Mechanics
Circular Motion
Contents
 1 Circular Motion
 2 The Conical Pendulum
 3 Motion around a Circular Banked Track
Circular motion describes motion of a particle around a circle. Circular motion can occur in various different ways, such as:
 A particle moves freely in a horizontal circle with no strings attached.
 A particle moves in a horizontal circle in contact with a smooth surface.
 A particle moves in a horizontal circle that is joined by a taut string/rod to a fixed point on a smooth surface.
 A particle moves in a horizontal circle inside a smooth hemispherical bowl.
Let’s now look at some terminology, formulas and forces that are used to analyse circular motion.
Angular Velocity
In uniform circular motion, constant angular velocity can also be defined as
Where,
Angular Acceleration
Angular acceleration is the rate of change of angular velocity with respect to time, i.e.
Linear Velocity
The linear velocity , also referred to as the tangential velocity or the instantaneous velocity of a point moving in circular motion.Consider, the rate of change of the arc length represents the tangential velocity .

Forces Acting on a Particle Moving in a Circle
The Mathematics Extension 2 syllabus requires students to be able to prove the tangential and normal components of the force acting on a particle moving in a circle of radius , with angular velocity . However, motion is restricted to uniform circular motion, i.e. constant angular velocity.
The tangential force on a particle moving in nonuniform circular motion
Consider,
Now by Newton’s 2^{nd} Law, we can derive the formula:
The normal force on a particle moving in nonuniform circular motion
Consider the variable point which makes an angle with the positive axis.
Now, applying implicit differentiation with respect to time


Since we are considering uniform circular motion, then.  
Applying vector addition,
Since,
Now by Newton’s 2^{nd} Law, we can derive the formula:
Now, by substituting , we have
Note: the normal force is also called the centripetal force.
The next examples demonstrate some of the concepts we have looked at so far.
Example 1
A particle of mass 6 kg is travelling at constant velocity of 2 round a circle of radius m. Find the rate at which the angle at the centre of the circle is changing if the centripetal force has a magnitude of 12 Newtons.
Solution 1
m
We also have
rad/s
Reaction Forces
Reaction forces are a result of Newton’s third law which states for every action there’s an equal and opposite reaction force. When a particle P is in contact with a surface, the force acting on P by the surface is called the reaction force. The reaction force exerts on a particle is always at right angle to the surface.
Tension Forces
Tension forces are also a result of Newton’s third law. When a particle is attached to a taut string, the string exerts a tension force on the particle. Also, when a particle attached to a taut string is moving in circular motion, the tension force is the centripetal force. Tension forces are of greater significance in the conical pendulum section.
Example 2
A particle of mass kg is attached to a taut string of length , which is fixed on the table. The particle moves in a horizontal circle in contact with the smooth table. If the maximum tension that the string can support without breaking is equal to a gravitational weight force on an object of kg, find the maximum number of revolutions per second that the particle can make without breaking the string.
Solution 2
The tension force is equal to the centripetal force.
We are given that the maximum tension is equal to the gravitational weight force on mass
rad/s
Example 3
A particle moves with constant angular velocity in a horizontal circle of radius on the inside of a fixed smooth hemispherical bowl of internal radius . Show that .
Solution 3
Resolving forces vertically:
Resolving forces horizontally:
Example 4
[HSC 2007, Question 3]
A particle of mass undergoes uniform circular motion with angular velocity in a horizontal circle of radius about . It is acted on by the force due to gravity, , a force directed at an angle above the horizontal and a force which is perpendicular to , as shown in the diagram.
(i) By resolving forces horizontally and vertically, show that
(ii) For what values of is ?
Solution 4
i)
Resolving forces horizontally:
Multiply by ,
Resolving forces vertically:
Multiply by ,
Solving simultaneously, we have:
ii)
rad/s
The Conical Pendulum
The Conical Pendulum describes the motion of a particle moving in (uniform) circular motion which is suspended from a fixed point by a light taut string. There are different forms of conical pendulum, for example:
 A particle is suspended from a fixed point by a light inextensible string.
 A particle is suspended from a fixed point by a light inextensible string. The particle is moving in uniform circular motion on a smooth horizontal table.
 A string passes through a smooth hole in a smooth table where each end of the string is attached to a particle. One of the particles is on the table moving in uniform circular motion and the other particle is below the table where its motion can be described by the motion of a conical pendulum.
When dealing with conical pendulum questions, students need to be able to:
(i) Understand all the different forces acting on the particle, e.g. tension force, centripetal force, gravitational force, normal force and reaction force.
(ii) Resolve forces in vertical and horizontal directions separately.
(iii) Solve the question by methods of substitution and elimination.
Example 5
[HSC 2012, Question 7]
A particle of mass attached to a string is rotating in a circle of radius on a smooth horizontal surface. The particle is moving with constant angular velocity . The string makes an angle with the vertical. The forces acting on are the tension in the string, a reaction force normal to the surface and the gravitational force .
Which of the following is the correct resolution of the forces on in the vertical and horizontal directions?
