cie-a2-physics-9702-theory-v2-znotes.pdf - TABLE OF CONTENTS 3 CHAPTER 1 3 CHAPTER 2 5 CHAPTER 3 6 CHAPTER 4 7 CHAPTER 5 9 CHAPTER 6 12 CHAPTER 7 13

cie-a2-physics-9702-theory-v2-znotes.pdf - TABLE OF...

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TABLE OF CONTENTS 3 CHAPTER 1 Motion in a Circle 3 CHAPTER 2 Gravitational Fields 5 CHAPTER 3 Ideal Gases 6 CHAPTER 4 Temperature 7 CHAPTER 5 Thermal Properties of Materials 9 CHAPTER 6 Oscillations 12 CHAPTER 7 Electric Fields 13 CHAPTER 8 Capacitance 14 CHAPTER 9 Magnetic Fields 18 CHAPTER 10 Electromagnetic Induction
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CIE A2-LEVEL PHYSICS//9702 P AGE 2 OF 37 20 CHAPTER 11 Alternating Currents 22 CHAPTER 12 Quantum Physics 24 CHAPTER 13 Particle & Nuclear Physics 25 CHAPTER 14 Direct Sensing 26 CHAPTER 15 Electronics 29 CHAPTER 16 Communication 32 CHAPTER 17 Medical Imaging
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CIE A2-LEVEL PHYSICS//9702 P AGE 3 OF 37 1. M OTION IN A C IRCLE 1.1 Radians Radian: one radian is the angle subtended at the center of the circle by an arc of length equal to the radius of the circle Angular displacement: the angle through which an object moves through a circle ± ² ³´ 1.2 Angular Velocity Angular velocity: the rate of change of the angular position of an object as it moves along a curved path µ ² ´ Period: the time taken by the body to complete the circular path once µ ² ¸ ² 2·¹ Relating angular velocity and linear velocity: º ² µ³ Example: The drum of a spin dryer has a radius of 20cm and rotates at 600 revolutions per minute. a. Show that the angular velocity is 63 rad s -1 b. Calculate, for a point on the edge of the drum, its linear velocity Solution: Part (a) Find rate per second 600rev ∶ 60sec 10rev ∶ 1sec Hence 1 revolution is 0.1sec Use angular velocity formula µ ² 0.1 ² 62.8 Part (b) Using relation between angular and linear velocity º ² µ³ ² 62.8 » 0.2 ² 12.6 ms -1 1.3 Circular Motion A body moving in a circle at a constant speed changes velocity since its direction changes. Thus, it is accelerating and hence experiences a force. Centripetal force: resultant force acting on an object moving in a circle, always directed towards the center of the circle perpendicular to the velocity of the object ¼ ² ½º ¾ ³ ² ½³µ ¾ Centripetal acceleration: derived by equating Newton’s 2 nd law and centripetal force ¿ ² ³µ ¾ or ¿ ² À Á  Example: A horizontal flat plate is free to rotate about a vertical axis through its center. A mass à is placed on the plate, a distance , 35cm, from the axis of rotation. The speed of rotation is increased from zero until the mass slides off the plate The maximum frictional force ¼ between the plate and the mass is given by the expression ¼ ² 0.72" Determine the maximum number of revolutions of per minute for the mass à to remain on the plate. Solution: The centripetal force on the particle is the frictional force so the max speed is when friction is at max Centripetal Force = Frictional Force ú ¾ ³ ² 0.72" Manipulating expression by adding µ and removing à Ã#µ³$ ¾ ³ ² 0.72Ã% µ ¾ ³ ² 0.72% Find the angular velocity µ ² & '.(¾»).*+ '.,- ² 4.49 rad s -1 Find radians covered in a minute using ratios 4.49rad ∶ 1sec 269.5rad ∶ 60sec Divide radians covered by to find revolutions 269.5 ² 42.9
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