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# Introductory Physics I PHYS 6

UCSC

GPA 3.78

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This 24 page Class Notes was uploaded by Ms. Schuyler Kozey on Monday September 7, 2015. The Class Notes belongs to PHYS 6 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 55 views. For similar materials see /class/182309/phys-6-university-of-california-santa-cruz in Physics 2 at University of California - Santa Cruz.

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Date Created: 09/07/15

Using MATHCAD to calculate the trajectory of a projectile taking into account air resistance D 001 Drag coef cient g 98 Acceleration of gravity m 05 Mass V 0 z 30 Initial speed and direction on 30 deg voxv0costx vox2598 X00 Initial velocity components voyvosintx voy15 y00 n 250 At 001 Number of iterations and the time step size i 1 n ti iAt Calculate the time of each step 1quot I X0 X 0 yo y 0 v V Initial position and velocity X 0X V y V for ie 1 n Loop over the time steps 2 2 V V x V y Calculate the speed D v v X ax m Calculate the acceleration in X D v v y ayeg Calculate the acceleration in y l 2 Xiexi 1 V x 3 aX At Add on the change in X position I 2 yieyi 1 V y A ay At Add on the change in y position v X v X a X At Add on the change in X and y velocity components vy vyayAt n0i quoti Save the results for plotting n1i yi Physics 6C Introduction to Physics Iquot Electricity and Magnetism Robert Johnson Professor of Physics rjohnsonscippucscedu November 22 2006 Physics 6C Lecture 24 AC Current and Voltage Graph of voltage versus time AC 9 R Vt 2 V0 cos wt 0 mean voltage is zero Graph of voltage squared versus time mean Vt2 2 V02 1 cos2at 0 1 V The rms voltage is Vms E ltV2gt To 0 0395 t1 1395 2 2 November 22 2006 Physics 6C Lecture 24 2 Phasors The emf oscillates a 5 as S Sncoswr Peak 39 80 emf 0 I I t T 2T 50 The oscillation period is T lf 2770 Copynghl e 2004 Pearson EducAlion Inc publishing a Addison Wesley b The length of 39The phasor rotates the phasor is Elyuquot ccw at angular frequency wt The phase angle is ml l m 39 K The instantaneous emf value E coswt is the projection of the phasor onlo the horizontal axis The tip of the phasor goes T once around the circle in time T Copyright e 2004 Pearson Education lncn puhlishing as Addison Wesley Represent the cosine function as the xcomponent of a rotating vector This provides a graphical means to keep track of both amplitude and phase Note for the mathematically inclined this is equivalent to representing the cosine function as the real part of a complex number Physics 6C Lecture 24 November 22 2006 AC Circuits Resistance Ac XIER Vt 2 V0 cos wt 10 V0 It cosat U R Vt R 0 Kirchhoff s rule V t R Energy is dissipated turned into heat by the resistor P av 2 P 12R 2 Vcos 002 2 2 Mi Vrms 2 R R November 22 2006 Physics 6C Lecture 24 Phasors for Resistive Circuit a 00 Voltage phasor length VR vR and iR VR vR VRcos wt IR iR IRcosmt 0 I I t 5quot T 39 quotR VR Instantaneous The resistor voltage and current and voltage current oscillate in phase Cupyngm lt0 2004 Pemon EducAtionr Int puhhshing as Addison Wesley Comrrghl a 21104 Mm amtmmn Inn publishing u Addison Wesley November 22 2006 Physics 6C Lecture 24 5 AC Circuits Capacitance Vt O Kirchhoff s rule L VtVocosat l I W idQ 1 1 d1 Cdt C Ht 2 Coil f Con sin wt 11 13C cosat Capacitive reactance XC