PHY 2048 Spring 2014 Acosta, Rinzler. Exam 2 Solutions

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1 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Exam Soluton Note that there are everal varaton o ome problem, ndcated by choce n parenthee. Problem 1 Four dentcal oda can ntally at ret have a recracker explode between them uch that one can recol agant the other three whch are tll tuck together. Wth V1 and V3 the velocte o the ngle and 3 can, repectvely, what the rato o the magntude o the peed o the ngle can to the magntude o the peed o the other three (V1/V3)? (1) 3 () (3) 4 (4) 1/3 (5) 1/ Problem Ue conervaton o momentum, ntal (where the can are at ret) equal nal: 0 = 3mV 3 mv 1 V 1 V 3 = 3 A cylndrcal bar algned along the x- ax ha a ma denty that ncreae lnearly along t length: λ(x) = (4 kg/m ) x. I the bar extend rom x=0 to x=1.5 m (or 0.5m or m), at what poton along the x- ax (n m) the center- o- ma o the bar? (1) 1 () ½ (3) 1/3 (4) 4/3 (5) 3/ To calculate the center o ma n x or a contnuou ma dtrbuton n 1D o length L ue: x com = 1 M 0 L xλ ( x)dx The total ma M gven by: L L M = λ ( x)dx = 4x dx = ( kg/m )L 0 0

2 PHY 048 Sprng 014 Acota, Rnzler Exam oluton x com = 1 M L 4x dx = L 3 L3 = L 3 So, /3 o the total length Problem 3 Conder a ball o ma 0.5 kg that travel only n one dmenon along the x ax. It ntal velocty at tme t=0 v= 5.0 m/. A orce then appled to the ball n the x drecton a a uncton o tme a hown by the graph. What the velocty o the ball along the x- ax, ncludng gn, at a tme o 0.4? (1) 3 m/ () 8 m/ (3) - 13 m/ (4) - 1 m/ (5) 0 m/ The mpule gve the change n momentum: J = Δp = F( t)dt The area under the curve thereore gve the momentum change: So Problem 4 F( t)dt = 1 ( 0) ( 0.1) + ( 0 ) ( ) + 1 ( 0 )( ) = 4 kg m/ Δp = m( v v ) v = J m + v = = 3 m/ In the gure, a horzontal orce Fa o magntude 0 N appled to a.5 kg book a the book lde a dtance d=1 m (or 4 m) up a rctonle ramp at angle θ=30. What the peed o the book at the end o the dplacement t tart ntally at ret? (1).0 m/ () 4.0 m/ (3) 7.5 m/ (4) 16.0 m/ (5) 1.0 m/

3 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Fnd the net orce along the nclne actng on the book: F net = F a coθ mgco 90! θ ( ) = = 5.07N The acceleraton thereore: a = F net =.0 m/ m And the nal velocty gven by: v = ad v =.0 m/ (or 4 m/) Problem 5 The gure gve the orce actng on a 5.0 kg partcle a t move rom ret along an x ax rom x = 0 to x = 8.0 m. What the partcle' peed and drecton (gve a potve anwer the partcle move along the x ax n the potve drecton and negatve otherwe) o travel when t reache x = 5 m (or 6 m)? (1) +7.1 m/ () +5.5 m/ (3) m/ (4) m/ (5) 0 m/ The work done by the orce change the knetc energy o the partcle: W = F( x)dx ( ) = ΔK = 1 m v v So the area under the graph up to the lmt o the travel hould be calculated. 5 W = F( x)dx = So the velocty gven by ( )( 1) + ( 50) ( 3 1) + 1 ( 50) ( 4 3) 1 ( 50) ( 5 4) = 15 J v = W m = 7.1 m/

