2. TRANSLATIONAL KINEMATICS NCEA Level 3 Physics

3. Vectors and Scalar Each quantity can have two types of component. They can either be a 1.Vector - A vector quantity is defined as a quantity that has both magnitude and direction. 2.Scalar - A scalar quantity is defined as a quantity that has magnitude only.

4. EQUATIONS OF MOTION As we learnt last year there are really 5 equations of motion. These are: v f = v i + at d = v i t + ½at 2 v f 2 = v i 2 + 2ad d = v i t - ½at 2 d = ½(v i + v f )t

5. Forces and Newton’s Laws There are three laws of motion: First Law An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. This law is often called "the law of inertia". What would happen if you don’t wear a seat belt and hit a boat?

7. Second Law Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object). However, the Second Law gives us an exact relationship between force, mass, and acceleration. It can be expressed as a mathematical equation: “F = ma”

8. Third Law For every action there is an equal and opposite re-action. Newton's Third Law. The rocket's action is to push down on the ground with the force of its powerful engines, and the reaction is that the ground pushes the rocket upwards with an equal force.

9. Exercise 1: Two tugboats pull a ship away from a wharf. The cables attached to the ship are at right angles to each other. The tension forces in the cables are T 1 = 36500 N and T 2 = 98500 N. The mass of the ship is 12000 tonnes and its acceleration away from the wharf is 0.0075ms -2. a.Calculate the combined pulling force on the ship. b.Calculate the direction of the combined pulling force as an angle to cable 2. c.Calculate the size of the force of the water resisting the motion of the ship. SOULTION: a. The resultant pulling force on the ship is the vector sum of the individual forces. F pull = 36500 +98500 F pull 2 = 36500 2 + 98500 2 F pull = 105000 N (3 sig fig) 36500N 98500N F pull

10. b. To find the angles between two directions, both directions must be either pointing towards each other or way from each other. The angle  is therefore the required angle: tan  = 3500  98500  = 20.3 o (round answer to 3 sig fig) c. The acceleration of the ship is caused by the total unbalanced force on it, which is in the same direction as F pull.  F net = ma = 12 x 10 6 x 0.0075 = 90000

11. The total unbalanced force is the vector addition of all the forces acting on the ship. The two pull forces have already been combined, and so the total unbalanced force will be the vector addition of the water resistance and the combined pulling force.  F net = F pull + F water F water = F net - F pull = (90000 – 105045) (use unrounded nos) = 15045 N “As the water resistance is a reaction force to the combined pull, it will act in the opposite direction.”

12. FORCE COMPONENTS Every vector is actually made up of two components which form together to produce a right angle triangle: A x y There is the horizontal (x) component and the vertical (y) component. These components can be anything from forces to velocities, momentum e.t.c. They are solved by combining both horizontal / vertical components for each vector i.e. all the horizontal vectors should add up to produce one overall vector as should all the vertical.

13. Example 2: A car of mass 920 kg is parked on a hill that makes an angle of 12 o with the horizontal. Calculate the brake force that is preventing the car from rolling down the hill. Solution: F Brake FgFg 12 O Sin12 o = F g (component)  F g F g (component) = 920 x 9.8 x sin12 o = 1874.5 N (down the hill) The brake force is preventing motion down the hill and so will act up the hill. It will balance the opposing gravity force component that acts down the hill. Thus :F Brake = 1900 N (up the hill)

14. CIRCULAR MOTION Centripetal acceleration. Centripetal force Summary Example exercise. Circular motion in a vertical circle. Example exercise.

15. v acac Acceleration a c in a circle is given by: a c = v 2 /r As an object moves around a circle they are accelerating as changing direction Linear Velocity v = d/t As d is in a circle 1 revolution equals 2  radians & size of circle is  radius ‘r’ then: v = 2  r/T CENTRIPETAL ACCELERATION

16. Newton’s 1 st Law “I want to go straight on!” Come back, I am a F v Circular motion requires a force, F, towards the centre of the motion F v A CENTRIPETAL FORCE Something has to provide it, e.g. the tension in a string

17. v v  Linear Velocity v = 2  r/T Centripetal Acceleration Centripetal Force SUMMARY

18. COMPLETE EXERCISE PAGE 48 – 49 RUTTER COMPLETE ACTIVITIES PAGE 49 RUTTER

19. F The necessary centripetal force, F is provided by the friction at the tyres Question If the track has a radius of 50m and the limiting frictional force is ½ of the car’s weight, find the car’s maximum speed before it slides off the track v = 15.7 ms -1 MOTION IN A HORIZONTAL CIRCLE

20. THE CONICAL PENDULUM The conical pendulum is made up of Mass tied to a length of string and swings in a circle. String produces a cone shaped path. 2 F acting weight force F g downwards & tension force T acing upwards in string. FgFg FTFT  i. Forces acting on the mass  FTFT F T cos  F T sin  ii. Resolving the tension force into 2 components m F T cos  F T sin  FgFg iii. Replacing the tension force with its 2 components

21. As m is not moving in a vertical direction, the 2 vertical forces must be equal and opposite. The resultant force on m is just the horizontal force. Since the resultant force acts towards the centre of the circle, it acts as a centripetal force, and causes the mass to move with a circular motion. m F T sin  Centripetal force is the horizontal component of the tension force

