Physics exam formula sheet

Physics exam formula sheet - Example 1 A heavy sphere with...

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Unformatted text preview: Example 1. A heavy sphere with mass M is attached to the end of a slender rod with mass m and length L The rod is pivoted at one end 0 and released from rest in the position shown, The size of the sphere is negligible compared with the length ofthe rod. Find the initial angular acceleration ofthe rod at the instant it is released Tin. W’q- 1, L rig-A5,)? 13 “iii'j 212.:2‘? (:ass)+l‘1j (Land) 5 Mimi's-259 +hV M : r‘ J , Z:-"mi.p‘+ MLL _ ,,,,,,,,,,, ti 3 ’14“ . 2, A block of mass M is attached to a cable that wraps around the pulley as shown. The pulley is pivoted at its center, and it has radius R and moment of inertia l. The block is released from rest on the ramp, andthe coefficient of friction is M. between the block and the ramp. Find the acceleration ofthe block and the tension in the cable/0 25 => 0 T N= I13 (usg =7 70“ :fl‘mjw’g zFx IMay ~M5~Me~r—Azn3msa = Mia. :49 , > I - ZT‘:Ia( 1. qlqiwkrétrfd‘; 4} 7-2:!e—Ee}; and —(:> [m 7476 ' ' ' ’ ’ ' /’ n=%eria=fg :>T7(~I%a \//\ IV 3. Two blocks A and B are connected by a massless cable over a pulley as shown. The radius of the pulley is R and it can be treated as a disk. The blocks are released from rest on the frictionless ramps. Find the accelerations ofthe blocks and tensions in the cable. ‘ A) 2F, :mfiiy B) Efx " "My W! B ““3 2 (0— rm m e” .7; = to“ I; - “454'“9a: mn‘L‘m T r v. ring "52947" w ’ l r- _ A: [Zix—{q my Z’Z‘HZ hot)“ (i J < v 2 ran, a,: rat, a” 2:192,- a :t/af + (1:: (m): + 0'me K[Ti/(j. "fl/it; ‘L who: : {23:} “js’lrafi WW: 3 ”ka1: WNS" ’ti9 42/wa Mmfim! {,err fi/q park’s/r a 7 (haw Why”? Vulat‘ r: dzsvlwswflzw‘wptw", 4:100 Ltczf (assert. ”Pt-filed: (I) 129% =_.Z;LNJ: El) Loflfsi‘vn ‘ mvl":.7:c<.) bjdzoé mgrzflaé‘mvfr => 3w“! 6) éavn‘pékbk (f) jo‘i’imosjk if FL, : GMT: , G : 6.67x10’” 141113ng 1" GM” GMC GM'E w=mg=m r2 =mm, wuzmgu=m R, m=mo+m . v E l 1 Us =7GMflm, Ug =_Gmlm3 =-ng 6 : 190 + was +303- r r Me =5.97x10“ Kg, Re :6380 km (atrflpgggwfl a)? = mg + 20509410) 1): GM“, r:Rg+h 1‘ 3 T=27r " GMv w GM Kzlmu“=GM"m, Ug:—GM“m, E:K+Ug:_ “m 2 2r ' r 2r U, usAiiz—E 2 GMml i WAZ=E2_EII ‘ [—7W} 2 ’ir: 03:45.“: MM" I” Characteristic: acceleration = — constant x position Angular frequency. frequency, and period: a) = JE,f = LJE, T = 27Z\/E m 271: m k Position, velocity, and acceleration: x = Acos(an‘ + $5), U = —aJA sin(a)t + p), a = 403A cos(a)z‘ + gt) Umnx = 60A! amax : (02A Amplitude and phase angle determined from the initial conditions: x0 and on: f 2 A: xnz-i- & , tan¢5=-i a) mxa Energy approach: 1 l E=2mug+21oc§=émuj+ilocz :ikA2 zémufim 7‘ @mir Q 053% MIA 1):» M 3. Two blocks A and l are connected by a cable over a pulley as shown. The system is released with an initial speed 00 for block A down the ramp. The radius of the pulley is R and it can be treated as a disk. Find the velocity of the blocks afier A slides a distance d along the incline. K1+U3. try; 4 M41: id“ M514 by; L f—LZWWEi “1+ &(£M§21)wal+ Mkjfismd,‘ Hiawflwwn 5i “flier; : it (mmfiua‘ + iéi Mpfinga‘ irmajfllo-V‘ BB {ofigpd ml} “135 B m3 Example 1. A sphere with negligible size is attached to the end of a slender rod as shown, The rod is pivoted at end 0, and the assembiy is released from rest at an angle 6 up from the horizontal direction. Find the speed of the sphere when the assembly swings down to the vertical position. %*M31 ‘Mkibljl D KL: i(fille+;—1ML1)OJ‘L ij; Myiugs =9 M3" "‘5 (fwgi‘LMfliiL/‘WL I“; Uio 4;" a — -1 x." 031‘ Mfif—flmigL) (9' ML H Hahn ;.L A a , _ W i‘ 3 a “9) shsirgnfilwMdéh'hlsi) m 3 The two blocks A and B are connected by a cable as shown, and the system is then released with an initial speed no for /\/\/\x\ ;, 1/1.. 