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Hmwk_Solutions_Chpt10

# Hmwk_Solutions_Chpt10 - Physics 212 Homework 1 Required...

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Unformatted text preview: Physics 212, Homework 1 Required Problems. Chapter 1U: 37, 95, 101, 1GB. lﬂﬁ Problem 3'? u i la. Two blocks with mass ml = ﬂﬂkg and m2 = ltlﬂltg sit on frictionless inclines of St} degrees and 66 degrees respectively, as shown below. They are connected by a string passing over a pulley with radius 0.25 In and moment of inertia I. .,. __ _ m». _ .__ l-p-,_-,:I.:. - .___.- ___ :- J- .-. (.5. — I..—r'L./'-. J (ali‘Draw a free—body diagram for each of the two blocks and the pulley. {b} Find Pm and PH, the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, 1 I. lc. We assume that the length of the string stays constant, so the blocks and pulley must all move together. This means that at any time the speed of both blocks (m and ﬁg) and the tangential speed {wit} of the pulley. Also ﬂ‘ff : ﬂﬁlf‘ﬁo‘in the same time interval the accelerations of both blocks and the tangential acceleration of the pulley must be equal, a“ = efﬁng. -—— i . . 5.9 2a. The balance of forces and Newton's second law (F = Ina] apply to both blocks. be- cause they show translational motion, whereas the balance of torques and r = la. apply to the pulley, because it only rotates. Q ti: II h I”; " J:- "\ I} a. ,, .,- 1: D k} '91. ,_.-"'-'_7 T i T —-_-:_| ,r .- n' - f",- "‘\I _ {1" k. '. ,. I“? s.— .i - ,' " ’ .- 0, I. I __g __ q“, I {WI-J [3 I ’ I:- II}!- I) «.7 r1 .- "r‘w bl“:- i'c 1. In,» 'In ”___. _: I I f . ' _ u. __ L r r .g 2 If.- 'r I i" r" ;. ,- ' l' . _ / -. ,_,_ ,J |—.;__j‘_:?.y:ii..>f . Let ':'_ _ _ - . - -_E J _ _ | T ee- 'I.' I v.7 1 Dr— F r r ' ";/ r- I ‘7 (1' Hi!“- r 'I_ _ _ =-r - >3" i.” o~ on x ,.o-. a: ./C'}I .1. r i _ w» or F 2: : ' '9 (3.2: .—___._ 4,.____,- .c J _. i I_ - :5. iii-rd!" l ' t J id] 5 TEL mf ﬁrﬁM 24" EFTTIF ’2 _5 22? SMCE Emlﬂﬁkbﬂ (gm '1": 0L: (1:: -r-_ 1"“: r'} K! U){O-Q%m}#(ﬁr?m(ovﬁ‘a w?) H W1 “11¢ #:pﬁ i 5 [ac JI ?:Id\ 1cm 3%. £170.“: {"3 OK. P..an Ill-bun? 4'0 13: mcciTJu'ai-m mrmﬂnﬁg MR3; WE Swﬂ-ﬁmﬁm Tn ﬁhﬁsknﬁ. .i...';.-E%: ;.rr~.?§;\ i‘; Fatah-~11"! 4123:) m Jr \$7.1 5F“ 5?an «H4 3., ﬁtm +9 "Hag, “(311% may! 5mm m3 Ta, More Mmﬂi..fﬁi‘ ﬂat!) 6' 5h.F~£. Aim, 'Hna {angina}; agar-n Pg“.§wﬁmL.-€ 1&53 11AM [ﬁng- ¥craﬁ G? 02"" {RH-C, Problem 95 la. A centrifuge starts at rest [w = H = ﬂ} and then accelerates. As it accelerates, it goes through 2E] revolutions in 1 min. Find: (a) Its angular acceleration, a (1)) its final speed, wf 1c. We assume that the acceleration is constant over the 1 min period. This will allow us to use rotational kinematics. 2a. Rotational kinematics apply. 35. 0" a. r.._, gel-p. .-'.i( rrs- ,' a, “WM _ "ifj’lc {3 {399* or .c/ (3-1“??? Elf-go g.“ + ’3 “LIPID ISA/"Ff": “l’irxJ-S'n- III Jug-'5' ID Kiln-"o! "’14 Hr IfggaL-n I “ 7' a): f I “I”? _'__ _ _ is}: {:3 “W’s- = i a e- ' - U {Tot Ai- ﬂxc and r; limo “is -:*r:/:n=.- r {he mm, L 301.- I “if” ’ ‘ ' 1. J. 1'“ c I; Problem ID] 1a. A roll of paper has radius 7".6 cm and moment of inertia l = 2.9 X lﬂ‘akg-mz. A force of 3.2 N is exerted on the end of the roll for 1.3 s, causing the paper to unroil (with no tearing). A constant friction torque of I11] m‘N is exerted on the roll, which gradually brings it to a stop. Assuming that the papEr’s thickness is negligible, calculate {a} the length of paper that unrolls during the time that the force is applied [b]: the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving lc- We assume that the thickness of paper is negligible, so the moment of inertia of the roll does not change as the paper enrolls. Now, the cit-I13.r effect of the entailed paper is to measure the amount that the roll turns. an. tam “H41 m “Esra mm inset 4L; I'ei'l, fr 15-! WI'CJ f'hii‘wi‘J—DMl li/E'FV‘H'Cn. til! ‘4': [—1232- T‘:II"II "' 'ﬁbr ‘ {l b. m f ,r’ " ' €33.11fo ﬁaumJELﬁ pal r. ' r r -| _ ‘- _ \ C30 'iﬂ‘ , the: t G" I' c.0— r'ip D r ,. gr [3.] I.LTMr-£J H-b bf??? 1r“ circieﬁ. e1? .ILEV'P-r a" ad'ntﬁ “Tint e er (‘3 “4"“3‘” "5 It. :2 ff} 5;"? ’5: 0» 01¢ w is r ' J a l ‘3 : gr.