Module M3: Relative Motion

Transcription

1 Mdule M3: Relative Mtin Prerequisite: Mdule C1 Tpics: Nninertial, classical frames f reference Inertial, relativistic frames f reference References: Fwles and Cassiday Analytical Mechanics 7th ed (Thmsn/Brks/Cle 2005) Chapter 5 Thrntn and Marin Classical Dynamics f Particles and Systems 5th ed (Thmsn/Brks/Cle 2004) Chapters 10 and 14 Taylr Classical Mechanics (University Science Bks 2005) Chapters 9 and 15 Read thrugh A Graphing Checklist fr Physics Graduate Students befre submitting any plts in this mdule. See the curse website fr the link.

2 MODULE M3 Name: Term: Prblem Max 1 st 2 nd Final 0 textbk TOTAL 80

3 I. Nninertial, classical frames f reference A. The general case What fllws is a cndensed versin f the theretical backgrund fr prblems in this sectin. Yu may find it useful t supplement this material by reading frm ne f the abve references. Cnsider tw crdinate systems, ne the unprimed system and ne the primed system. Assume initially that their crdinate axes are parallel t each ther but that their rigins, O and O, d nt necessarily cincide. Let r be the vectr that pints frm O t a given pint P; let r " be the vectr that pints frm O t the same pint P, and let R be the vectr that pints frm O t O. That is, a persn lcating the pint P in the unprimed system wuld use r while a persn lcating P with respect t the primed system wuld use r ". R describes the relative psitin f the tw crdinate systems. Assuming that measuring devices behave the same way in bth frames f reference (an assumptin that we will questin in the next sectin), vectr additin implies: r = r " + R (1) Differentiating twice with respect t time, again assuming that differentiatin is viewed the same in each frame f reference, v = v " + V ( 2) a = a " + A 3 where v = d r a = d v ( ) v " = d r " a " = d v " V = d R A = d V Let us assume that the unprimed frame f reference is inertial (it des nt accelerate). Newtn s 2 nd Law applies t inertial frames f reference, and hence F = ma = m a " + ma (4) Equatin (4) applies in a straightfrward manner as lng as the tw sets f crdinate axes remain parallel. If the primed system rtates with respect t the unprimed system, things get a bit mre cmplicated. T see this, suppse that the rigins f the unprimed and primed crdinate systems cincide, and they remain the same, but that the primed crdinate system rtates with respect t the unprimed system. Let ω be the angular

4 velcity f the primed system. Fr instance, imagine painting a crdinate system n a disk and then setting the disk spinning with angular velcity ω. The disk wuld represent the primed frame f reference while the grund wuld be the unprimed frame. Since the rigins f the crdinate systems cincide, R = 0, s that by (1) r = r ". That is, xˆ i + yˆ j + zk ˆ = x " i ˆ " + y " ˆ j " + z " k ˆ " Once again, assuming the unprimed crdinate system is inertial (and hence nt rtating), then differentiating yields dx i ˆ + dy ˆ j + dz k ˆ = d x " i ˆ " + d y " ˆ j " + d z " k ˆ " + x dˆ i " " + y dˆ j " dˆ " + z k " " v = v " + x " dˆ i " + y " dˆ j " + z " d k ˆ " Here v is the velcity f pint P as measured in the unprimed frame f reference; v " is the velcity f that same pint but measured in the primed (rtating) frame f reference; and the final three terms in the equatin crrect fr the fact that the measurement is made in a rtating frame f reference. In a prblem belw, yu will shw that dˆ i " = ω i ˆ " and similarly fr the ther unit vectrs, s that v = v " + ω r " (5) Repeating this prcess yields the acceleratin transfrmatin: a = a " + d ω r " + 2 ω v " + ω ω r " ( ) (6) The left hand side represents the acceleratin as measured in the nn-rtating (inertial) frame. The first term n the right hand side represents the acceleratin an bserver in the rtating frame wuld measure, the secnd crrects fr the pssibility that the rtatin rate f the frame is nt cnstant, the third term is the Crilis term, and the furth term is the centripetal acceleratin term. These last three terms can be thught f as the price yu pay fr measuring the acceleratin with respect t a rtating set f crdinate axes. We can cmbine equatins (4) and (6) t arrive at a general transfrmatin between crdinate systems, where the primed system is bth accelerating and rtating with respect t the unprimed system: a = a " + d ω r " + 2 ω v " + ω ω r " ( ) + A (7).

