PHYSICS HSC NOTES
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Space Weight force on an object due to gravitational field π€πππβπ‘ = πππ π × 9.8 Projectile Motion Galileo’s analysis of projectile motion 1. Vertical motion is independent of horizontal motion 2. Horizontally the speed is constant 3. Vertically acceleration is uniform ( = 9.8m/s/s) Analysing and calculating projectile motions Step: 1. Work out horizontal and vertical components of initial velocity Calculate vertical flight time using π£ 2. Vsinπ V π = π’ + ππ‘ Vcosπ u = Vsinπ v= 0 a = 9.8m/s/s t = ½ of flight time Using Vertical flight time to calculate range using π = π£ × π‘ t = flight time v = Vcosπ d = horizontal distance object travels. 3. Maximum range Vertical Analysis π£ = π’ + ππ‘ 0 = ππ πππ − ππ‘ ππ‘ = ππ πππ π‘= Vsinπ π = 9.8π/π /π π‘ = π‘πππ π‘ππππ π‘π ππππβ max πππ’π βπππβπ‘ ππ πππ π π‘π‘ππ‘ππ = 2ππ πππ π‘π‘ππ‘ππ = ππππβπ‘ π‘πππ π Horizontal analysis πππ π‘ππππ πππππ = π‘πππ π ππππ π = 2ππ πππ π = π»ππππ§πππ‘ππ πππ π‘ππππ(πππππ) π π = ππππ π × π = π£ 2 π ππ2π π 2π£π πππ π = π£ 2 2π ππππππ π π (π ππ2π = 2π ππππππ π) πππ₯. π£πππ’π ππ π πππ = 1 π€βππ π = 90° π‘βπππππππ 2π = 90° π = 45° πβπππππππ πππ₯. π ππ π€βππ π = 45 ° π£2 π = π π ππππ max ππ sin 2π = 1 V π Vcosπ 2 Angles of projection for all other ranges other than max range π£ 2 π ππ2π π = π π π π ππ2π = 2 π£ −1 π π sin π£2 π= 2 Parabolic path of projectile Vertical Analysis 1 π = π’π‘ + ππ‘ 2 2 1 π¦ = ππ£πππ‘ π‘ + 2 (−9.8)π‘ 2 y = vertical distance Horizontal Analysis π₯ πβπππ = π‘ π₯ π‘= πβπππ Therefore π¦ = ππ£πππ‘ π₯ πβπππ π₯ = βππππ§πππ‘ππ πππ π‘ππππ – 4.9 2 πβπππ π₯2 Therefore path is parabolic (only for short distances due to curvature of earth surface) Space Important Values G = gravitational constant 6.67 × 10−11 Radius of earth = 6.4 × 106 π Mass of earth = 5.9742 × 1024 kilograms “G force” ππππππππ‘ π€πππβπ‘ π(π + π) π + π “π” = = = πππ‘π’ππ π€πππβπ‘ ππ π Example 2: 65kg physics student shot out of a cannon in a lift and is falling back to earth g= π(π+π) ππ = π(9.8+ −9.8) ππ =0 Example 3: 65kg physics student goes on circular rollercoaster loop and is upside travelling 15m/s what is his apparent weight 15m/s 11.25m/s/s acceleration due to loop Radius = 20m 9.8ms/s/s gravity g = 9.8m/s/s π= π£2 π = 152 20 = 11.25m/s/s ππππ = 11.25 – 9.8 = 1.45m/s/s F = ma = 65 1.45 =94.25N Circular motion Orbital velocity = v Centripetal force, πΉπ = ππ£ 2 π Centripetal acceleration/ acceleration of circular motion = π£2 π Escape velocity = speed of launch so that the speed when the object is a very large distance away is zero (relative to earth/ object launched from) Rockets Balloon (similar of like rocket combustion chamber) Air Particles Zero Net force Net forward push The removal of a force from a previously balanced system >>> net force >>> acceleration The force of air particles impacting on front end of balloon causes balloon to be propelled forward Combustion chamber Exhaust Fuel Oxygen Exhaust particles impact on front end of combustion chamber propelling rocket forward Due to conservation of momentumβπ × πππ₯βππ’π π‘ = π × βπππππππ‘ πΉπππ π= π βπ ππ£ − ππ’ πΉπππ = = βπ‘ π‘ πΉπππ × π‘ = ππ£ − ππ’ Thrust of Rocket = rate of change of momentum of rocket βπ = ππ ππππππ‘ βπ‘ = πβπ£ βπ‘ βππ£π = ( βπ × πππ₯βππ’π π‘ = π × βπππππππ‘ ) βπ = π£π βπ‘ βπ βπ But = − βπ‘ βπ‘ βπ‘ Therefore thrust on rocket = βπ βπ‘ = π£π Force of Gravity