Homework 8: Rigid Body Dynamics Due Friday April 21, 2017

Transcription

1 EN40: Dynacs and Vbraons Hoework 8: gd Body Dynacs Due Frday Aprl 1, 017 School of Engneerng Brown Unversy 1. The earh s roaon rae has been esaed o decrease so as o ncrease he lengh of a day a a rae of approxaely.3 llseconds per cenury. Take he earh o be sphercal wh a radus 6371k and ass 6x10 4 kg. Calculae he rae of change of angular oenu of he earh and hence esae he agnude of he orque acng on he earh (neglec coplcaons lke he earh s precesson). We can use he angular oenu equaon 1 0 IG= 1 0 = 5 k ( π / ( ) π / (4 3600) ) h IG ( ) = = kg s d d 16 The orque s equal o he rae of change of angular oenu so Q = N [3 PINTS]. The fgure shows a spool (e.g. a yo-yo) wh ouer radus, ass and (D) ass oen of nera IGzz = / resng on a able. The hub has radus r. A (known) horzonal force P s appled o he yo-yo srng. The goal of hs proble s o () fnd a forula for he (horzonal) acceleraon of he spool, and () fnd a forula for he crcal value of P ha wll cause slp a he conac beween he spool and he able. r P.1 Draw a free body dagra showng all he forces acng on he spool (assue no slp a he conac) r g N P T [3 PINTS]

2 . Wre down he equaons of oon (Newon s law, and he angular oenu equaon s eases o apply he oen-angular oenu forula abou he conac pon) F= a ( P + T ) + ( N g) = agx dh r F+ Q= ( r) Pk = agx+ IGzzazk d 1 ( r) Pk = agxk+ azk.3 Wre down he kneacs equaon relang he acceleraon and he angular acceleraon of he spool agx = a z [1 PINT].4 Hence, calculae he acceleraon of he spool (does he spool accelerae o he lef or rgh?) Usng.3,.4, noe ha 1 ( r) P ( r) P = agxk agx agx = 3 Snce a Gx > 0 he spool acceleraes o he rgh..5 Calculae he reacon forces a he conac. If he coeffcen of frcon a he conac s µ, calculae he crcal value of P ha wll cause slp a he conac. ( P + T ) + ( N g) = agx shows ha ( r) P P r N = g T = P = (1 + ) 3 3 P A he onse of slp T = N (1 + r / 3 ) = g P = 3 g / (1 + r / 3 ) 3

3 3. The fgure shows an pulse haer ha s used o srke he ground n a sesc experen. I consss of a slender rod wh a haer-head a s end. The rod has ass, lengh L and ass oen of nera L /1 abou s cener of ass. The haer head has ass and has neglgble ass oen of nera abou s cener of ass. The rod pvos freely a. I s released fro res wh he slender rod vercal. The goal of hs proble s o calculae () he speed of he haer-head when us hs he ground and () he reacon forces acng a a he nsan us before he haer-head srkes he ground. =0 L Jus before pac 3.1 Fnd he oal ass oen of nera of he syse (he rod ogeher wh he haer-head) abou The oal ass oen of nera abou s L /1 + ( L / ) + L = 4 L / 3 3. Usng energy conservaon, show ha he angular speed of he rod us before he haer-head srkes 3 he ground s = / g L 1 3 PE + KE = cons gh = I L = gl + gl / = g / L Hence, fnd he speed of he haer head us before hs he ground 3 v = L = gl [1 PINT] 3.4 Fnd he dsance d of he cener of ass of he syse fro, n ers of L. d = ( L + L /)/( ) = 3 L /4 G d [1 PINT] 3.5 Fnd a forula for he acceleraon of he cener of ass of he syse a he nsan when he shaf s horzonal n ers of d, he angular velocy and acceleraon α. Express your answer as coponens n he, bass shown. The rgd body kneacs forula gves G= α G / + G / = dα + d α k r k k r

4 3.6 Draw a free body dagra showng he forces acng on he syse on he fgure provded below. Assue ha he haer-head has no ye h he ground. d g x (also K o draw gravy on he head and shaf separaely) y 3.7 Usng he rgd body dynacs equaons (.e. he equaons relang angular acceleraons and oens and/or Newon s laws) show ha he angular acceleraon of he haer a he nsan 9 g us before srkes he ground s α = 8 L g Iαk = M L α = g L α = L 3.8 Fnd a forula for he reacon forces acng a a he nsan us before he haer-head hs he ground, n ers of and g (no oher varables should appear n your soluon). Express your answer as coponens n he, bass. 3 9 g 3 3 F= αg x+ ( y g) = ( dα+ d) = L + g / L L 4 8 L x = g y = 1 g = g [3 PINTS]