5 Prblems: 1. A pendulum is held at rest with respect t a subway car. When the car accelerates, it is bserved that the pendulum is deflected backward by an angle, θ. a. Shw that Newtn s Secnd Law des nt apply when measurements are made in the frame f the subway car. b. Determine the acceleratin f the subway car as a functin f g and θ. 2. Fill in the gaps in the derivatin f the equatins fr measuring velcity and acceleratin in rtating crdinate systems. In particular, a. shw that dˆ i " = ω i ˆ ". b. derive equatin (5). c. derive equatin (6). Hint: the derivatin in Fwles and Cassiday is, in my pinin, uncnvincing since it relies n an undefined term and an unprven intermediate step. It is straight frward t prve equatin (6) by ging thrugh the same type f direct differentiatin that is necessary t derive equatin (5). 3. A rtating restaurant is typically set t cmplete a revlutin in ne hur. Cnsider a patrn walking 1.2 m/s radially ut frm the center at a distance f 10m frm the center. Estimate the size f the centripetal acceleratin and the Crilis acceleratin fr the patrn. Will either acceleratin likely be nticed by the patrn? On what basis can yu decide that? B. Prjectile Mtin Fllwing the apprach in Fwles and Cassiday s Analytical Mechanics, nte that fr an bject suspended by a string and at rest in a frame f reference attached t the surface f the earth, Newtn s 2 nd Law cmbined with equatin (7) yields F = ma T + mg = ma T + m( g A ) = 0 where g is the acceleratin due t gravity at the earth s surface as cmputed by the Universal Law f Gravitatin and T is the tensin in the string. Had we ignred the fact that that the earth s surface is nn-inertial, we wuld have written (as is cmmn in first year physics bks) T + mg = 0 Thus what we typically measure as the acceleratin due t gravity is g = g A Frm nw n, we will use g t represent the measured acceleratin due t gravity. It has incrprated in it a crrectin fr the fact that any pint n the earth s surface is accelerating. Treating the center f the earth as inertial, then A just represents the centripetal acceleratin f a pint n the earth s surface, whse directin pints in

6 twards the axis f rtatin (nt generally the center f the earth) and whse magnitude is ω 2 ρ where ω is the rtatinal velcity f the earth and ρ is the distance frm the pint n the surface t the axis f rtatin. Nw cnsider a prjectile near the earth s surface. Neglecting air resistance, F = ma mg & = m a # + d ω r # + 2ω v # + ω ( ω r #) + A ) ( ' + * g A ( ) = a # + d ω r # + 2 ω v # + ω ( ω r #) g = a # + d ω r # + 2 ω v # + ω ( ω r #) The secnd term n the right hand side is zer, assuming the earth rtates at a cnstant rate. The last term n the right hand side is generally smaller than the thers and hence can be drpped. Thus we have a " = g 2ω v " (8) The angular velcity vectr fr the earth pints frm the suth ple t the nrth ple. Hwever, when making measurements n the surface f the earth, we generally use a lcal crdinate system. Let the x axis pint east, the y axis pint nrth (bth tangent t the earth s surface) and the z axis pint vertically upward (equivalently, utward alng a radius f the earth). Let λ be the latitude f the rigin f ur primed crdinate system (that is, ur bservatin pint n the surface f the earth). In terms f this angle, ω = ω csλˆ j $ +ω sinλˆ k $ s that (8) can be rewritten ( ) x " = 2ω z " csλ y " sinλ y " = 2ω x " sinλ z " = g + 2ω x " csλ (9) Fr an bject launched frm the rigin with initial velcity v " = x " ˆ i " + y " ˆ j " + z " ˆ k ", these equatins may be integrated nce with respect t time t yield ( ) + " x " = 2ω z " csλ y " sinλ y " = 2ω x " sinλ + " y z " = gt + 2ω x " csλ + " x z (10) Using equatins (10) as substitutins int equatins (9) and drpping terms f rder ω 2, equatins (9) can be integrated twice t yield