πΉππππ£ = πΊππ d = distance from centre of 1st mass to centre of 2nd mass π2 Acceleration due to gravity π= π π ππ = π= πΊππ π2 π πΊπ π = πππππππππ‘πππ ππ’π π‘π ππππ£ππ‘π¦ π2 Orbital Velocity of Satellite πΊπ ππ = π2 π£2 π= π 2 π£ πΊπ = 2 π π π£2 = πΊπ π πΊπ π£= √ π ππ’π‘ π£ = 2 π£ = >>>>> Orbital Velocity of Satellite 2ππ π 4π2 π2 π2 2 2 ∴ 4π π πΊπ = 2 π π 4π2 π2 = ( πΊπ ) π 3 >>>>>>Orbital period formula Low earth orbit 250km – 1000km high enough to avoid atmospheric drag while avoiding Van Allen radiation belt Geostationary Orbit An orbit at an altitude at which the period of the satellite’s orbit = period of orbit of earth (1 sidereal day 3mins 56 secs shorter than 24 hour day) If this orbit is over the equator the satellite would remain ‘parked’ over a fixed point on the earth’s surface. Useful for comm. Sat. as a receiving dish will only need to point in one direction. The radius of the orbit is calculated using Kepler’s law of periods and = 42 167 km (approx. 35800 above earth surface) Effect of the Earth’s motion on a launch To increase efficiency, reduce fuel consumption etc. Orbits of the earth other planets can be used to increase the speed of satellites/rockets etc. 1. Earth surface rotates at 1700km/h relative to sun and orbits it at 10700km/h – rockets are launched to the east to give them a boost – timed to give maximum boost in desired direction 2. Slingshot effect – a space probe is given a boost as it passes by and is pulled in by the planet’s gravitational effect; it acquires the speed of the planet relative to the sun Effective velocity Planet Motion of planet Probe travelling past planet Elliptical Orbits Most orbits are elliptical Gravitational Potential energy of orbiting Satellite π¬π = GPE = Energy of mass due to its position within a gravitational field. πΊππΈ = ππβ π = πππ π π = 9.8 β = βπππβπ‘ In falling objects KE gained = grav. PE lost 1 ππ£ 2 = ππβ 2 ππ£ 2 = 2ππβ π£ 2 = 2πβ π£ = √2πβ >>>> final velocity of falling object prior hitting ground The zero reference point of GPE is distance away because Fgrav = 0 at Work must be done to move objects altitude away from earth distance Therefore GPE at =0 Therefore at point near earth GPE < 0 Mathematical derived >>> πΊππΈ = −πΊ Mm π M = mass of planet; m = mass of object ; d = distance seperating masses (m) ππβ = ππ × β =πΉ ×π = π€πππ ππππ πΊππ πΊππ πΊππ π€πππ = × βπ + … .. 2 × βπ + 2 × βπ + π1 π2 π32 =∑ πΊππ π2 =∫ × βπ πΊππ × βπ π2 Kinetic energy of orbiting satellite πΎπΈ = 1 ππ£ 2 2 πΎπΈπππππ‘πππ π ππ‘πππππ‘π = 1 πΊππ × 2 π ππ π£πππππ‘ = √ πΊπ π πΎπΈπππππ‘πππ π ππ‘πππππ‘π = πΊππ 2π Total energy of orbiting satellite Total energy = = − πΊππ 2π −πΊ Mm π 1 πΊππ 2 π Therefore for escape velocity (πππ’πβ) 1 πΊππ ππ£ππ ππππ 2 − = 0 + 0 (Energy at very large distance away) 2 π 1 πΊππ ππ£ππ ππππ 2 = 2 π π£ππ ππππ 2 = π£ππ ππππ 2πΊπ π =√ 2πΊπ π Changing Satellite orbits Total final energy (KE + GPE) = total initial energy (KE + GPE) + energy provided by fuel πΎπΈππ’ππ = πΉππππ π‘ππ‘ππ ππππππ¦ − ππππ‘ππ π‘ππ‘ππ Example: Raising satellite from 300km to 500km orbit above earth using 141kJ/gram hydrogen fuel − 1 πΊππ 2 6.9 ×106 = − 1 πΊππ 2 6.7 ×106 + 141 × 103 × βπβπ¦ππππππ Magnetism Characteristics of Magnets 1. Like poles repel (N and N, S and S) 2. Unlike poles attract (N and S) 3. Strongly Magnetic metals (metals greatly affected by magnets) are Iron, Nickel and Cobalt. These can be made into