Propeller Angular Acceleration - Mechanics - Solved Past Paper, Exams of Mechanics

Download Propeller Angular Acceleration - Mechanics - Solved Past Paper and more Exams Mechanics in PDF only on Docsity! Physics 15] Roster No.:. SOLKUTIO NS Due: Friday, April 20, 2007 . Score: Fart A* 2 - poseible Take-Home Midterm Exam #3, Part A Total: SO pe. pombe NO exam time limit. Calculator required. All books and notes are allowed, and you may obtain help from others. Complete all of Part A AND Part B. For multiple-choice questions, circle the letter of the one best answer (unless more than one answer is asked for). For fill-in-the-blank and multiple-choice questions, you do NOT need to show your work. Show your work on all free-response questions, Be sure to use proper units and significant figures in your final answers. Ignore friction and air resistance in all problems, unless told otherwise. Physical constants: \t’s an open-book lest, so you can look them up in your textbook! Useful conversions: It’s an open-book test, so you can fock them up in your textbook! L. (4 pts.) Convert the following quantities into the given units. Fill in the blanks. (You do NOT need to show your work.) Use scientific notation where appropriate (very large or very small values), and express all fina! values to THREE significant figures. 4 los \ flmw) 2 age mW (5 «0 pad) iN ASS ww. LOC AID koh (+00 1% pe) Ca ey Lag. (aeons (2022 lo a. 9,85 x 107 MW = b. 3.00 x 10° unv/ns = c. $2,500 tpm = 0.20% ku rpm” =rev/min) = lot * ot be “Na, Boo tee. ev . ( ise) = 20g EY = 20¢ Ug fens om d. 19,300 kg/m? = 1%, 3 a tte mgm? (this is the density of gold) ‘ 1O-te, OS ke = 0,2 Nia,2008 (PELL ( at 19.3 May 2. A 430-g soccer ball flying directly to the left at speed v, strikes the head of a soccer player. The entire impact iasts 15 ms. Afterward, the ball rebounds directly to the right at the same speed v,. Express al] answers in MKS units: a. (2 pts.) If ¥,=21 m/s, what is the total impulse that the ball receives during the impact? ig 42 or N ad 3 b. (2 pts.) What is the average force that the ball exerts on the player’s head during part (a)? LZ¥*1O°N ee [200 N ¢. (2 pts.) The player’s head has a mass of 4.8 kg. The player wilt experience a concussion (brain injury) if his head undergoes an average acceleration greater than 75 gees. (1 gee = 9.80 m/s”. Ignore the head’s attachment to the rest of the player's body; just consider the head alone.) Assuming thai an impact always lasts 15 ms regardless of the value of v,, what is the minimum ball speed ¥, that will cause a concussion? 6 2 “/g (Fortunately, speeds this large are nor normally attained by soccer balls during play.) (a) Choose +irdweetion te the regbety Vea mit ~ Mb » Vibe fn “UY. Sapie BE MET mg oe) = me (oxy - Cog) > Apem AY, = (ome Uyy(a(2i"4) = 16.06 9% - (> By? ae \ By Newle’s 3% Law, Fy ow bead = Fay ou ball, and Tmpulte received by head bt 16 Same for tot head and ball, ave. fone acting oy ead wap of Wad = bp of ball. 2.(b) continued : 4 p 8.0b gm Fao = PPieat 2 AP teu . s = (204 N & 12x:0°A) OM head at at i) —=> (e) “Take. lmitng Case of wikimun . aecelgration ad force 40 Cause Concussion: : = : = 4.60 mist = mT fous: “i WM Bay 2 (4b (25 gees (AEC) = 9526 Ni (niinan Se) “Lupule: OR ay At =(3528 vy (¢. 1S s) = 5242 “gs Lupulse of a and impulse of ball muct be egual ancl opposite, due +o Rewtods 3% Las. “Thar: AP bat F mye UY (ve pork () 52.42. 4 = (OF bg) 2- Vy P Vee OLS Ye - (his is apprey. 13S mpl F Fortunately , seceer balls are Utuelly much slower.) Altematve Solution: Buy = OX fer head Pp s . = a.gomle o\e= Vest = Gay bt = Cs gee) pert (0.018 s) ILO “Ye (40 +e left) Courerv. of momentum fa ball [uead collision: MA Var + MVE, = ML, + Mb tbe (Choose +X fo the right) > -vL y% PMV MU, = “LVL + My vy ” ~My DOV = Zap ° Vee Th A Vneee Am, 2 7 (4.4 ) (-tho “hs . a Shey 7 bn Ve = 61.5%, — same answer ag above, v HF sun ¥max comet's orbit 5. Halley’s Comet (mass = 1.7 x 10’ kg) has a very elliptical orbit: it varies from a distance of 0.586 AU (at closest approach) to 35.1 AU (at greatest distance) away from the Sun. (Recall: 1 AU = average Earth—Sun distance = 1.50 x 10" m.) Assume that the comet is a “point mass” in space. Kepler's 2 Law tells us that any body orbiting the Sun should have greatest velocity when closest to the Sun and slowest velocity when farthest. Kepler discovered this by trial-and-error, but Newton later showed that this law can be derived as the result of conservation of angular momentum: a. (3 pts.) If the comet’s speed is 54 km/s at its closest approach to the Sun, find its angular momentum at that BO 1m Bl x10” point. Express your answer in MKS units: Lo, 4" b.