top of page  # 6.1. Circular Motion

So we have already looked at a lot of mechanics of motion in a straight line. This chapter begins to introduce the idea of rotational motion. It is really important before starting this stuff that you are fully up to speed with the Mechanics we did in Chapter 2, so make sure you have a look back over that stuff (particularly forces and free body diagrams) if you are rusty.

When we started learning about forces we learnt that they can do 3 things:

• Change an object's speed

• Change an object's direction

• Change an object's shape

It's the second of these that we will be looking at throughout this chapter. Imagine a bucket being whirled around your head on some string - the bucket's speed may be constant, but its direction is constantly changing. From what we know about vectors, this means its velocity must also be constantly changing.

I quite like this video as a starting point with this topic, with a sprinter trying to complete a loop the loop on foot.

The chapter has been divided up as follows:

Angular Motion

#### Angular Motion

We are very familiar with our equation for linear velocity, that is: Now we are looking at motion in a circle we introduce a new term called angular velocity. Angular velocity is a measure of how quickly something moves through a certain angle, measured in Radians per Second, rads-1.

N.B. A radian is another way to measure angle, other than degrees. It is essential you confident converting between degrees and radians for this chapter. If you are unfamiliar with the concept, spend some time brushing up on this topic here.

Our angular velocity is defined using the following equation: Crash Course have a nice overview of circular motion which is worth watching.

PHET have a nice simulation on Revolutions (downloading the Java version seems to work best). Select the tab called 'Rotation' and give the turntable an initial angular velocity. The simulation demonstrates quite nicely that the angular velocity is the rate at which the turntable moves through an angle (also notice the position graph at the bottom, and the similarities with SHM).

Some of the terminology we previously used when looking at waves can also be used here:

• Frequency - the number of rotations per second, in Hz (though often presented in rpm - revolutions per minute)

• Time period - the time in seconds to complete one revolution

These are linked by the equation f = 1/T. An object moving in a complete circle moves through an angle of 2π radians, leading to alternative equations for angular velocity. Next, we can look at the link between angular and linear velocities. Mathematically, the length of an arc of a circle is linked to the angle through the equation s=rθ. At this point, it's probably worth spending some time practicing some of the basics. Isaac Physics have a nice selection of test-yourself questions to make sure you are confident.

## Video Lessons

 Chris Doner Uniform Circular Motion IB Specific Khan Academy Introduction to Angular Velocity Relating angular and linear motion Science Shorts Circular Motion Degrees or Radians Study Nova Circular Motion Circular Motion (Lecture)

## Resources

 IB Physics Topic 6 Notes IB-Physics.net Chapter 6 Summary IB Revision Notes Isaac Physics Radians Maths Recap Circular Kinematics Isaac Physics Circular Kinematics Mr. G 6.1 Teaching Notes 6.1 Student Notes Physics and Maths Tutor Further Mechanics Definitions Further Mechanics Key Points Further Mechanics Detailed Notes Further Mechanics Flashcards A Level Resources - content slightly different

## Questions

 Cambridge University Press Topic 6: MCQs CUP Website Link Freely available online Dr French's Eclecticon Uniform Circular Motion Uniform Circular Motion Solutions Link to Dr French's Site Extension: Pre-University Material Grade Gorilla 6.1 (Circular Motion) MCQ Topic 6 (Gravity/ Rotation) End Quiz Quick IB Specific Mixed MCQs Isaac Physics Units of Rotary Motion Mr. G 6.1 Formative Assessment Topic 6 Summary Qs IB Specific Questions Physics and Maths Tutor Circular Motion (AQA 1) Circular Motion MS (AQA 1) Circular Motion (AQA 2) Circular Motion MS (AQA 2) A-Level Qs: overlapping content
Centripetal Force

#### Centripetal Force and Acceleration

Now we have a few of the basics out the way, let's look at a few key ideas towards motion in a circle.

Let's start by looking at motion of the Earth around the sun as shown below.

The direction of the linear velocity (shown in blue) is always changing. As velocity is a vector quantity, this means there must be an acceleration. (Note, the velocity is constantly changing, but the speed is constant).

• The linear velocity is always tangential to the circle.

The only force acting on the Earth as it orbits is the gravitational force of attraction towards the sun. As the resultant force is towards the centre of the circle, the acceleration must also act towards the centre (N2L - the acceleration is always in the same direction as resultant force), and this is known as the CENTRIPETAL ACCELERATION.

• The (centripetal) acceleration always acts towards the centre of the circle (and is perpendicular to the velocity). This is a key characteristic of objects moving with circular motion: there is always acceleration, therefore the object can never be in equilibrium.

If an object is moving in with uniform circular motion, there is always a resultant force acting, directed towards the centre of the circle. This resultant force is known as the centripetal force.

## Equations of Circular Motion

The centripetal acceleration is a vector quantity that always acts towards the centre of the circle. Mathematically, the equation for centripetal acceleration as follows (derivation for these can be found here) By using our equation v = rω, we can derive a second equation for our centripetal acceleration in terms of our angular velocity. Our centripetal force is linked to our centripetal acceleration by Newton's Second Law, i.e. F = ma. By multiplying through by mass we get: ## Worked Example - a banking plane

Once you've got to grips with the basic formulae, it's important to have a bit of practice at applying these. One of the trickiest examples to get your head around involve resolving multiple forces acting on an object to find centripetal force. See a worked example below for a banking plane.

