Discussion Question 1A P212, Week 1 P211 Review: 2-D Motion with Uniform Force

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1 Discussion Question 1A P1, Week 1 P11 Review: -D otion with Unifor Force The thetics nd phsics of the proble below re siilr to probles ou will encounter in P1, where the force is due to the ction of n electric field on chrged prticle. A point prticle of ss trvels freel in the -direction with unifor velocit v 0. At = 0, it enters region between two pltes oriented perpendiculr to the -is; the plte spcing is w, nd then plte length in the -direction is L. The prticle enters on the id-plne = 0. While between the pltes, it eperiences constnt, sptill unifor force F in the +-direction. After eiting the pltes the prticle gin oves freel. v 0 +w/ L F w/ () Our first tsk will be to obtin n epression for the -coordinte of the point t which the prticle eits the pltes. We will ssue tht the plte spcing is wide enough tht the prticle never strikes either plte. But before we strt, consider these possible solutions: F (1) (v0 L) () FL v 0 (3) Fw v 0 Could n of the be correct? Wh or wh not? Reeber, units nd liiting behvior! In fct, fro onl those two considertions, ou cn write down the correct nswer to within fctor of without using n foruls t ll. Wnt to give it tr? (This procedure is clled diensionl nlsis nd phsicists use it ll the tie, especill when developing new theories.) None of the three nswers is correct. As ou ll see in the net proble, this diensionl nlsis hs lost given us the coplete nswer, nd hs provided vluble check on our upcoing work. But to get the coplete solution we hve to perfor clcultion. (b) Now go hed nd clculte the correct epression for. pge 1

2 = 1/ t where = F / nd t = L / v 0 therefore, 1 F L v 0 (c) Net, find n epression for the iu vlue of the force F for which the prticle psses through the force region without striking either plte. = w / nd = F L / ( v 0 ) therefore F = v 0 w / L (d) For the conditions of prt (c), find n epression for tn where is the ngle of deflection t which the prticle eits the force region. (Did ou drw sketch? Did ou check our units?) tn v / v. v = t tn = w / L. pge

3 Discussion Question 1B P1, Week 1 P11 Review: Grvittionl Forces nd Superposition F 1 G 1 r 1 ˆ r 1, 1 r 1 ˆ r 1 This proble develops skills ou will need in P11 in finding the electric fields creted b sets of point chrges. Four prticles of equl ss re fied t the corners of squre with sides of length. A fifth prticle hs ss nd oves under the grvittionl forces of the other four. X () Find the - nd -coponents of the net grvittionl force on due to the other four sses when is locted t the center of the squre (left-hnd figure). Hint: Drw sketch! Use superposition nd drw vector digr consisting of four vectors, ech representing the force eerted b one of the corner prticles on. For ese of reference, lbel the four (equl) corner sses 1,, 3, nd 4. Lbel the corresponding force vectors F 1, F, etc. With the vector digr in hnd, it is vstl esier to clculte the requested coponents of the totl force. A sketch ieditel revels tht the forces on the test ss cncel, giving net force of zero. pge 1

4 (b) Find the - nd -coponents of the net grvittionl force on when it is locted t the center point of the right-hnd side of the squre (iddle figure). Use the se solution procedure tht ws recoended bove for prt (). F = F 3 + F F 3 = F 4 = G / r, with r = (/) + = 5 /4 4 F 3 = F 3 cos, where cos The finl nswer is: F = 0 nd F = -16 G / (55 ) (c) Check our nswer to prt (b) b testing t lest 3 eples of liiting behvior. Do ou get the results ou epect? Now, n iportnt point: suppose ou were given nuericl vlues in this proble: = 3 kg, = 1 kg, nd = 5 c. If ou hd plugged those nubers into our equtions right fro the strt, ou d get the finl result F = N. Would ou be ble to check the liiting behvior of this nswer? No! We ve lerned n iportnt lesson: Never plug in nubers until the end of our clcultion. F goes to infinit s or go to infinit (infinite sses eert infinite force) F goes to zero s or go to zero (without ss, there is no grvittionl force) F goes to infinit s goes to zero (infinite force eists when the sses re right on top of ech other) pge

5 F goes to zero s goes to infinit (no force if the sses re infinitel fr prt) (d) When ss is locted on the -is distnce X lrge copred to (right-hnd figure), one cn use siple phsicl rguent to see tht the net force on due to the other four prticles is pproitel F CGX nd F 0 where C is nuericl constnt. (This is long-distnce pproition: it gets better nd better s X increses.) Wht is the pplicble phsicl rguent? Use it to find the vlue of the diensionless constnt C. The squre of sses will look like single object with ss 4. The constnt C in the pproite forul is therefore 4 (- sign needed since the force is ttrctive). pge 3

6 Discussion Question 1C P1, Week 1 Review: Potentil Energ Four prticles of equl ss re fied t the corners of squre with sides of length. A fifth prticle hs ss nd oves under the grvittionl forces of the other four. Consider the grvittionl potentil energ U of the ss s function of its position (,). This potentil energ p U(,) is n etreel useful w of representing the effect of the grvittionl force on the ss : If is held t rest t soe point ( 1, 1 ) nd ou let it go, it will lws ove towrd point of lower potentil energ. You cn therefore think of the function U(,) s topogrphicl p: the prticle will lws roll downhill. If the ss oves fro point ( 1, 1 ) to point (, ), the potentil energ difference U( 1, 1 ) U(, ) tells ou ectl how uch work ws done b grvit, nd ectl how uch kinetic energ the prticle gined s result. Let s visulize ll this. The figures below show the potentil energ p U(,) for the squre configurtion of four sses illustrted bove. The height of the surfces indictes the gnitude of U t ech point. Which one is correct? The correct p is the botto iddle one. The top three cn be eliinted since the indicte higher potentil energ ner the sses. Tht cn t be: the grvittionl force ttrcts sses towrd ech other, so the potentil energ ust be lower closer to the sses. The botto left 1