(1 pt.) In which direction does the comet’s angular momentum vector point? . righ+- howd rule + curl right doawal A. into the page ® out of the page C. direction varies during orbit Fagere in dteechow af stom (Cew) > Heumls pons ¢. (2 pts.) Knowing that the comet’s angular momentum remains constant as it orbits the Sun, find the comet’s out of of page. speed when it reaches its farthest point from the Sun, in [km/s]: 0.40 km/s (@ Le mover: smd where f= (angle betscen Font T= 40" ot bot points show above At closest 1s ene *) » hoe more 2 (W710 54 000 \(o.s%b aperaaot L. more © Cite ks)( 1 £)(6 fa Tae L= $.07 x10 Kagem ©) L comet remams Conclent Haroug hout orbit F L tose > Le fattest E ctosect = OV Dectese’ 30 Iegeu? - $0 Tv 10 awe = (ictxto® 'g) wn fae (35:1 at): gone =P Ve, = 2% = 0,90“ . iu . : ~ - Alternative Solucten ; > Pate ne nook to b eegect = L fantuege Vac = Page * Vetse cowvert wats Ph Vatese Vesose. = PU Vase Mage ar 2 CA SBGAK cy ke = 9,49 HA for “Bs( AS S  > : é bo 44a toads 6. A shelf is supporied by two posts at a distance Z apart from each other. Three canisters, each of mass M, are located at distances of (1/5)L, (2/5)L, and (3/5)L from post A. The shelf itself also has mass M. The entire configuration is at rest in equilibrium. Express your answers below ONLY in terms of M, L, g, and numerical constants: 23 M9 or 2.3 Me a, (2 pts.) Find the upward force exerted by post A on the shelf: io \7 “D . Or . . b. (2 pts.) Find the upward force exerted by post B on the shelf: oM 4 © ! 7 M 4 The shel€ is tn 4) 5 Sache equilibrinm: F =0 ako z O about amy pivot (cis of rotations). ( ay = Ath - F-FL-F- F, O = the Meg Mig Mag Meg > a ee Mg @ (2):We con A uot anywhere alma shelt... cheose permt A” (elt end). then: ZU T, - 4, -T1-U, -T, + Te O R-5F-2%-48 -nR +h -/{E 2 = D -[ap-Eaires)- (eovy [ag +o PER LM. befye EM g(b+Z+4.8) > Fe Ww Ww (7 to MS TM - ' , 7 Suvelihee tats back rete (Sb): B+ oM4)= 4g ” Fy = a M4 » 2.3M4 Alternmtt. solution: Can choose pret at pert “3 [reget end), or ceuter of Hef, or amy other pent, and still got same fnel answert for Fi and F osition, Bo (Note Heat, by choosing pivot right at a. partialer force tat Gere thers Lerp torque, and drone out of the equostton, SOLUTIONS Score: fart B 1g pts. possible Physics 15} Roster No. Spring 2007 Take-Home Midterm Exam #3, Part B 1. Two unequal masses m, and m, are located in a large, frictionless bowl, and both start at rest at the bottom of the bowl, touching each other. An explosion between the two masses pushes them apart horizontally in opposite directions. Mass m, slides to a maximum height H up the left side of the bowl. (8 pts.) For both parts of this question, your answers will be algebraic expressions. You may solve parts (a) and (b) in whichever order you wish. For both parts: + Fxpress each final answer ONLY in terms of m,, ms, H, g, and any necessary mathematical constants. * Show all steps of your derivation, + Simplify each final answer to its most compact algebraic form. a. To what maximum height does mass m, slide up the right side of the bow!? b. Find the total amount of kinetic energy gained by both masses in the explosion. (0) Conservation of momentuur iA explosion : Zpi = Zee (metmn) ¥ = Mm Ne + ML Vg WA Vip + My Ve 7 “ee * GaleMe) + () “then, Couservotion of energy 0% stes side up romps: ExaZ, Eto (Us. © ectin® (Uy + Bag eat lw = = MG tap + L Lug? 2h bottom = 4 Wing continued on next page.. nus fe z Seed Do Mee > Vee ae 12 gh, &) y