Circular Motion
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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 ππ₯ π ππ π Vf Vi ππ Vf ππ Vi ππ₯ π ππ π Vf ππ Vf ππ₯ π= ππ‘ ππ₯ = π. ππ‘ ππ₯ π π. ππ‘ ππ = π ππ = ππ ππ = π ππ π. ππ‘ = π π ππ π2 = ππ‘ π π2 π= π ππ 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. π= π£2 π = 4π2 π π2 πΉ = ππ 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 1 What is the speed of the car? (2 marks) Question 2 What is the acceleration of the car as it goes around the corner? (2 marks) Exam Question (VCAA , 2009) Question 3 Draw 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 FN FN Ff Fg Ff On the diagram, draw the forces acting on the car. Remember the car is travelling in a circular path. Centre of circular path FN FN Ff Fg Ff 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 π½ Ball on a string π½ Fg Ball on a string π½ Ft Fg Ball on a string π½ Ft Fg Ft π½ Fg Ball on a string π½ Ft π½ Ft π Fg Fg Ball on a string π½ Ft Ft π Fg π Fg πΉ = πΉπ × πππ(π) πΉπ πΉπ‘ = πΆππ (π) 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? πππ Banked Corners Banked Corners Fg Banked Corners FN π½ Fg Banked Corners FN π½ Fg Fg π½ FN Banked Corners FN π½ Fg Fg π½ FN π Banked Corners FN π½ Fg πΉ = πΉπ × πππ(π) ππ£ 2 = ππ × πππ(π) π Fg π½ FN π Banked Corners FN π½ Fg πΉ = πΉπ × πππ(π) ππ£ 2 = ππ × πππ(π) π Fg π½ FN π Banked Corners FN π½ Fg πΉ = πΉπ × πππ(π) ππ£ 2 = ππ × πππ(π) π π£= ππ × πππ(π) Fg π½ FN π Exam Question: VCAA 2010 Question 4 On the diagram, draw an arrow to indicate the direction of the acceleration of the rider (1mark) Exam Question: VCAA 2010 Question 5 The 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 FN π½ Fg 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? Fg π½ FN π Leaning into corners Leaning into corners FN πππ π = ππ = π Ff Fg