7 ( ) = 1 3 ωgt 3 csλ ωt 2 " x " t ( ) = " y " t y t ω x " t 2 sinλ ( z csλ y " sinλ) + x " t (11) ( ) = 1 2 gt 2 + " z " t z t +ω x " t 2 csλ Prblems: 4. Derive equatins (11) frm equatins (9) and (10) as utlined abve. 5. Suppse we were t cnstruct a 100m tall twer and drp a rck frm the tp f the twer. The Crilis effect will cause the rck t land ff-center. Hw far away frm the center f the twer base will the rck land, and in which gegraphic directin? In additin, prvide a qualitative explanatin fr why the directin is what it is. Nte that the latitude f Ypsilanti is abut Inspectin f equatins (11) shws that the maximum height achieved by a prjectile is influenced by the Crilis effect. a. In which directin (nrth, suth, east r west) shuld an bject be launched in rder t maximize the peak f the trajectry, and hw high will it rise? Assume the angle f launch (with respect t the hrizntal) and the launch speed are nt changing. b. In which directin shuld an bject be launched in rder t minimize the peak f the trajectry, and hw high will it rise? c. Fr an bject launched with a speed f 50 m/s and at an angle f 45, what is the difference in heights between (a) and (b)? Take the latitude at the launch pint t be 42, and neglect air resistance. II. Inertial, Relativistic Transfrmatins I will nt re-derive sme f the basic equatins here, nr g in any detail int hw the pstulates f relativity came t be. This material is easily fund in almst every Mdern Physics textbk as well as in many intrductry textbks. The purpse f this sectin is t intrduce the transfrmatins in the cntext f 4-vectrs and t lk at energy and mmentum relatins in mre detail. Cnsider tw frames f reference, S and S, such that their rigins cincide at t=0 and S mves in the +x directin at a velcity V with respect t x. An event bserved t take place at time t and psitin (x,y,z) in frame S will be bserved in frame S t take place at time t and psitin (x, y, z ), where

8 ( ) x " = γ x Vt y " = y z " = z % t " = γ t Vx ( ' * & ) γ = 1 c 2 1 ( V c) 2 (12) If these equatins are nt familiar t yu, nw wuld be a gd time t get a Mdern Physics text bk and read its intrductin t Special Relativity. The rest f this may nt make much sense therwise. I will nw recast these equatins in terms f the cmpnents f the spacetime 4-vectr: x 1 = x x 2 = y x 3 = z x 4 = ct (13) It is als useful t intrduce β=v/c s that γ = ( 1 β 2 ) 1 2. The transfrmatin equatins are nw x 1 " = γx 1 γβx 4 x " 2 = x 2 x " 3 = x 3 x " 4 = γβx 1 +γx 4 (14) I will intrduce the fllwing ntatin: x refers t the cnventinal 3-cmpnent psitin vectr. x refers t the 4-vectr defined by " x 1 % $ ' x 2 x = $ ' $ x 3 ' $ ' # & x 4 Equatins (14) can then be written cmpactly as x " = Λx (15) prvided we define

9 & γ 0 0 γβ) ( Λ = ( + ( ( + ' γβ 0 0 γ * (16) We will adpt as ur mre general definitin f a 4-vectr any quantity which, when measured by bservers in frames S and S as defined abve, transfrms accrding t q " = Λq (17) where Λ is as given by equatin (16). Fr a cnventinal vectr, we calculate its length by a mdified Pythagrean therem: x = x x x 3 2 A prperty f this definitin f length is that it is invariant under rtatins. We can view equatin (17) as the 4-vectr generalizatin f a rtatin. In this case, hwever, the spatial axes are nt being rtated. Rather, the #1 and #4 axes are being rtated. It can be shwn that if we define the 4-vectr length as q = q q q 3 2 q 4 2 (18) then this length is invariant under the transfrmatin described by equatin (17). The final negative sign makes this equatin lk a little dd. As a result, sme authrs will incrprate a factr f i int the 4 th cmpnent, which then prduces a negative sign when squared. This gets rid f the negative sign in (18) but requires yur 4 th axis t be imaginary. Prblems: 7. Verify that q " = q fr any tw 4-vectrs related t each ther by equatin (17). 8. Suppse yu set up an bservatry in space with tw synchrnized clcks m apart. A miniature prbe has been cnstructed s that it will emit a signal nce every µs. This prbe is launched such that it passes parallel t the line determined by the clcks, and it des s at a cnstant speed f 0.200c. Hw many times will the prbe emit a signal as it passes between yur tw clcks? Answer this questin using the transfrmatin equatins (14) (nt using simplistic time dilatin arguments).