7 figure corresponds to the potentil round single point ss, not four of the. Finll, we cn see tht the botto iddle figure is the correct one b considering superposition: the potentil energ p for the four sses is just the superposition (su) of four ps for single ss, like the botto-left figure. The botto right figure shows dditionl fetures (ridges) in the p tht would not be cused b our siple collection of 4 sses. Now let s perfor soe clcultions using potentil energ. Here s the ster forul for the grvittionl potentil energ between two sses 1 nd seprted b distnce r 1 : U 1 G 1 r 1 () Show tht the net grvittionl potentil energ of prticle when t the origin is U(0,0) 4 G /. Hint: The eqution bove gives the grvittionl potentil energ of ss 1 in the presence of ss, or vice vers. Use it nd superposition to find the net potentil energ of ss in the presence of the four fied sses. As long s the sses reined fied, it is not necessr to consider their utul potentil energies. B superposition, the potentil energ of in the presence of the four sses (which we ll cll 1 through 4 ) is just the su of the potentil energ due to ech individul ss. In sbols, U = U 1 + U + U 3 + U 4. We used superposition of forces in proble 1 notice how uch esier it is with potentil energ since tht quntit is sclr (just nuber), not vector, nd so does not involve n coponents. Fro the left-hnd digr, we see tht the distnce r between nd ech of the s is the se: r = (/) + (/) = /, nd so r = /. Fro the potentil energ forul, we get U = 4.U 1 = -4 G /. (b) If the prticle is relesed t rest fro infinit nd psses through the origin, clculte the gnitude of its velocit there. If the prticle flls fro infinit to the origin, it loses potentil energ in the ount U tht ou clculted bove. This energ hs to go soewhere it is converted to kinetic energ = v /. Rerrnging the ters in -U = v / we cn find the prticle s finl velocit: v 8 G

8 (c) Clculte the net potentil energ of when it is t the id-point of the right-hnd side of the squre. Agin b superposition, U = U 1 + U + U 3 + U 4. Since ll the sses re the se, the onl difference between the U i s is the distnce fro the ss i to the test ss. For the two righthnd sses, r 1 = /. For the two left-hnd sses, r = (/) + = 5 /4, nd so r = 5/. U = G / r 1 G / r = -4 G / (1 + 1/5) (d) If the prticle is relesed t rest fro the id-point of the right-hnd side of the squre, does it rech the origin? Support our nswers with phsicl rguents. Fro prt (), U t origin = -4 G / = G /. Fro prt (b), U on right-hnd side = -4 ( 1 + 1/5) G / = G /. The potentil energ is thus lower on the right-hnd side thn t the origin. Tht ens tht the ss prefersto be on the right-hnd side thn t the origin. Further, it cnnot rech the origin unless it strts with enough kinetic energ to overcoe the potentil difference between the two positions. Since our prticle is strting fro rest, no it will not rech the origin. (e) In Discussion Question B, ou clculted the force on the ss when it is t the id-point of the right-hnd side of the squre. If ll went well, ou found tht this force points in the direction, i.e. towrds the origin. Given this fct, plus our nswer to prt (d), cn ou ke rough sketch of the potentil energ function U() for points long the -is? Give it tr! U / / 3

9 Discussion Question 1D P1, Week 1 P11 Review: Unifor Circulr otion F 1 G 1 r 1 ˆ r 1, 1 r 1 ˆ r 1 In P11 ou will encounter probles where chrged prticles ove in unifor circulr otion. The forces involved be electric or gnetic in nture. The nswer to prt(e) contins the secret of the cclotron. Kepler s Third Lw (K-III) for plnetr otion bout the sun for circulr orbits is T CR 3 where T is the period, R is the rdius of the plnet s orbit nd C is constnt. v R () Derive K-III for circulr orbit nd in the process find n lgebric epression for C in ters the ss of the sun S, the universl grvittionl constnt G, nd nuericl fctors. F = -G S / R F = = - v / R Set equl to ech other nd substitute period for velocit. T = 4 R 3 / (G S ) (b) Using our nswer fro prt (b), re-epress K-III s reltionship between the ngulr frequenc of the otion nd the rdius R of the for: = f(r, G, S ). = G S / R 3 pge 1

10 (c) Consider unifor circulr otion of bod of ss bout centrl force which depends on velocit: F Dv R b where D is constnt nd nd b re known eponents. Derive K-III for this force, gin epressing our nswer s reltionship between nd R, (The constnt D will necessril pper in our finl epression, but v ust not.) - = D R + b 1 / (d) For the cse = 0 nd b =, verif tht our nswer to prt(c) collpses to tht for prt (b) (this is n ecellent liiting behvior check). (e) Evlute our nswer to prt (c) for the cse = 1 nd b = 0. How does the ngulr frequenc depend on rdius for force of this nture? = D/, ngulr frequenc independent of rdius. pge