“ :\ an, .-~5—-1 c‘ I g: f- r;- I; I '1. :'/{~..-’ [aer—J Fllgq Jl lip—'9 3 c- was mt) ’j'r'e age is irww ' If]: : L‘s-7R. ~ Cit-H =16): trip-z Tia wlsui- = Cilia”? .6 Inf-Er: .- C ITwill u J—Qf‘l “I”: fr: g Cpl-j.""Diﬂttlﬁsaqrngf/s‘lYl-ESB: 59'?rﬁd/g 0/ j??- “Eels 7+ Ji—raacia t'ﬁ'i'pﬂ- Per-V"? "i"? “tr/M? :0.) ‘-- _ (bl; I DJ..ﬂ '-" C") O: (EQ,E 'Jg*1[-F1:,ﬂ\ie ..—5 a": ﬁllrﬂfe-ﬂi E‘ f-ft} ‘3' SI é Mil; fﬁiinﬂr iI-Hﬂl-E‘I'Ilfdr Problem IDS 1a. A person of mass 375 kg stands at the center of a rotating men'ngo—round platform of radius 31) m and moment of inertia loos kg‘mE. The platform rotates without friction with angular velocity 21] rad!r s. The person walks radially,»r to the edge of the platform. {a} Calculate the angular velocity when the person read‘tes the edge. (b) lCompare the rotational kinetic energies of the System of platform plus person before and after the person's walk. 1c. We are told that the person walks radially to the edge of the platform — this means we can assume that the act of walking does not exert a torque on the platform. So.r the only effect of the person’s change in position is that the moment of inertia of the merry-goer changes. ﬁrs PH: ﬁyftcf'ﬂﬁ-I rg‘nes. ' Slit. rAEII-J- ErraLﬂ-Ap‘ wsﬂﬂr‘lﬁ/Vﬁx I '3'? be: Ira an? - _ to m - ~-——F— Km \ p 1—35 kg. TVS:- F pafﬂ— -" 3 rm:?#:ﬁﬁ*' r1: sﬂﬁ'anﬁ: ‘9 EH- rm}: Ln;1 A —- GH- ﬁQr-“é'ga IFEFhmTI‘E'K ;\ #- TF '2 ..__ La_rﬂ’}> .l—Dwn- rd. (J (lomto M’l' r; .l ,9! _ “’1” la”: _' 1' liltli/‘l £321.: 7—11 Him-t a} Lina-n; sweet: g: {Tug 5.3”“: r .Eac E “a”. J 1 KsT-ttwﬁ "J II kP—__T: £411.13 _ i ' . . . r . __ _ .. iow * “ﬂew in a...» an no team l'""*£ﬁ: I . I . I / r. Egg-J girl-CE,- - Jl a lﬂ Eight EELS I. —-ll-.IL-.I fl; JFK-15‘ LE 3 (“ZEN f V TH I yum}: rgé'f'5l'“ O~~.T1...Je~rpf or a Motrin fl 0 _ r W“: .. I “t” ire-:2. Em: L A I Jr ' j" C I -. I _ Obs. l -..'J - Ci ‘0‘!- Inn-.1 H'Ol-Cli lr-“t' Eu{.5r '/"~ §__ dtrcc‘h; (Jr! “on-"hr:- d‘f I '5'”- t C l,“ 5’.”’“-.5+‘3 t" e rras, a“: t - was NJ : 11 sale. «that; ﬁwelﬂ‘l can be wheel be?“ «Estes other} bah». weft-ads ewe-ﬁne- Problem 105 la. A cord connected at one end to a block which can slide on an inclined plane has its other end wrapped around a cylinder resizing in a depression at the top of the plane as shown below. Determine the speed of the block after it has traveled 1.312} m along the plane, starting from rest. Solve the problem both (a) if there is no friction (b) if the coefﬁcient of friction between all surfaces is p = DIES. 1c. We know that the cord is wrapped around the cylinder. This means that any time the block travels a certain distance clown the incline, it unwraps an amount of string of equal length. So, the distance traveled by a point on the edge of the pulley [or R times the angular distance 9} is equal to the distance traveled by the block. We also assume that the depression in the inclined plane is exactly the right shape so that the string pulls the block only parallel to the incline, and pulls the pulley only tangentially to its surface. QCA ' bﬂjgﬁeﬁ Twig-r5 -i; ﬁg [Flu-r magi/1- E i.. . I. _,l. _ u Lg} l€ r1 [hm-n.3- ;_ L. P'q‘. III} { k (a?! . .a' --I’ .I.L .‘g—‘ .r 5-5,. I d l” a l; l , L y’l’l: -_.")L1_""{‘h v _ g x _ C -. f 4e: x/ (ﬂ 1 K j “xi—l _|'_ R a If 2 ‘ - 3 CHU- F " Jr's High Q ' '30.]? ,Itpr \\ s ,m, 2w ~ o .. g x- r: (.J m burger“ Jan was 5,: no:-:ff:~,L-J{er‘:{.s";eﬁrs3 I' ,.. . / nor-5. ﬁg?— Q‘IEJE -: .__r /.-+ EILDJ- Matt}— __!l “GLIT— H} = f I-_,. .u w-a . it: .25: "J___ L-I c__. 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