8A Review Solutions Roer Mon Ferury 24, 2007 Question We ein y doin Free Body Dirm on the mss m. Since the rope runs throuh the lock 3 times, the upwrd force on the lock is 3T. (Not ecuse there re 3 pulleys!) Hence F net = 3T m, or T = 3 m. The force required is the tension of the rope, so F = 3 m. The mount of work done is just W = F d = ( 3 m) (l) = 3 ml. c The mount of work done on the mss BY the tension is F d, with the force is tripled ut the displcement ein l, such tht the totl lenth of the 3 rope doesn t chne. Hence W rope = ml, which demonstrtes tht enery 3 is conserved, even thouh force isn t. The work done y rvity is ml. 3 Question 2 There re two types of ccelertion to wtch for: The centripetl ccelertion from movin in circle v2, nd the tnentil ccelertion from the increses r in speed d v. dt
At the top of the pendulum, the speed v = 0, so there is no centripetl ccelertion. However, it is speedin up, so there must e tnentil ccelertion. Doin the Free Body Dirm of the mss, we require tht the net force lon the direction of the rope to e zero, since its instntneous ccelertion is perpendiculr to the lenth of the strin. We hve T = m cos θ. At the ottom of the pendulum motion, the only two forces in the Free Body Dirm is the tension nd rvity, hence its horizontl ccelertion is zero. As result, its ccelertion is just the centripetl ccelertion, usin Newton s Second Lw F = m we et mv2 = T m or T = mv2 m. r r We hve to solve for v, usin Enery-Work theorem. The chne in kinetic enery is the work done y rvity: K = W 2 mv2 f 2 mv2 i = mh Where the h is the heiht fllen h = mr ( cos θ), r is the lenth of the strin. Finlly: 2 mv2 = mr ( cos θ) Question 3 2mr ( cos θ) T = m r = m (3 2 cos θ) Doin Free Body Dirm, there re 3 forces: rvity, norml force, nd friction. At the mximum nle θ 0, the friction is t its mximum, such tht the lock is on the vere of slippin. The net force is zero, which ives followin two equtions: F netx = m sin θ 0 F f = 0 2
Since F f = µ s F n, we et tn θ 0 = µ s. F nety = F n m cos θ 0 = 0 There re two wys to do this prolem, one y usin forces nd kinemtics, the other usin forces nd enery. Method We first drw the Free Body Dirm for the system. At the end of the dy, you should hve F n = m cos θ, F net = m sin θ µ k (m cos θ). Since F net = m, = (sin θ µ k cos θ). Usin the kinemtics eqution: vf 2 v0 2 = 2d, where the distnce trvelled d =, solvin for finl speed, h sin θ Method 2 v f = 2h ( µ ) k. tn θ Usin the work-enery theorem, we strt with K = W + W f. Work of rvity is mh Work of friction is F f d = mh µ k. tn θ We rrive t the sme result s efore. This should e expected. Question 4 You should e thinkin out Free Body Dirms instntly with this quesiton, since it involves forces. We strt with the Free Body Dirm of ojects with the lest mount of forces on it. Doin the Free Body Dirm of m, you should et tht the tension in the rope T equls m to prevent the mss from fllin. Doin the Free Body Dirm of m 2, you should et tht the tension in the rope mkes the mss ccelertion; T = m 2. So = m m 2. Finlly, doin the Free Body Dirm of the whole thin (includin the pulley nd ll the msses), you should et tht F is responsile for the whole 3
thin ccelertin. (Note, you cn do this ecuse m is not fllin; so ll the msses move toether with the sme ccelertion.) You re finl result should e: F = (m + m 2 + m 3 ) m m 2 Question 5 The plne trvels in projectile motion; it ccelertes t constnt downwrds. Hence you, the plne, nd everythin else is free fllin, which simultes zero-rvity experience. In projectile motion, the horizontl nd verticl componenets ehve seprtely. They re relted y the nle of 45 derees, which mens tht v 0y = v 0 sin 45 The ircrft mkes clim of 8000ft, or 2438m. Usin the micl eqution: vf 2 vi 2 = 2d c v 2 0y = 2( 9.8m/s 2 )(2438m) v 0y = 29m/s The weihtlessness ends with the plne imin down t 30 deree. Keepin in mind tht the horizontl velocity component remins constnt, we hve v 2y = v 2x tn 30 = 26m/s. Its netive ecuse the plne is now imin downwrds! Usin v f = v i + t, t = 35s. (Astronuts trin for zero-rvity ord NASA s ircrft, which mkes continous free-fll prols nd clims, vryin from 0 to 2 every minute. Hence the plne erned the nicknme: Vomit Comet) 4