magnets. Other metals are only slightly magnetic 4. Magnetically effected metals become Induced Magnetic Dipoles in a magnetic field therefore they are attracted by the magnet. Magnetic metals align to the magnetic “field lines” of a magnet 5. Direction of “magnetic” flow is from north to south on a magnet 6. Strength of Magnetic field is measured in Teslas (T) 7. Earth act’s like a giant magnets and compasses point north as they align with magnetic feild 8. S magnetic pole is near north pole, N magnetic pole is near south pole N end of magnet points near north andS end of magnet points close to south 9. Magnetic field of earth flips every few million years – evidence in solidified lava containing magnetic minerals surrounding volcanoes 10. All Magnetic fields are created by moving charges, Wires have magnetic field with no N and S Important Conventions Right hand thumb rule Thumb points in Direction of current Fingers curl in direction of magnetic field Right Hand Force rule Fingers in direction of magnetic field Thumb pointing direction of current in wire Palm “pushes” in direction of resulting force Right hand push rule – for generators Fingers in direction of magnetic field Movement of positive charge in wire Direction of force on positive charge in wire i.e. direction of induced current 3D drawing conventions Direction: Out of page – Arrow head shot through page from behind Direction: Into Page – Arrow shot into page feathers showing on back of arrow Magnets in Detail Inside magnets the magnetic field flows from S to N , outside the field flows from N to S. Forming “loops” Magnetism Terminology Faraday Called this “Magnetic Fluid” / “Magnetic Flux” symbol = ∅ π΅ = π»ππ€ ππ’πβ "magnetic flow" per square meter = ππππππ /π2 ∅ = π΄ = π ("πππ πππ ") Solenoids A Solenoid is a coil of conducting wire rolled into a helical spiral. Characteristics They generate magnetic fields with a north and south pole. The magnetic field inside is relatively uniform ππππ π’πππ ππ π€πππππ ππ Electric Motors Magnetics interact with the magnetic field generated by a solenoid causes the motor to turn Efficiency of a motor is improved by using multiple coils – which increases the time which magnetic force acts to rotate shaft. Using magnets with rounded ends to create pseudo radial pattern – changes the direction of force so that it acts more at a tangent to the circular path of the coil which increases the effective force on the rotation. Commutator Mechanical switch used the current every half revolution so the coil keeps rotating in a constant direction. Shaft The solenoid, Commutator are attached to the shaft and is used to transfer the kinetic energy so it can do work. The Commutator and solenoid are aligned so the gap between the two halves of the Commutator is perpendicular to the solenoid. Ammeter A motor with no Commutator and spring attached to shaft Magnetic force of 2 wires πΌ1 πΌ2 π πΉπππ ∝ π πΉπππ πΌ1 πΌ2 =πΎ πΎ = 2 × 10−7 (π΄ππππππ πΆπππ π‘πππ‘ ) ; π π π = πππππ‘β ππ π€πππ (πππππ‘β ππ π πππ‘ππππ πππππππππ) ; π = πππ π‘ππππ πππ‘π€πππ π€ππππ Amperes constant ‘K’ was defined and not experimentally determined – Used to determine 1 amp NOTE: If the two wires that are parallel are different in length the shorter length is taken – this is because the section of wire that has no wire parallel to it does not experience a magnetic force as the magnetic field of that section of wire does not