22. Example 3: The left hand side diagram shows a conical pendulum formed from a mass of 0.20kg moving in a circle of radius 0.40m, at a constant speed. The string makes an angle 30 o with the vertical. The forces on the mass are shown in the right- hand side diagram. Calculate the speed of the mass. 30 o 0.20kg 0.40m 30 o FgFg FTFT Solution: The gravity force F g is given by F g = mg F g = 0.20 x 9.8 = 1.96N Resolving the tension force into components shows that the vertical component of F T =F g 30 o F g = 2.0N FTFT F T cos 30 F T sin 30

23.  F T cos 30 = 1.96 N F T x 0.866 = 1.96 F T = 1.96  0.866  F T = 2.263N The resultant force is just the horizontal component of F T.  F T sin  = 2.263 x sin 30 = 1.132N This is the centripetal force which causes the mass to move in the circle. The speed of the mass is found from the formula F = mv 2 /r 1.132 = 0.20v 2  0.40 v 2 = 1.132 x 0.40  0.20 v 2 = 2.264 Taking the square root gives the speed of the mass as 1.5ms -1.

24. MOTION IN A VERTICAL CIRCLE FORCES ACTING: 1. Weight 2. Tension in the string mg T v Where F is the resultant force towards the centre At the bottom the tension has to provide a centripetal force AND support the weight

25. E K gain = E p lost gives velocity at the bottom At the bottom mg R v At the top mg R

26. Example 4: A girl swings a bucket containing 8.0kg of water in a vertical circle of radius 0.40m. She swings it at the slowest speed which keeps the water in the bucket. At the top of the swing: a.What is the gravity force of the water? b.What is the centripetal force acting on the water? c.What is the speed of the swing? Solution: a.F g = mg = 8.0 x 9.8 = 78N b. At the top of the swing, when the speed is the slowest necessary to keep the water in, the centripetal force is equal to the gravity force, 78.4N

27. c. The centripetal force is given by: F = mv 2  r 78.4 = 8.0 x v 2  0.40 v 2 = 78.4 x 0.40  8.0 v =  3.92 v = 2.0ms -1 This is the slowest speed at which the water will remain in the bucket

28. COMPLETE EXERCISES PAGE 48 – 50 RUTTER

29. Energy of Conservation In any object undergoing vertical circular motion the speed at which it travels is dependent upon the amount of energy is put in to it to keep it moving. In a roller coaster the energy needed to keep it going is given to it at the start and so its vertical circle is travelled without any outside injection of energy. The roller-coaster must be travelling fast enough at the bottom that the kinetic energy changed to gravitational potential energy as it goes up leaves it with enough kinetic energy at the top to give it the minimum speed it needs to keep moving in a circle

30. Example 5: A roller-coaster carriage, of total mass 560kg, travelling in vertical circle of radius 15m. a.At the top of the circle, what forces act on the carriage and in which direction do they act? b.Calculate the minimum speed needed to keep the carriage moving in a circle. c.Calculate the kinetic energy of the carriage when it is travelling at this minimum speed. d.Calculate the minimum kinetic energy the carriage must have at the bottom of the circle if it is to have enough energy at the top to keep going.

31. SOLUTION: a.Gravity force and the reaction force of the rails on the carriage both act vertically downwards at the top of the circle. b.At the minimum speed, the reaction force is zero and only the gravity force provides the centripetal force. F C = F g = mv 2  r mg = mv 2  r v 2 = 9.8 x 15 [the masses cancel out] v = 12ms -1 c. E K = ½mv 2 = 0.5 x 560 x 12.1244 2 [unrounded value of v used] = 41000J

32. d. As the carriage travels from the bottom of the circle to the top, it rises a vertical height of 30m and kinetic energy is changed to gravitational potential energy. E P(gained) = mgh = 560 x 9.8 x 30 = 164640J At the top of the circle, the carriage must have the kinetic energy of its minimum speed as well as the gravitational potential energy it has gained. It must therefore have all this energy as kinetic energy at the bottom of the circle. E K(bottom) = E K(top) + E P(top) = 41160 + 164640[use unrounded values] = 210000J

33. BANKING Driving a car or riding a bike around a bend is an example of circular motion. The bend is part of a circle and in order to go around the bend safely without sliding, there must be a sufficient centripetal force toward the centre of the circle. On a flat road with no banking, the centripetal force is provided by the friction between the tyres and the road. Depending upon friction to maintain movement round a corner is not safe. Water, ice and poor tyres reduce friction and result in skidding. A safer way is to bank the road which allows part of the reaction force from the roadway tp provide the centripetal force. This can be seen on cycle tracks, indoor athletics tracks and car raceways.  FgFg Resultant force = centripetal force Reaction force

34. Example 6: A plane turns horizontally with the wings banked at 45 0. a.Draw vectors to show the lift force and the weight acting on the plane. b.Sketch a vector triangle to show these two forces and their resultant. c.The mass of the plane is 1500kg. Calculate its weight force. d.Calculate the lift force. e.Use the vector triangle to calculate the centripetal force on the plane. f.Calculate the speed needed for a horizontal turn with a radius of 150m.

35. SOLUTION: FwFw Lift a. b. FwFw Lift Resultant c. F w = mg F w = 1500 x 9.8 F w = 14700N d. 45 o 14700NLift Lift = 14700  cos45 Lift = 20800N e. Resultant = 14700  tan45 F C = 14700N f. F C = mv 2  r =  (14700 x 150  1500) v = 38ms -1

36. COMPLETE EXERCISE PAGE 50 – 52 RUTTER