E i Energy is stored NOT dissipated max O C November 22 2006 Physics 6C Lecture 24 6 Phasors for Capacitive Circuit a C if peaks flquot before vc peaks We 521 that the current atS the voltage b 90 y b IC The current phasor leads a o vC and 1C u the voltage phasor by 90 C Voltage v wt 177 vC C 2 Voltage phasor VC wt 0 II I t lk I F 5T T 1C VC ZC These are the instantaneous C CLUTth C Copyright 20m Mmquot Edueulion 1m pummuiug n Addison Wesley November 22 2006 current and voltage Physics 6C Lecture 24 Copyright 0 2004 Pearson EdLeaLionr Inc publishing as Addison Wesley Exercise 359 A 20 nF capacitor is connect across an AC generator that produces a peak voltage of 50 V a At what frequency f is the peak current 50 mA b What is the instantaneous value of the emf at the instant when Vt 50cos27rft c Notation our textbook uses 1 to represent the timedependent current and I to represent the amplitude 1 C 2 1t C cosat November 22 2006 Physics 6C Lecture 24 RC Circuit Kirchhoff s loop rule holds at every instant in time i gt R v 8tVRtVct We want to find the resulting current 6 5 50005 aquot which generally will not be in phase V C AAAA with the voltage i l C 1 t maX cosat We have to find both the amplitude max and the phase VR 2 ImaXR cosat The resistor voltage is in phase with current 1 7r The capacitor voltage lags behind VC 2 max wC COS a j the current by 90 degrees November 22 2006 Physics 6C Lecture 24 Physics 6C Introduction to Physics Iquot Electricity and Magnetism Robert Johnson Professor of Physics rjohnsonscippucscedu November 27 2006 Physics 6C Lecture 25 RC Circuit Charging R 9 W 1 T C E 05 Qt CV0 1 e RC l l This is an example of a transient 0 1 2 3 4 5 response The current changes t with time after the switch closes 1 and eventually reaches a constant l l l l value of zero For AC circuits we are not 1t 05 interested in the transient response but instead we consider only the behavior after the circuit has been on for a long time 0 0 1 2 3 4 5 November 27 2006 Physics 6C Lecture 25 RC Circuit with AC Source Kirchhoff s loop rule holds at every instant in time 1 gtR V gt 51 W W gt i R We want to find the resulting current 6 5 5000st which generally will not be in phase 39th th It T C VC WI e vo age I 1 t maX cosat Assume that the source has been We have to find both the amplitude max turned on for a long time so all and the phase transients have died out VR 2 ImaXR cosat The resistor voltage is in phase with current 1 7r The capacitor voltage lags behind VC 2 max wC COS a j the current by 90 degrees November 27 2006 Physics 6C Lecture 25 AAAA quotW k lt x RC Circuit C rr 50 VRfVcf 7f 80 cos cot ImaXR cosat 5 max Ecoskot 15 3 This equation can be solved for both max and by using trig identities but it is easier to do it graphically using phasors The algebra then just looks like vector addition 50 all1 ch ImamR2 1aC2 can 1 max tan VC r gtR VR coRC 7C 8 As in a circuit with just a capacitor the voltage lags behind the current but by 2 2 less than 90 degrees gt 0 Note Z 1 R XC November 27 2006 Physics 6C Lecture 25 Problem 3536 II 159 uF a Evaluate VR at emf frequencies 100 300 1000 3000 and 10000 Hz V b Graph VR VS frequenCY 10 V colgwt 6 WM 1mnwm This is an example of a high pass filter Note that for f0 DC the current must be zero and therefore Vout is 0 November 27 2006 Physics 6C Lecture 25 5 159 F Vin 10 VCosmt 100 V out The output