Q. A plane of mass 10 000 kg is banking at 30° in a horizontal circle of radius 500 m. Work out the size of the centripetal force acting and the velocity of the plane.

For this sort of question, it is essential to start with a diagram. The centripetal force acts towards the centre of the circle, and is the resultant force acting on the plane.  We are looking for the resultant force (i.e. centripetal force), so we must consider the forces acting on the plane. Therefore we must draw a free body diagram. The weight force acts downwards, while the lift force acts perpendicular to the wings. We know that the resultant force acts towards the centre, so we should now resolve the Lift force into its x- and y- components.

The resultant force in the y-direction is zero, therefore:

W = L cosθ

10 000 kg x 9.81 = L cos(30)

∴ L = 113 000 N

The resultant force (i.e. centripetal force) on the plane the same as the x-component of the lift, so:

Fcent = L sin θ

= 113 000 cos(30)

∴ = 56 600 N

We can then substitute this into our equations for centripetal force to calculate the plane's linear velocity:

Fcent = mv²/r

56 600 = 10 000 v² / 500

v = 53 msˉ¹

Isaac Physics have several questions looking at centripetal force, have a bit of practice applying these ideas.

## Video Lessons

 Chris Doner Circular Motion (Centripetal force) IB Specific Khan Academy Centripetal Force Intuition Centripetal Force Problems Science Shorts Circular Motion Study Nova Centripetal Acceleration and Force Circular Motion (Lecture)

## Resources

 IB Physics Topic 6 Notes IB-Physics.net Chapter 6 Summary IB Revision Notes Isaac Physics Circular Dynamics Mr. G 6.1 Teaching Notes 6.1 Student Notes Physics and Maths Tutor Further Mechanics Definitions Further Mechanics Key Points Further Mechanics Detailed Notes Further Mechanics Flashcards A Level Resources - content slightly different

## Questions

 Cambridge University Press Topic 6: MCQs CUP Website Link Freely available online Dr French's Eclecticon Uniform Circular Motion Uniform Circular Motion Solutions Link to Dr French's Site Extension: Pre-University Material Grade Gorilla 6.1 (Circular Motion) MCQ Topic 6 (Gravity/ Rotation) End Quiz Quick IB Specific Mixed MCQs Isaac Physics Centripetal Acceleration Mr. G 6.1 Formative Assessment Topic 6 Summary Qs IB Specific Questions
Motion in Vertical Circles

## Motion in a Vertical Circle

The trickiest questions in this topic involve motion in vertical circles. Things like buckets being whirled on a string, cars going over hump-back bridges, or rollercoaster loops as shown below.

The most important thing to do when solving these problems is to draw a Free Body Diagram, labelling all the forces acting, and remembering that the centripetal force acts towards the centre of the circle.

In the example below we have Free Body Diagrams for a rollercoaster completing a loop the loop at a constant speed (N.B. in this example some work must be done to maintain a constant speed).

• At the top of the loop, the reaction force acts outwards from the track. This means that the reaction and weight forces act in the same direction. The centripetal force is therefore the sum of the the two forces' magnitudes.

• At the bottom of the loop, the weight still acts downwards, though now the reaction force acts in the opposite direction (outwards from the track). The centripetal force still must act towards the centre, therefore the reaction force must be much larger than the weight force. ## Worked example - minimum speed to complete loop

One common question that the IB likes to ask is how slow does the rollercoaster need to travel in order to just lose contact with the track.

Q. A rollercoaster has a loop diameter of 30 m. Calculate the minimum speed at which the rollercoaster will complete the loop.

The rollercoaster will lose contact when upside down at the top of the loop (left picture above). The slower it travels, the lower the reaction force from the track keeping the coaster moving in a circle. At the point where it just loses contact, the reaction force will be equal to ZERO.

Therefore at this point Fcent = W. By equating our weight to our equation for centripetal force we can solve for v.

v = √(gr)

= √(9.81 x 15)

= 12 msˉ¹

## Video Lessons

 Chris Doner Problem Solving in Circular Motion IB Specific Khan Academy Vertical Bowling Ball Loop Vertical YoYo Mass in a Horizontal Circle Science Shorts Circular Motion (Vertical) Study Nova Vertical Circle Example

## Resources

 IB Physics Topic 6 Notes IB-Physics.net Chapter 6 Summary IB Revision Notes Mr. G 6.1 Teaching Notes 6.1 Student Notes Physics and Maths Tutor Further Mechanics Definitions Further Mechanics Key Points Further Mechanics Detailed Notes Further Mechanics Flashcards A Level Resources - content slightly different

## Questions

 Cambridge University Press Topic 6: MCQs CUP Website Link Freely available online Grade Gorilla 6.1 (Circular Motion) MCQ Topic 6 (Gravity/ Rotation) End Quiz Quick IB Specific Mixed MCQs Mr. G 6.1 Formative Assessment Topic 6 Summary Qs IB Specific Questions