MTH6P2B: CIRCULAR MOTION

                             Introduction

You will have seen and experienced examples of circular motion throughout your everyday life. Theme park rides such as ‘The

Wave Swinger’, ‘The Pirate Ship’, ‘Loop-the-Loop’; cornering on a bicycle or in a car; household equipment such as washing

machines, tumble and spin driers, salad driers; specialised devices such as the centrifuge in the chemistry laboratory; the governors of a steam engine. These are just a few of the many examples which you will have come across or been a part of.

The purpose of this chapter is to enable you to analyse circular motion, to help you model some of the situations noted above and be able to predict and explain observations which you can make.

The philosophers of the ancient world considered circular motion to be a natural motion. The heavenly bodies, the planets and the stars, moved in circles around the Earth. Once set upon their paths by the Gods, these bodies continued to move in circles without any further intervention. No force was required to sustain their heavenly orbits.

However, from Newton’s point of view, circular motion requires a force to sustain it. Recall Newton’s First Law from Chapter 2,

A body remains in a state of rest or moves with uniform motion, unless acted upon by a force.

A body describing a circle is certainly not at rest. Nor does it move with constant velocity, since, whether the speed is changing or constant, the direction of the motion is changing all the time.

From Newton’s First Law, there must be a resultant force acting on the body and Newton’s Second Law equates this force to the mass  times acceleration, so the body must be accelerating.

An example of this is the case of the Moon (M in diagram opposite) orbiting the Earth (E). The Moon describes a path around the Earth which is approximately circular. The force which the Earth exerts upon the Moon is the force of gravity and this pulls the Moon towards the Earth.

If this force of gravity did not act, the Moon would move off at a tangent to its path around the Earth. Indeed Newton envisaged the motion of the Moon around the Earth as a series of steps, a tangential movement followed by a move in towards the Earth. If you make the time intervals sufficiently small, this saw-tooth curve becomes a circle.

A further consequence is that since the force upon the Moon is directed towards the Earth so the acceleration is also in this direction.

A body moving in a circle must

  • from Newton’s First Law, have a resultant force acting on it;
  • from Newton’s Second Law, be accelerating in a specific

Angular velocity and angular speed

The relation between velocity and angular velocity

For this activity you will need a piece of polar graph paper, a ruler, a protractor, or a piece of polar graph paper and a programmable calculator.

A piece of polar graph paper can be used to represent the path of the particle moving in a circle with constant angular speed, the radius of the circle sweeping out equal angles in equal intervals of time.

The magnitude of the average velocity of the particle is

Now investigate the average velocity of the particle for different times t.

Copy and complete the table below, using the diagram opposite.

approaches the velocity of the particle at P.

What is the magnitude of the velocity at P?

Is there a relation between this magnitude, w , and r, the radius of the circle in metres?

In which direction is the velocity at P?

Describing motion in a circle with vectors

then has magnitude r and is in the direction of the unit vector

The velocity, v, of the particle is then given by

The results you should have obtained from Activity 1 are therefore validated.

In the introduction to this chapter you saw that any particle moving in a circle must be accelerating. Using the system of unit vectors which has just been set up, this acceleration can now be found.

From equation (3) above,

Example

A car has wheels which are 75 cm in diameter and is travelling

at a constant speed of 60 kph. If there is no slipping between the wheels and the road, calculate the angular speed and the acceleration of a stone wedged in the tread of one of the tyres.

Solution

Converting the speed to ms-1 gives

How might your answer be changed if there is slipping between the wheels and
the road?

Hence the acceleration is 8t2 ms-2 directed radially inwards and 4 ms-2 tangentially in the direction of increasing q.

Motion in a circle with constant angular speed

In section 7.2 you saw that the acceleration of a particle P moving in a circle of radius r metres with constant angular speed

Investigating the forces in circular motion

For this activity you will need a circular cake tin or a plastic bucket and a marble.

Place the marble inside the cake tin.

Holding the cake tin horizontally, move it around so that the marble rolls around the sides of the tin, in contact with the base.

You should be able to get the marble going fast enough so that for a short period of time you can hold the tin still and the marble will describe horizontal circles with uniform angular velocity.

What do you feel through your hands?

What are the forces acting on the marble?

Draw a force diagram for the forces acting on the marble.

What is the magnitude and direction of the resultant force acting on the marble?

Force  and acceleration

For each of the following situations, mark on the diagram the forces which are acting on the particle, the direction of the resultant force and the acceleration.

Example

The marble of Activity 2 has mass 10 g and the cake tin a radius of 15 cm. If the marble rolls round the bottom of the tin in contact with the side at an angular speed of 20 rad s-1, what are the normal contact forces of the side and base of the tin on the marble?

Solution

The forces on the marble are: its weight 0.01 g N vertically downwards; the normal contact force B newtons of the base on it; the normal contact force S newtons of the side on it.

The acceleration of the marble is radially inwards and has magnitude

Since there is no acceleration vertically,

B – 0. 01 g = 0,

so                 B = 0.1 N.

