Circular Motion

Introduction• What is Newton’s First Law how does it relate to

circular motion?

• How does Newton’s second law relate to circular motion?

Acceleration

Vi

Vf𝑑𝜃

𝑑𝑥

𝑟𝑟

𝑑𝜃Vi

Vf𝑑𝑉

Vi

Vf𝑑𝜃

𝑑𝑥

𝑟𝑟

𝑑𝜃Vf

𝑑𝑉

𝑉=𝑑𝑥𝑑𝑡

𝑑𝜃=𝑑𝑥𝑟 𝑑𝜃=

𝑑𝑉𝑉

𝑑𝑥=𝑉 .𝑑𝑡 𝑑𝜃=𝑉 .𝑑𝑡𝑟

𝑑𝑉𝑉

=𝑉 .𝑑𝑡𝑟

𝑑𝑉𝑑𝑡

=𝑉 2

𝑟

𝑎=𝑉 2

𝑟

Acceleration• In uniform circular motion, which direction is the

acceleration? o There is no component of the net force adding to the speed of the

particle.o Therefore, the net force must always be perpendicular to the Velocity

Vector.

• The acceleration of a particle in uniform circular motion is always towards the centre of the circle.

• The particle is always deviating from it’s straight line path towards the centre of the circle.

Exam Question (VCAA 2010)

A racing car of mass 700 kg (including the driver) is travelling around a corner at a constant speed. The car’s path forms part of a circle of radius 50 m, and the track is horizontal. The magnitude of the central force provided by friction between the tyres and the ground is 11 200 N.

Question 1What is the speed of the car? (2 marks)

Question 2What is the acceleration of the car as it goes around the corner? (2 marks)

Exam Question (VCAA , 2009)

Question 3Draw an arrow to show the direction of the net force on the motorcycle.

On the diagram, draw the forces acting on the car. Remember the car is travelling in a circular path.

Centre of circular

path

On the diagram, draw the forces acting on the car. Remember the car is travelling in a circular path.

Centre of circular

path

Fg

FNFN

FfFf

On the diagram, draw the forces acting on the car. Remember the car is travelling in a circular path.

Centre of circular

path

Fg

FNFN

FfFf

Since the vertical forces are balanced, the net force (which we call centripetal force) is the sum of the sideways frictional forces.

Ball on a string

𝜽

Ball on a string

Fg

𝜽

Ball on a string

Fg

Ft

𝜽

Ball on a string

Fg

FtFg

Ft𝜽 𝜽

Ball on a string

Fg

FtFg

Ft

∑ 𝑭

𝜽 𝜽

Ball on a string

Fg

FtFg

Ft

∑ 𝑭

𝜽 𝛉

∑ 𝐹=𝐹𝑔×𝑇𝑎𝑛(𝜃)𝐹 𝑡=

𝐹𝑔

𝐶𝑜𝑠(𝜃)

Example: Ball on a string

A ball of mass 250 g is attached to string in a game of totem tennis. The string makes an angle of 40o to the vertical pole. Calculate: a. the net force on the ball b. the tension in the string c. the length of the string in terms of it’s speed, v?

𝟒𝟎𝒐

Banked Corners

Banked Corners

Fg

Banked Corners

Fg

FN

𝜽

Banked Corners

Fg

FN

𝜽

𝜽FNFg

Banked Corners

Fg

FN

𝜽

𝜽FNFg

∑ 𝑭

Banked Corners

Fg

FN

𝜽

𝜽FNFg

∑ 𝑭

∑ 𝐹=𝐹𝑔×𝑇𝑎𝑛(𝜃)𝑚𝑣2

𝑟=𝑚𝑔×𝑇𝑎𝑛(𝜃)

Banked Corners

Fg

FN

𝜽

𝜽FNFg

∑ 𝑭

∑ 𝐹=𝐹𝑔×𝑇𝑎𝑛(𝜃)𝑚𝑣2

𝑟=𝑚𝑔×𝑇𝑎𝑛(𝜃)

Banked Corners

Fg

FN

𝜽

𝜽FNFg

∑ 𝑭

∑ 𝐹=𝐹𝑔×𝑇𝑎𝑛(𝜃)𝑚𝑣2

𝑟=𝑚𝑔×𝑇𝑎𝑛(𝜃)

𝑣=√𝑟𝑔×𝑇𝑎𝑛(𝜃)  

Exam Question: VCAA 2010

Question 4On the diagram, draw an arrow to indicate the direction of the acceleration of the rider (1mark)

Exam Question: VCAA 2010

Question 5The circular path of the bicycle has a constant radius of 120 m, and the bicycle will be travelling at a constant 9 m s-1. What should be the value of the angle of the bank, θ, so that the bicycle travels around the corner with no sideways frictional force between the tyres and the track? (3 marks)

Banked Corners

Fg

FN

𝜽

𝜽FNFg

∑ 𝑭

The force diagram doesn’t consider friction.

Challenge: What would the force diagram look like if we considered friction? In which direction would the net force be?

Leaning into corners

Leaning into corners

Fg

FN

Ff

∑ 𝑭=𝑭 𝒇=𝒎𝒗𝟐

𝒓