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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