10 Other 4-vectrs Starting frm the transfrmatin, % t " = γ t Vx ( ' * & ) c 2 we cnsider tw nearby pints in spacetime: % d t " = γ Vdx ( ' * & c 2 ) If in the unprimed frame f reference, dx=0, then this frame f reference wuld crrespnd t ne in which the tw events take place at the same psitin. Such a situatin wuld arise if yu are lking at tw events which take place n the same bject, and the bject is at rest. In this case, d t " = γ prper = d t " γ = 1 β2 d t " Time measured in the rest frame f an bject is knwn as prper time. This time is relevant t a clck munted n a mving bject (the clck reads the bject s prper time) as well as t the time that gverns the decay prbability fr an unstable particle. It can be shwn that if yu differentiate the spacetime 4-vectr f a particle with respect t its prper time, the result is a new 4-vectr: # γv x & % ( % γv y ( % γv z ( % ( $ γc ' If we multiply this by the mass f the particle, we btain the relativistic 4-mmentum: # γmv x & % ( γmv y p = % ( % γmv z ( % ( $ γmc ' The first three cmpnents are γ times the nn-relativistic mmentum. In the limit that v<<c, these reduce t the n-relativistic mmentum. The furth cmpnent can be interpreted as the particle energy, E, divided by c. In particular, if yu d an expansin fr v<<c, yu find that E=γmc 2 has as its lead term mc 2 and its secnd term, the classical kinetic energy, 1/2 mv 2. The lead term mc 2 is interpreted as the rest energy f the particle, r its mass-energy equivalent. Anything in excess f that is the particle s kinetic energy. Thus we have: p = γmv (19) E = γmc 2 (20)

11 KE = ( γ 1)mc 2 (21) With these definitins, then the principles f mmentum cnservatin (in the absence f unbalanced external frces) and energy cnservatin apply. These will be explred in prblems belw. Prblems: 9. Shw that fr v<<c, the energy E can be expanded in the frm E = A 1 + A 2 v 2 + A 3 v 4. Determine the cnstants A 1, A 2, A 3, and cmment n the significance/interpretatin f each term. 10. Shw that a. p p = - mc ( ) 2 ( ) 2 + p 2 c 2. This equatin implies that fr a massless particle, E=pc, a result b. E 2 = mc 2 that will be helpful in prblem # A prtn is accelerated in a lab t the pint that its kinetic energy is 537MeV. a. Calculate γ fr this prtn. b. The prtn cllides with a secnd prtn that is at rest in the lab frame. Shw that the center f mass frame f reference, defined as the ne in which ttal mmentum f the system is zer, has a velcity with respect t the lab frame given by v T = γ 1+γ v x. Here v x and γ refer t values assciated with the accelerated prtn in the lab frame (that is, values yu dealt with in part a). Nte that yu cannt slve this prblem by calculating the center f mass velcity in the nnrelativistic apprximatin. Yu shuld begin by seeing what it will take fr the ttal mmentum t be zer. Als nte, that by the time yu cmplete the next part f this prblem, there will be several γ s and β s invlved, depending n the particle and frame f reference. An additinal set is assciated with transfrming between the tw frames f reference. Identify thse with the T subscript. c. Hw much kinetic energy des each prtn have in the center f mass frame? d. Discuss why the ttal kinetic energy in the center f mass frame puts an upper limit n the energy available t create new particles, and why this limit is better than using the kinetic energy in the lab frame. 12. A particle f mass m mves in the +x directin with a speed f 0.75c. Determine its speed in a frame f reference mving in the x directin at 0.50c. D this by calculating the 4-mmentum in the riginal frame f reference, transfrming the 4-mmentum int the new frame, and determining the speed frm the transfrmed 4-mmentum.

12 13. A pin (π - ) can decay int a mun (µ - ) and a mun antineutrin (ν µ ). The rest energy f the pin (mc 2 ) is 140 MeV, while that f the mun is 105.7MeV. The mun antineutrin can be treated as massless (see prblem 10). a. Shw that principles f energy cnservatin and mmentum cnservatin require that in the center f mass frame, the energy f the neutrin must be 30.1 MeV. Make sure yu use the relativistic energy expressin when applying energy cnservatin. b. Suppse a pin is mving at 0.68c in the +x directin in the labratry frame f reference when it decays. The mun is bserved t leave at an angle f 12.2 belw the x axis. In what directin (in the lab frame) will the neutrin be traveling? Hint: there are tw pssible slutins. As a check, verify the cnservatin f the 4-mmentum vectrs in the lab frame. 14. In this mdified versin f the previus prblem, cntinue t assume the pin is mving in the +x directin at 0.68c, but d nt assume anything is knwn abut the directin f the utging mun. a. Determine the range f pssible utging angles fr the mun. Yu may either d this analytically r numerically, but whichever way yu apprach it, make sure yu explain yur reasning. b. Plt the energy f the utging mun as a functin f its utging angle, as measured in the lab frame.