4 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Problem 6 A hgh jumper o ma 70 kg run toward a hgh jump bar wth a peed o 6.7 m/. What would be the maxmum bar heght the center- o- ma o the jumper could clear he/he were able to convert all o h/her ntal knetc energy nto a vertcal jump? (1).3 m () 4.6 m (3).4 m (4) 0.4 m (5) 1.5 m Ue conervaton o mechancal energy: Problem 7 E mec = K +U = K +U K = U 1 mv = mgh h = v g =.3 m A 0.5 kg ma attached to the end o a relaxed, vertcal hangng prng and the ma let go rom ret. What the maxmum dtance the prng tretche downward ater the ma releaed the prng contant k=50 N/m? (1) 10 cm () 30 cm (3) 5 cm (4) 0 cm (5) 40 cm Apply conervaton o mechancal energy, but there are two ource o potental energy: gravtatonal and elatc: E mec = K +U = K +U E mec = K +U,grav +U,prng = K +U,grav +U,prng U,grav +U,prng = U,grav +U,prng 0 = mgy + 1 ky y = mg k ( )( 9.8) = = 0.1m (nce the ma at ret ntally and nally) Note that unle you do work on the ytem and take away ome o the ntal energy, the dtance NOT gven by Hooke law: y = mg = 5 cm. Rather the k prng tretche arther to 10cm beore toppng, and n act wll ocllate.

5 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Problem 8 A roller coater to go through a vertcal loop a hown. What the mnmum peed at the bottom o the loop (Vb) neceary o that the car do not all o o the track at the top o the loop the radu o the loop R=10 m? Treat a roller coater car a a pont ma gong through the loop. (1) m/ () 0 m/ (3) 10 m/ (4) 17 m/ (5) 500 m/ Problem 9 Not only do we need enough ntal knetc energy to reach the top o the loop, we need enough centrpetal acceleraton or the car to tay on the track! Namely t hould be at leat a large a g: a c,t = V t R g Now ue conervaton o mechancal energy: E mec = K +U = K +U 1 mv b = 1 mv t + mgy 1 mv b = 1 m gr V b = ( ) + mg ( R ) 5gR = m/ Two tcky equal ma ar hockey puck collde and tck together at the orgn o the rctonle x- y plane hown n the gure. The hown angle are θ1=30 and θ=60. The ntal peed o both puck are equal: v1 = v = m/. Determne the outgong angle θ3 or the combned ytem o two puck, meaured counter- clockwe wth repect to the x- ax. (1) 15 () 60 (3) 0 (4) 75 (5) 30

6 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Snce th an nelatc collon, we can only ue conervaton o momentum, whch we apply eparately n x and y: î : mv 1 coθ 1 + mv coθ = mv 3 coθ 3 ĵ : mv 1 nθ 1 + mv nθ = mv 3 nθ 3 Take the rato (and note that v1=v) tanθ 3 = v nθ + v nθ 1 1 = nθ + nθ 1 1 = + 3 v 1 coθ 1 + v coθ coθ 1 + coθ θ 3 = 15! = Problem 10 The ytem depcted n the gure n equlbrum. A block o ma M=300 kg hang rom the end o a male trut whch xed to the ground at the other end and make an angle θ=30 (or 60 ) wth repect to the horzontal. Fnd the tenon T n the male, horzontal cable attachng the trut to the wall. (1) 5100 N () 900 N (3) 1700 N (4) 5900 N (5) 3400 N Balance the torque rom the ma and the tenon T on the trut about an ax at the end xed to the loor (L the length o the trut, but t cancel): LT nθ LMgn ( 90! θ ) = 0 T nθ = Mgcoθ T = Mg tanθ = 5100 N

7 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Problem 11 A ma attached to, and hang rom, two rope whch are xed at ther other end n the celng, a hown n the gure. The let rope make an angle o 60 wth repect to the horzontal, and the rght rope 30. What the rato o the tenon n the let rope to that o the rght (TL / TR)? (1) 3 () 1/ 3 (3) 1 (4) 3/ (5) Let ue orce balance n the x and y drecton: î : T L co60! + T R co 30! = 0 ĵ : T L n60! + T R n 30! Mg = 0 The rt equaton enough or the rato: T L co 30! = T R co60 = 3 /! 1/ = 3 Problem 1 A 5 kg object upended rom the celng by a cylndrcal copper wre wth a dameter o 1 mm. I the wre ntally ha a length o m, what the ncreae n t length n mllmeter (mm) ater the object attached the Young' modulu o copper N/m? (1) 1.0 () (3) 0.5 (4) (5) Ue the tre- tran relaton or tenle tre: F A = E ΔL L So olvng or the change n length: ΔL = FL EA = MgL EA = 1.0mm