(A) and
(B) and
(C) and
(D) and
Solution 5
Answer: (A)
Example 6
[HSC 2008, Question 3]
A particle of mass is attached by a string of length to a point . The particle moves with constant angular velocity in a horizontal circle with centre which lies directly below . The angle the string makes with is .
The forces acting on the particle are the tension, , in the string and the force due to gravity, .
By resolving the forces acting on the particle in the horizontal and vertical directions, show that
Solution 6
Resolving forces horizontally:
Resolving forces vertically:
Solving these simultaneously, we have:
We also know that .
Example 7
[HSC 2011, Question 5]
A small bead of mass is attached to one end of a light string of length . The other end of the string is fixed at height above the centre of a sphere of radius , as shown in the diagram. The bead moves in a circle of radius on the surface of the sphere and has constant angular velocity . The string makes an angle of with the vertical.
Three forces act on the bead: the tension force of the string, the normal reaction force to the surface of the sphere, and the gravitational force .
(i) By resolving the forces horizontally and vertically on a diagram, show that
and
(ii) Show that
(iii) Show that the bead remains in contact with the sphere if .
Solution 7
i)
Resolving forces horizontally:
Note that the centripetal force is directed towards the centre (and is the net force). From the diagram above, the force F is contributing positively whereas the force N is opposing the overall force towards the centre. (It may help to draw a line perpendicular to the net force direction, which in this case is a vertical line. Any force to the left of the vertical line in this case would contribute positively to the centripetal force, whereas any force to the right would be negative).
Resolving forces vertically:
Since the bead is not moving up or down, the net force vertically is 0. Again, the line perpendicular to the vertical motion is a horizontal line, and clearly, F and N are above the horizontal whereas the weight force is below.
ii)
We need to eliminate F.
Consider
Multiply both sides by and make the subject.
Similarly, for the other equation, multiplying by sin ,
Now consider (1) – (2)
(
Divide both sides by
iii)
The bead remains in contact when
Note that and that .
Substituting this in, we have:
Example 8
[HSC 2009, Question 4]
A light string is attached to the vertex of a smooth vertical cone. A particle of mass is attached to the string as shown in the diagram. The particle remains in contact with the cone and rotates with constant angular velocity on a circle of radius . The string and the surface of the cone make an angle of with the vertical, as shown.
The forces acting on the particle are the tension, , in the string, the normal reaction, , to the cone and the gravitational force .
(i) Resolve the forces on in the horizontal and vertical directions.
(ii) Show that and find a similar expression for .
(iii) Show that if then
(iv) For which values of can the particle rotate so that ?
Solution 8
i)
Resolving forces horizontally:
Resolving forces vertically:
ii)
iii)
iv)
Note that and
Therefore, .
Example 9
[HSC 2006, Question 5]
A particle, , of mass is attached by two strings, each of length , to two fixed points, and , which lie on a vertical line as shown in the diagram.
The system revolves with constant angular velocity about . The string makes an angle with the vertical. The tension in the string is and the tension in the string is where and . The particle is also subject to a downward force, , due to gravity.
(i) Resolve the forces on in the horizontal and vertical directions.
(ii) If , find the value of in terms of and .
Solution 9
i)
Resolving forces horizontally:
Resolving forces vertically:
ii)
Substitute
Consider ,
Motion around a Circular Banked Track
In this section, we will study the forces of a particle travelling at constant velocity around a banked circular track. Since the track is banked at an angle to the horizontal to minimise the risk of slipping, the force of friction between the particle and the track needs to be considered. Optimum speed occurs when the friction force equals to zero. When dealing with frictional forces, it is required to direct them in the opposite direction to the sliding movement of the particle, in order for the particle to not slip. Let’s now derive the following results as required by the syllabus.
Now analyse the forces on a particle of mass traveling at constant velocity around a horizontal circular arc with radius of curvature on a track that is inclined at an angle to the horizontal, as shown below.
Example 10
[HSC 2005, Question 3]
The diagram shows the forces acting on a point which is moving on a frictionless banked circular track. The point has mass and is moving in a horizontal circle of radius with uniform speed . The track is inclined at an angle to the horizontal. The point experiences a normal reaction force from the track and a vertical force of magnitude due to gravity, so that the net force on the particle is a force of magnitude directed towards the centre of the horizontal circle.
By resolving in the horizontal and vertical directions, show that
Solution 10
Resolving forces horizontally:
Resolving forces vertically:
Consider
Also consider ,
Using a triangle, we see that:
Substituting this into the above we have:
Example 11
[HSC 2010, Question 4]
A bend in a highway is part of a circle of radius , centre . Around the bend the highway is banked at an angle to the horizontal.
A car is travelling around the bend at a constant speed . Assume that the car is represented by a point of mass . The forces acting on the car are a lateral force , the gravitational force and a normal reaction to the road, as shown in the diagram.
(i) By resolving forces, show that .
(ii) Find an expression for such that the lateral force is zero.
Solution 11
i)
Resolving forces horizontally:
Resolving forces vertically:
Consider
ii)
If lateral force force is zero, then