interact with other magnetic fields. A current carrying wire only experiences a magnetic force when the magnetic field is perpendicular to the wire. Strength of Magnetic Field πΉπππ ∝ π΅ ; πΉπππ ∝ π (ππ’πππππ‘ ππ π€πππ); πΉπππ ∝ π (length of wire) πΉπππ = π΅ππ πΉπππ π΅= ππ π΅ = ππππππ‘ππ πππππ π π‘πππππ‘β ππππ ππ’ππ ππ πππ πππ (π) πΌ π΅ = πΎ π1 For around a current carryin gwires When wire is not perpendicular to magnetic field πΉπππ = π΅ππ π πππ β πβπππ π = πππππ πππ‘π€πππ πΌ πππ π΅ Generators Conv. I Fmag I e V e e e e e e e e Lenz’s Law An induced current will produce a magnetic field which will interact with the external magnetic field to produce a magnetic force opposing the change which caused it. I Fmag V Principle can be used to determine the current in wires moving through magnetic fields etc. by using right hand force rule with force in opp direction of movement. Generator Formulas πΉπππ = ππ£π΅ π = πβππππ π£ = π ππππ β« πΉπππ = π΅ππ ππ πππ ππ’ππ π π€πππ ππππ π‘π πππ£π π ππππ 1 πππ ππ π€πππ π‘π ππππ‘βππ = πΉ × π = πΉπππ × π π = ππ£π × π π€πππ π΅ππ‘π‘πππ¦ ππππ‘πππ = πππ’ππππ ππ£π × π = π = π£ππ πΈ. π. πΉ(πππππ‘πππππ‘ππ£π πππππ) = π£π΅π sin π πΈππ = π£π΅π π£π΅π × π‘ = π‘ (π£π‘ × π ) × π΅ = π‘ π΄ ×π΅ = π‘ π΅π΄ = π‘ ∅ ∅ = ππ π΅ = π‘ π΄ π = πππππ πππ‘π€πππ π£ πππ π΅ πΈππ = − β∅ βπ‘ πΉππππππ¦π πΏππ€ Back emf Every motor acts as a generator once in motion as it is a coil moving past magnet This causes a current to be generated in the opposite direction of the supplied current hence name Back emf. πππ‘ π£πππ‘πππ = π − ππππ’πππ ππππ πππ ππππ’πππ πΈππΉ ≤ π Effects: 1. Controls spin rate 2. Prevents motor burn out 3. If motor jams > no back emf > huge current > damages motor > burns insulation 4. Start up high current can burn motor > solution is variable resistor gradually opened Transformers Primary Coil = Input Iron core = Flux linkage Step down transformer Voltage in primary coil is higher than voltage in secondary coil Current in Secondary coil determines current in Primary ππ ππ = πππ πππππ π‘ππππ πππππππ ππ ππ ππ = ππ πππ πππππ π‘ππππ πππππππ ππ πΌπ = ππ πΌπ πΌπ ππ ππ = = πΌπ ππ ππ Principle of Conservation of energy ππ ≥ ππ πππ ππππ π‘ππππ πππππππ ππ = πΈπππππππππ¦ ππ πππππππ‘πππ × ππ Electrical Distribution Networks AC Generator Step up transformer Transformers High Voltage Transmission lines Local Network Power generated at 11000 V stepped up to 500 000V High Voltage wires are insulated from poles using ceramic cups 1. Cup shape to keep conducting rain off Why do this? 1. Especially with non-ideal transformers Can Transmit same power at: 2. High V, Low I 3. Low V, High I (π = ππΌ) Series of Step down High V and low I preferred as: 1. Decreases πΌ 2 π Losses in transmission lines π(πππ€ππ ππ βπππ‘ ππ π€ππππ ) = π × πΌ π2 = π π2π = 2 π = πΌ2 π Series of Step down transformers for safety in suburban areas Insulated Laminations ο· Iron cores of transformers – made from thin sheets of metal sandwiched with thin insulation ο· Reduces eddy currents ο· Heat ο· Power losses ο· Small sheets result in smaller eddy currents ο· Results in less power loss 3 phase power supply 3 generators in 1 – each coil connected to own set of slip rings – resulting in 3 currents Torque – Twisting force (angular accerlation) π = πππππ’π = πΉ⊥ × πππππ πππ£ππ‘ ππππ‘ππ = ππ (πππ€π‘ππ πππ‘πππ ) ππππ‘ππππππ = ππ΅πΌπ΄ πΆππ π π = ∠πππ‘π€πππ πππππ ππ ππππ πππ π΅ πππππ πππππ π΄ = ππππ ππ ππππ