The rotor ride

What are the forces acting on the people in the ride shown opposite?

Is there a critical angular speed at which this ride must operate to achieve the effects on the people which the diagram shows?

Modelling the ride

Set up the model by considering each person as a particle, and looking at the forces acting upon it. You looked at a physical situation very similar to this in Activity 2, so you should have some idea of the forces acting.

The forces acting are:-

  • the weight, mg, acting vertically downwards;
  • the force of friction, F, acting vertically upwards and preventing the particle sliding down the wall;
  • the normal contact force, N, exerted radially inwards by the wall on the

The particle is moving in a circle, radius r, with constant angular speed w , so there is an acceleration rw 2 directed along the radius towards the centre of the circle.

Considering the forces in the two mutually perpendicular directions, vertically and radially, and applying Newton’s Second Law in each direction you have,

Therefore there is a critical angular speed,

which is independent of the mass of the particle or person. This is clearly shown by the people on the ride itself. No matter what their size is, they all behave in exactly the same way. The angular speed, w , must be greater than the critical angular speed above, for the effects to be observed.

For a particular ride, the diameter of the drum is about 10 metres and the coefficient of friction is 0. 5. If g is taken to be 10 ms-2 , then the critical angular speed

Therefore the ride will have to be rotating in excess of this angular speed for the effect of people ‘sticking’ to the wall to be observed.

The conical pendulum

For this activity you will need either a mass on the end of a piece of string or a mechanics kit containing a conical

pendulum.

A bob is made to describe circles which are in a horizontal 

plane. If you have access to a suitable mechanics kit you can set it up as shown opposite. If not, then you can do much the same thing using a mass on the end of a piece of string. This particular form of motion is often referred to as the ‘conical pendulum’.

 

 

If the angular speed increases, what happens to the bob? Does the mass of the bob have any effect?

Is it possible to get the bob to describe:

 

  • a circle so that the string is horizontal?
  • a circle above the point of suspension rather than below?

 

To model this situation, what assumptions would you make?

Modelling the conical pendulum

Set up the model

To begin, the following assumptions simplify the problem;

  1. the bob is a particle;
  2. the string is light and inextensible;
  3. the point of suspension is directly above the centre of the horizontal circle described by the bob;
  4. the angular speed is

With these assumptions the problem becomes one of motion of a particle in a circle of radius r about the axis of rotation ON, with constant angular speed w . The length of the string is l and it makes an angle q to the axis of rotation.

Validating the solution

The questions raised in Activity 4 , relating to the conical pendulum, can now be explored mathematically using the solution in equation (3).

The conical pendulum revisited

Do your answers to Activity 3 agree with the mathematical model?

The chair-o-planes ride

Look carefully at the picture of the chair-o-planes ride. What might happen if the solution depended on mass?

How would you amend the model of the conical pendulum to model this ride?

Find a formula for the angle that each supporting rope makes with the vertical.

It may be possible to visit a fair or a park and, by calculating some distances, estimate the angular speed of the ride.

Motion on a banked curve

Why do birds need to bank in order to change the direction of their flight?

What advantages are gained?

Going round the bend

The diagram shows a car taking a bend at a constant speed v.

What is the greatest speed at which it can take the bend on a horizontal road?

Do you think that the car can take the bend at a greater speed if the bend is banked, than if it is not?

Modelling the motion of a car on a banked road

Assume that the car is a particle, the bend is a part of a circle

and the road is inclined at an angle  a to the horizontal.                                            N

The forces acting on the car are its weight mg, vertically

downwards, the normal contact force, perpendicular to the road surface, and the friction force F. However there is a need to

stop and think for a moment about the friction force, F.

In which direction does it act?

In which directions can it act?

Is it possible for the friction force to be zero?

If the car were on the point of sliding down the incline of the bend, F would act in the direction up the incline.

If the car were on the point of sliding up the incline, F would act in the direction down the incline.

Between these two extremes, F can be zero. This is important because of the reduced wear on the car tyres and the greater comfort for passengers and driver.

Consider the critical speed when friction is zero.

Taking the components of the forces acting on the car in two, mutually perpendicular directions, vertically and radially, and applying Newton’s Second Law in each case,

Substituting for N in equation (2) from equation (1) gives

For a bend of given radius, this equation defines the correct angle of banking so that there is no friction force for a given speed. Alternatively, for a given angle of banking it defines the speed at which the bend should be taken.

Including the effects of friction

Suppose that the coefficient of sliding friction between the car and the road is m.

Find an expression for the constant speed of a car on a circular banked track of angle a to the horizontal:

  • if the car is on the point of sliding up the incline;
  • if the car is on the point of sliding down the
When a railway engine takes the bend on a railway track, what is the effect on the rails of the flanges on the wheels?

ASSIGNMENT : CIRCULAR MOTION ASSIGNMENT MARKS : 20  DURATION : 3 hours

 

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