which means that it is provided by real force(s).

For example:

When an object is attached to a rope and swung in circles, it undergoes circular motion. If the motion has constant speed, the circulation motion is uniform. The centripetal force in this example is provided by tension in the rope, which has a direction towards the centre of the circular motion. Learn more here
When a car is moving around a horizontal circular bend, it experiences centripetal force provided by friction between the car’s tyres and the ground.
When a spacecraft or satellite is orbiting the Earth, gravitational force provides the centripetal force for circular motion.

Factor Affecting Centripetal Force

$$F_c = \frac{mv^2}{r}$$

From the formula for centripetal force, we can deduce the following:

Centripetal force is directly proportional to:

• Mass of the object undergoing circular motion

• Tangential velocity of the object squared

Centripetal force is inversely proportional to the radius of circular motion

Therefore, a greater centripetal force (net force towards the centre of circular motion) is required to maintain an object’s uniform circular motion if:

• its mass increases

• its tangential velocity increases

• its radius of circular motion decreases 

 

Conversely, changes in centripetal force can affect velocity and radius (mass does not change)

• If centripetal force increases, velocity needs to increase in order to maintain the radius of circular motion, otherwise the object spirals inwards to reach a new circular path with smaller radius.

• If centripetal force decreases, velocity needs to decrease to maintain the radius of circular motion, otherwise the object spirals outwards to reach a new circular path with larger radius.

 

Change	Condition	Effect
Increase in centripetal force	Velocity remains constant	Radius decreases
Increase in centripetal force	Velocity increases	Radius remains constant
Decrease in centripetal force	Velocity remains constant	Radius increases
Decrease in centripetal force	Velocity decreases	Radius remains constant

 

Period and Frequency

The period (`T`) of circular motion is the time it takes for one complete revolution. It's a measure of how long the object takes to go around the circle once. In simple terms, it’s the time for one cycle.

Period of circular motion is given by:

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

where:

• `r` is the radius of circular motion
• `v` is the linear velocity of the object

 

Frequency (`f`), on the other hand, is the number of complete revolutions per unit time. It's typically measured in cycles per second or Hertz (Hz).

Frequency is the reciprocal of the period:

$\$\$f\; =\; \backslash frac\{1\}\{T\}\$\$$

$\$\$f\; =\; \backslash frac\{v\}\{2\backslash pi\; r\}\$\$$

Angular Speed

Angular speed (`\omega`) is a measure of how fast the angle changes as an object moves along a circular path. It's often measured in radians per second (rad/s).

Angular speed is given by the angle of one revolution (`2\pi`) divided by the period (time taken to complete one revolution):

 

$$\omega = \frac{2\pi}{T}$$

 

Alternatively, angular speed is given by the change in angle divided by the time taken:

 

$$\omega = \frac{\Delta \theta}{t}$$

 

Angular speed can also be expressed in terms of linear velocity:

 

$\$\$\backslash omega\; =\; \backslash frac\{v\}\{r\}\$\$$

 

where:

• `r` is the radius of circular motion
• `v` is the linear velocity of the object

Example 1

A car rounds a bend on a road that follows the arc of a circle with radius r. The car has mass m and is travelling at a velocity v. Explain the following situations:

(a) Why are drivers advised to slow down during wet weather, specifically when they are making a bend.

(b) Assuming the friction between the tyres and the road does not change, describe the path of a car with mass 2m when it rounds the bend at velocity v?

(c) A motor cyclist rounds the same bend at velocity 2v. If the mass of the motorcycle is 0.25m, what would be different about the centripetal force acting on the motorcycle compared to that on the car? 

 

Solution to part (a):

During wet weather, the kinetic friction between a car's tyres and the ground is reduced. This means the centripetal force acting on the car during its bend is reduced. As a result, velocity needs to decrease to maintain radius of the curvature.

Solution to part (b):

Since centripetal force remains constant, the radius of curvature is doubled. This means for a car with mass 2m travelling at the same speed v, it requires a greater distance to complete the bend. 

Solution to part (c):

 

As seen above, substituting 0.25m and 2v into the equation `F_c=(mv^2)/r` will yield a magnitude of centripetal force identical to one with mass m and velocity v. This means the centripetal force acting on the motorcycle remains unchanged and so does its radius.

Example 2

A 1000 kg mass travels in a circle of radius 5 m and completes 4 revolutions in 2 seconds. Calculate the period, frequency, angular speed and linear speed and centripetal acceleration acting on the mass.

Example 3

A 500 kg mass travels in a circle of radius 2 m and completes 10 revolutions in 2 seconds. Calculate the period, frequency and centripetal force acting on the mass.

 

  

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