Question 6 A mss m = 400 is dropped from heiht h = 30cm on top of sprin tht is initilly relxed. The sprin hs stiffness k = 20N/m. Solution This prolem requires n pplyin the conservtion of enery. We re interested in the initil nd finl stte of the system. At mximum compression, the speed is zero. Usin this s the finl stte, nd initil stte ein from where it ws dropped: K + U + U s = 0 2 mv2 f 2 mv2 i + m( h) + 2 kx2 f 2 kx2 i = 0 The chne in heiht h is netive (since it ws dropped), h = h y, where y is the compression of the sprin. There re no kinetic enery t the einnin or end, nd the initil sprin is in equilirium, so 2 kx2 i = 0. The finl enery of the sprin 2 kx2 f = 2 ky2. Hence the conservtion of enery ecomes: m(h + y) + 2 ky2 = 0 This is qudrtic eqution in y nd cn e solved y the qudrtic formul. You cn two nswers with numers iven: y = 58cm or y = 9cm. Only the possile nswer mkes sense, since we ve defined y to e the compression of the sprin. The velocity t mximum compression will e zero. The ccelertion cn e computed y the forces ctin on the mss; hence Free Body Dirm is required. m = kx m, so = kx m = 29m/s2 9.8m 2 = 9.2m 2 upwrds. 5
c The mss will e propelled ck up to the sme heiht (i.e. 30cm) y the conservtion of enery. It cnnot o hiher ecuse tht mens enery comes from somewhere, it does not o lower unless enery is dissipted elsewhere. Question 7 Accelertion down the rmp is = sin θ = 5, the distnce trvelled is d = h 5, so totl time t = 2d = 0 h 3.2 h. For this one you do ech of the rmp sement seprtely. The first prt should et you t = 2 h. The finl speed is 2h, so the second prt tkes t 2 = h. Totl time t = ( ) 2 + h 2 2 2.7 h. The second pth, thouh loner, tkes less time! Question 8 v 0 = 2h Question 9 (This question quite difficult.) A few concepts is required: In order to support the lrer mss, the tension in the rope hs to e T = M. The smller mss undero circulr motion; this mens tht its net force F net = mv2 r. 6
The rdius of the circulr motion r = l sin θ, where θ is the nle the rope mkes with the verticl. Doin Free Body Dirm on the smller mss, nd usin the ove fcts, T cos θ = m, nd T sin θ = mv2 r. The rev time = Time it tkes to do one revolution = Usin the ove points, we cn solve for rev = M. time 2π ml M, s the lenth l increses, the revo- ml From the ove eqution rev lution slows down. Question 0 = time 2π v. 2πr Drw trinle, one side should e 20km, the second side should e 00 mt, s the third side should e 30 mt, the nle opposite to the 00t should e s 0. Usin the sine lw: sin A = sin B = c sin C We et tht () The plne should e flyin N46.4 E, () It will tkes 400s 23min for the trip. Question This prolem involves severl distnces nd speeds, it is importnt to know which one to use for wht sitution! Also, 72km/h = 20m/s, 20km/h = 33.3m/s. Wht s importnt is tht the cr must stop efore 30m, rerdless of how lon it tkes or wht the trin does. We know tht: Initil Speed v 0 = 20m/s. 7
Distnce covered d = 30m. Finl Speed v f = 0. (So the cr stops) Unknown: ccelertion. Solvin for ives 6.7m/s 2. Wht s importnt in this cse is tht the ck end of your cr must cler the intersection efore the trin rrives. Initil Speed v 0 = 20m/s. The trin to rrive t the intersection t t = Distnce to cover d = 30m + 2.5m + 3m = 35.5m. Unknown: ccelertion. solvein for ives 6m/s 2. Question 2 40m =.2s. 33.3m/s At Equilirium, we hve the followin condition (equilrium t x): kx = m Applyin the conservtion of enery. (Extr compression y = cm, nd it oes up to z ove the equilirium): 2 k(x + y)2 my = 2 k(x z)2 + mz Expndin nd usin the first eqution: 2 k(x2 + y 2 ) = 2 k(x2 + z 2 ) y 2 = z 2 y = ±z Wht this mens is tht the mss will o up cm ove equilirium, nd then fll ck down cm elow equilirium, nd ck up, oscilltin. 8
Question 3 Losin 36% of the enery mens tht the velocity is reduced to 80% = 4 5 fter ech ounce. The first ounce tkes 2h time. The next ounce tkes 2 5 4 2h (up nd down), with ech ounce tkin 4 of the time from the 5 previous ounce. The totl time it tkes is: 2h t = + 8 ( 2h + 4 ( 4 2 ) 5 5 5) + +... Question 4 2h = + 8 2h 5 5 2h = 9 Just keep intertin, strtin from crcle c(t) = d5 x dt 5 = c. x(t) = x 0 + v 0 t + 2 0t 2 + 3! j 0t 3 + 4! s 0t 4 + 5! ct5 9