1/1/71 . ’ 0 :m (41% M1, = «1;: WM5A wxiNQSB A"; \/\/\/\/‘g .‘x =7‘s(»a3wsa.)al- flu; («‘3 (159,04: K ' 3 e '/'r l/ig t—a 1_,. 1/ 1/lr'L/iq / BOW ma.) u. 4- m1: (at 6M9) #1:.[mI3L060J11‘AK (“:3 ”‘93-“ _ m . 1( new)” +m;[email protected])} +‘u60t‘kéiudcs Q‘iUANmJ-‘Mfi e VINE; spew 4. The small glider (mass m) is held at rest against the spring (spring constant k) as shown. The glider is released and then slides along the track The track is frictionless except for a horizontal rough section with length d (a) Find the minimum initial compression in the spring, xmm, required for the bglider to complete the loop (b) If the velocity of the glider at point B 15 ‘\ l) entering the =lojp find the for aactin £32]: slideicb afizjack at point C. {1 my“? kid \mn", i: A13 Musk mm‘tmin Lan‘ac XM=>N,.— 1) [Win u N10 Eamon ¢nflq y‘V‘ '1. l-WH‘ _._§—~d.m—-———~) fulfil”: K “41‘ (gummy... camel“ ““he) I‘m/‘1 o =- -\ mv "an ‘11 w._'.. 4,4,0”th = were: A r=F a": F- --r sing}: Moment of inertia — Special cases: +___ _,_ fl [4 (l) Slender rod (mass— — Mand length— L) pivoted at the center: 1— ML' 1’12 ctr-*9 (2) Slender rod (mass— = Maud length= I.) pivoted at the end: I= ML [3 HM M (3) Circular disk or cylinder (mass = Mai-id radius = R) pivoted at the center: 1= MHZ/2 @ (4) A point mass Mof negligible size. I= Mr‘, 2 r— — distance from the pivot to the particle @3125 — Parallel-axismorem:10:10". + mad, d_= distance between the pivot point 0 and cm. 2. The small metal sphere with masstg is attached to two linear springs (k: . N/m) and spins horizontally aroiirid the vertica rod as shown. Efthe upper spring stretches twice as much as the lower one find (a) the elongation ofthe lower spring, and (b) the period of the rotating red. I ‘= M "E’s: Ks 1ka 94° choe fl ‘ 'Fgm, ‘9 4“}: r: = (1116‘ ‘—'_""‘_ :Hfln =P‘\ lens-fl 'Sin 9 6- p“ ?- ic-. F / Phanfl'mnwmém‘) (4) ‘ K i T1 3. Two blocks on the inclines are connected as shown and released from rest. Find the acceleration of the blocks and the tension in the cable. “A 1 [7‘1 : 0 FL): “rpm 91 FF =MK\(M‘:1‘5M 91) ~43 Mamie ”T" MiMKfljmeT-H‘o‘ M”) FM: “1%?" e1, 1) :r-mwai- M¢Mpflfiwt91 =f’\v,_0s. Fa: M1411 Waging?) ‘ 3. The frictionless track is made ofa snaight vertical section and a half circle as shown. Slider A (M,.: m) is released from rest at the top ofthe track and makes an ith the stationary slider 8 (M5— 2m) at the bottom ofthe track. Find the normal force from the track on Her .’ w en it reaches point C. ‘LD 28 : man: (3) mm Hg”; : mm: mm ,V‘ o “if/“*3“ lfiylkj; m 11:33 = ‘yiv: +Vng}4 3R = o 3 :1" V,» i r! (‘SQ (:9) “6 :Q'QELE +7 [KW/fin; t;"‘-ai/1.«= VJVWC :7“? : Li R i = W1 +312 'ms 4 Two carts A (mass — MA) and B (mass= MB) are moving along the frictionless tabletop as shown, with DA > u”. A linear spring (spring constant: k) is attached to the front of can A. Find the maximum compression in the spring during the collision. ,_ V l I<r+l$ZiClL3~HA41 J‘ ft L‘ x2. + ‘ K ., fififivnlili’iuvn‘ __, inky: +l MB 1 hi! M 1m. Mm we "eve ' Mail MW M. : MVL VmL -‘ Va: =V4 , m, em, 2m. 0, =———'vl +—‘—*L)2 m. + m2 ml + m2 2m ml 1); =——'~—Ul +7 m? LL. ml +m, m L+m2 ‘ Work, Dim: f: ~QAM$IL§' (All: :flws¢ .94 dds/F’KLLL , wtjwwfimw NF: 7: 5 f , piggy?) KmMWlwlil Ml; 9‘ (19“ W X-4‘19a I.) U ‘ WkN- S A Iv. 1 *'~ w m 599 (a) W M “ V ““1"“ Wmmtu W3 5;”‘g 3114942) bf ’ ~' ' ‘5’ “J; fi‘k(rfljtart}'0)5 awe CL ?' F’v . COM: mIUI +m303 = miui +m20i P = T ' 45'.le : "E; Rotational kinetic energy: K = (11’2)1a;", Gravitational potential energy: UR = mgcmw = 1; . 5 jofgd—L. ? f: F If gWJfaerm =3 ngfiaj’L ”fl—Em" 71;?)61 P'UdTP'J'L‘ Loni. T- : it): if}; 1': Fr'rex , r 14/69 a: we COM: mat), + mgr)2 : (m. + mflur 9.3 common velocity after collision ...
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