8 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Problem 13 The graph how the potental energy o an object acted upon by a conervatve orce a a uncton o t poton x. The potental value ndcated are: UA = 0 J, UB=5 J, UC=40 J, and UD=50 J. At whch o the lted poton n x the orce larget n the +x drecton? (1) m () 7.5 m (3) 0.5 m (4) 5 m (5) 6.5 m The relaton between orce and potental energy : F x = du dx So we need the larget negatve lope o the potental energy curve to have the larget orce pontng n the +x drecton. That occur or x=m. Problem 14 The ollowng object () a old phere, () a phercal hell and () a cube all have the ame ma. The object all tart to move, rom ret, rom the ame elevaton on an nclned urace (ame nclne), at the ame tme. The phere each roll wthout lppng whle the cube lde wthout rcton. The order n whch they get to the bottom o the nclne, atet rt, : (1) (), (), () () (), (), () (3) (), (), () (4) (), (), () (5) (), (), () Soluton: Snce the object all rom the ame elevaton they convert the ame amount o potental energy to knetc energy. For the cube whch lde reely, wthout rcton, all the potental energy become tranlatonal knetc energy o t wll be atet arrvng, at the bottom rt. The rollng phere wll convert ome o ther ntal potental energy nto rotatonal knetc energy, leavng le tranlatonal knetc energy or each o them. That 1 rotatonal knetc energy K = Iω. Snce th depend lnearly on the rotatonal nertal I, the one that ha the greater rotatonal nerta wll have converted more o t ntal potental energy to rotatonal energy leavng t the le tranlatonal knetc energy o t wll be lowet and arrve at the bottom lat. For the ame ma the rotatonal nerta o a

9 PHY 048 Sprng 014 Acota, Rnzler Exam oluton phercal hell greater than that o a old phere o t come n lat. The order, rt to lat, down the nclne thu cube, old phere, phercal hell or (), (), (). Problem 15 A wheel ntally ha an angular velocty o (30, 35, 40) rad/ but lowng at a rate o rad/. By the tme t top, the number o revoluton t wll have turned through : (1) 36 () 49 (3) 64 (4) 81 (5) 5 Soluton: Ue the knematc expreon: θ θ : ω =ω o + α( θ θ o) olved or o ω ωo ( θ θ o) = α where, ω= 0, ω o = (30, 35, 40), α=. The reult wll be n radan and mut be rad converted to revoluton by multplyng wth the converon actor 1 Rev/π rad. Gvng oluton (36, 49, 64) revoluton, repectvely. Problem 16 A thn hoop o radu R and ma M welded to a thn unorm dk o the ame ma and the ame outer radu uch that the hoop and dk are coaxal (dotted lne). The rotatonal 3 7 nerta or th object about an ax that d = ( 4, 8)R rom the dotted ax, a hown n the ketch : (1).63MR () 3.03MR (3).06MR (1).7MR (1) 1.63MR Soluton: From the Table the rotatonal nerta o a dk and a hoop about ther relevant center o 1 ma (com) axe are I = MR, and I= MR, repectvely. Snce rotatonal nerta about the ame axe are mply addtve, the rotatonal nerta o the hoop and dk 3 welded together Icom = MR. To nd the rotatonal nerta about the new, parallel, 3 7 ax a dtance h = (, )R away rom the rt, we ue the parallel theorem: 4 8 = com + (where we mut ue M M I I Mh = a the total ma n the nd term nce the hoop and dk each contrbute a ma M) o that: I = MR + (M)( R, R) = [ + (, ) ]MR = [ + (, )]MR = (.63,3.03)MR