voltage across resistor is in phase with the current which leads the input voltage by an angle Mt atan ZnfRC l l i Low f the capacitor impedance dominates and Y current leads voltage by 900 l l 7 03 i High f the reSIstor 02 impedance dominates and current is in hase with Vin 01 p U l 0 2000 4000 8000 110 6000 f November 27 2006 Physics 6C Lecture 25 Problem 3536 R E 100 V 10 C 1591076 m 1H m R Vent Vin39 2 R24 1 1 2nfc vout100 0994 vout1000 7068 vout10000 995 vout300 2871 vout3000 9486 High Pass Filter 10 V0mm 5 4 0 0 2000 4000 f6000 8000 1104 6 Problem 3536 m 100 lgn15910 6 Vin 10 n 2711500 con nued Mt Vin0030 t This plot demonstrates for one frequency 1m L I MK I 31716ng 1500 Hz how the lR2 Lg 3RC resistor and capacitor wCJ voltages add together 10 ImaXcosoat 4 at all times to yield the voltage of the source 1 N 71 cosmt I DC 2 VRt ImaXRcosoat 4 VCt ImaX 10 VRO To C VRtVCt V02 10 0 210394 410394 610394 810 4 November 27 2006 Physics 6C Lecture 25 Problem 3537 a Evaluate VC at emf frequencies 1 3 10 30 and 100 kiloHertz 16 I b Graph VC vs frequency Vin 10 V coswt This is an example of a low pass filter Note that forf0 DC the current must be zero and therefore Vout is Vin Vout 39210 uF Copyngm lt9 2004 l39eman Education Inc puhliahing an Addtsnn Wesley November 27 2006 Physics 6C Lecture 25 10 Vc05wt v9 AIMA VVVV 160 in Vout 10 MP The voltage across the capacitor lags behind the current by 90 50 Vout has a phase angle M 39iannfRc 2 Low f the capacitor looks like an open circuit so Vout is the same as Vin 7 the current is in phase with Vin while Vout lags by 90 i High f the resistor dominates so November 27 2006 V Problem 3537 V0m1000 995 Vom10000 7052 Vom100000 099 V0m3000 9574 Vom30000 3147 Low Pass Filter Vout 5 HE U 0 2104 4104 6104 8104 1105 f Physics 6C Lecture 25 9 1 7 2751RC E Problem 3537 Mi 2 atan I 10 100 111603 1104 110 It is very common to plot frequency on a log scale In that case the same phaseangle plot looks as shown here The point where the phase angle is 45 is called the crossover frequency It occurs where 1 a E6251ltHzorf10kHz C November 27 2006 Physics 6C Lecture 25 AC Circuits Inductance d1 Kirchhoff s rule V0 8 0 8 2 LE V Lg AC L VtV0 coswt l I L 1 1r Z der V0 1t T Icoswtdt Ht 2 iV0 sin wt wL V0 It cos wt 1 0 M lt 2 Energy is stored NOT dissipated November 27 2006 Physics 6C Lecture 25 11 Phasors for Inductive Circuit 9 L a iL peaks iT after vL peaks b VL and iL We say that the current lags 5 the voltage by 90 VL VL IL Voltage phasor cot 77 0 I T L VL IL II VL Oltage v quotThe current phasor lags L the voltage phasor by 90 C anm Q 1004 mmquot Edum m hmquot WHEN M Maw Wm Cupmgm 0 20m Peumm Education 1m pnblnllinu 15 Amman wtalgy November 27 2006 Physics 6C Lecture 25 12 Reactance All 3 elements provide an impedance Z to the flow of current but one has to specify a phase difference between current and voltage as well as a change in amplitude Assume that the voltage is given by Vt 2 V0 COS wt 1 max cosat V W 102E0cosmt ZRand 0 Reactance V0 7 1 4Hi It 1aCcosat ZZXCZEand Zg wL November 27 2006 Physics 6C Lecture 25 13 RL Circuit with AC Source Problem 3547 The current is lagging behind the V voltage negative phase lt 0 L 0V0 vt V0 cos wt VR z39t max cosat V0 quotVRZ VL2 mam122 coL2 1VL 710L 2 2 tan tan N t Z1iR X VR R 0 e L November 27 2006 Physics 6C Lecture 25 14

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