10 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Problem 17 A old unorm phere o ma 0.4 kg and dameter 10 cm roll rom ret wthout lppng down a 30 nclne. Ater rollng a dtance (8.0, 10, 1) m meaured along the nclne, t rotatonal knetc energy : (1) 4.48 J () 5.60 J (3) 6.7 J (4) 7.84 J (5) 8.96 J Soluton: Mechancal energy conerved uch that the phere change n knetc energy ater rollng a dtance L along the nclne wll equal the change n t potental energy o havng allen through heght h = Ln θ. Snce the phere tart rom ret t ntal knetc energy zero o we have that, K + K =Δ U = mgl n θ, or tran, rot, 1 1 mv + Iω = mgl n θ. Snce the phere roll wthout lppng we can wrte that v = rω, alo ung the rotatonal nerta or a old phere I= mr. Then r 5 r gl n (1 + 5)r ω = gl n θ 7 ω = θ m(r ω ) + ( mr ) ω = mgl n θ ω + ω = θ r gln 5 5 gln θ ω =. The queton ak or the rotatonal knetc energy whch 7 r gln θ Krot, = Iω = 5mr ω = mr = mglnθ, or nce the angle 30 o 5 7 r 7 and n(30 o ) = 0.5 we have, 1 1 m Krot, = mgl = (0.4kg)(9.8m )(8 m, 10 m, 1 m) = 4.48J, 5.60J, 6.7J 7 7 Problem 18 A dumbbell rotor cont o a male rgd rod ree to pvot around a vertcal ax through the mddle o the rod. Two mall, equal mae, are attached to the rod at a dtance R rom the rotaton ax. The rotor rotate wth an angular velocty o (1.0,.0, 3.0) rad/. An nternal mechanm pull the mae n ymmetrcally untl they are each at (½)R rom the ax. The angular velocty o the rotor ater th move (n rad/) : (1) 4 () 8 (3) 1 (4) 1/4 (5) 1/8

11 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Soluton: Snce no external torque act on the ytem angular momentum conerved and requre I that L = L or Iω = Iω o that, ω = ω. Snce the mae are mall we treat them a I I= mr here we have ntally ( ) 1 1 mr ω = ω, or 1 4 mr rad rad rad rad rad rad 4(1,,3 ) 4, 8,1 pont partcle or whch I = m R = mr o that, ω = = I ω = ω then = mr and nally Problem 19 A phercal hell ree to rotate about t vertcal ax havng rotatonal nerta I = 40 N m about that ax. A motor caue a torque that gve the hell a contant angular acceleraton o (1., 1.4, 1.6) rad/. In rotatng the hell through 15 revoluton the motor doe an amount o work (n J) equal to, (1) () (1) (1) (1) Soluton: θ The work done by the motor wll be, W = τdθ, Snce torque τ= Iα, both o whch are here contant, we can wrte, θ1 θ θ W= Iαdθ= Iα dθ= I α( θ θ 1) = IαΔθ. The 15 θ1 θ1 rad rev rad 3 revoluton mut be converted to radan o that Δθ = (15 rev)( π ) = 94.5 rad then, W= I αδθ= (40N m )(1.,1.4,1.6) (94.5) = (4.5,5.3,6.0) 10 J Problem 0 Tarzan, havng a ma o 88.0 kg, wng on a 10.0 m long vne uch that jut beore gettng to the bottom o the wng h angular velocty 1.40 rad/. At the bottom o the wng he pck up Cheetah (ma o 0.0 kg) who wa watng there at ret. Jut ater the pck-up ther angular velocty n (rad/) wa (hnt: treat both Tarzan & Cheetah a pont mae): (1) 1.14 () 1.30 (3) 1.47 (4) 0.98 (5) 0.81

12 PHY 048 Sprng 014 Acota, Rnzler Exam oluton Soluton: At the bottom o the wng the orce o gravty perpendcular to the drecton o the moton (contraned to an arc by the vne) o t generate no torque there. Wth no external torque the angular momentum conerved and requre that: L = L or I Iω = Iω o that, ω = ω. Wth Tarzan and Cheetah pont partcle I = mtarzanrvne I and Tarzan vne Cheetah vne Tarzan Cheetah vne I = m r + m r = (m + m )r. Then m r m ω = ω = ω Tarzan vne Tarzan (mtarzan + m Cheetah)rvne mtarzan + mcheetah 88 kg ω = (1., 1.4, 1.6) = (1.14, 1.30, 1.47) 88 kg + 0 kg rad rad