5 4. A sold cylnder wh ass and radus ress on a conveyor bel. A e =0 boh cylnder and bel are saonary. The bel hen sars o ove horzonally wh speed V. 4.1 Snce he cylnder s a res, and he bel s ovng a =0, slp us nally occur a he conac beween he. Draw a free body dagra showng he forces acng on he cylnder C V g C N T 4. Wre down he equaons of ranslaonal and roaonal oon for he cylnder (use he D equaons) [3 PINTS] Translaonal oon T+ ( N g) = ax oaonal oon (oens abou ) Tk = IGzzαzk 4.3 Use he frcon law and he soluon o 4. o calculae he lnear and angular acceleraon of he cylnder durng he perod of slp g T = µ N because we have slp so solvng 4. gves ax = g az = IGzz 4.4 Hence, calculae how long akes before he cylnder begns o roll whou slp on he bel. Fnd he lnear and angular velocy a hs e. g The velocy of follows as v = ax = µ g The angular velocy s z = αz = IGzz A he onse of rollng he conac pon C has o have he sae velocy as he bel. The velocy of C s g vc = v + z = g + IGzz A onse of rollng V I 1 Gzz V g + = V = = I Gzz g + I 3g Gzz for IGzz 1 = Velocy and angular velocy follow as V V v = z = 3 3 [3 PINTS]

6 k Back Fron Torsonal acuaor 5. Ths sarer every day clp helps explan he physcs of he ca rghng reflex. The goal of hs proble s o esae he necessary o coplee he aneuver. A ca can be dealzed as wo cylnders wh reracable fron and back legs, conneced by a orsonal acuaor. We wll assue ha The acuaor exers an equal and oppose orque on he fron and back cylnders Wh legs exended, one cylnder and a par of legs have a cobned axal ass oen of nera I xx ax ; wh legs reraced hey have a ass oen of nera I xx n The cener of ass of ass of boh fron and back (ncludng legs) les on he axs, and does no ove when he legs are exended or reraced. The ca flps self uprgh n he followng sequence (1) The ca sars a res n an nvered poson. () Iedaely afer s released, he ca reracs s fron legs and Q 0 exends s back legs. (3) The acuaor hen apples a posve orque Q= Q0 o he fron cylnder (and an equal and oppose orque o he back) for a e -Q 0 nerval (see he fgure), followed by a orque Q= Q0 for he e < < (4) The ca hen reracs s back legs and reracs s fron legs (5) Durng he subsequen e nerval < < 3 he acuaor apples a negave orque Q= Q0 o he fron cylnder and an equal and oppose orque o he back, and fnally for 3 < < 4 he acuaor exers a posve orque θ back Q= Q 0. k Afer hs sep he ca should be facng wh legs down. The goal of hs proble s o calculae he agnude of he orque Q 0 and e θ fron nerval. Q() 5.1 Consder oon of he syse durng sep (3). Use he equaon of roaonal oon o fnd he angular acceleraon and hence fnd forulas for he angular veloces xfron, xback and he angles θ fron and θ back a e =, n ers of Q0, I, xx ax, Ixx n. I xxax I xxn The rgd body roaon forula gves

7 Q() Q () = IGα+ IG Here roaon s always abou he x axs (whch s a prncpal axs of nera) so we can use Q () = IGxxα x We can negrae he acceleraon graphcally: he angular velocy s he area under he acceleraon curve; he angle s he area under he angular velocy curve. Ths gves Q 0 -Q 0 () fron = 0 Q0 θfron = Ixx n back = 0 Q0 θback = Ixx ax [3 PINTS] 5. Consder oon of he syse durng sep (4). Fnd forulas for xfron, xback and he angles θ fron and θ back a = 4. θ () Q 0 /I xxn Q 0 /I xxn The sequence repeas n he oppose drecon, so 1 1 fron = 0 θfron = Q0 Ixx n Ixx ax 1 1 back = 0 θback = Q0 Ixx n Ixx ax 5.3 Esae and Q 0 fro he followng daa: The hgh-speed vdeo suggess ha Gg s able o rgh herself over a drop dsance of abou 4 fee. You can use hs o calculae. Ixx ax = Ixx n = kg The e aken o fall 4f can be calculaed usng he sragh-lne oon forulas 1 / 1. / sec g = x = x g = Hence = 0.5 / 4 = (1/ 8) sec The ca urns hrough 180 degrees n hs e, so Q = π Q 0 = 8 π I xx ax = 3 I N xx n Ixx ax

8 6. The fgure shows a desgn for a vbraon solaon plafor. The wo roller bearngs have radus and ass ; he plafor self has ass. The goal of hs proble s o calculae a forula for he naural frequency of vbraon. 6.1 Suppose ha he plafor s dsplaced by a dsance x fro s sac equlbru poson, as shown n he fgure. Wre down he oal poenal energy of he syse n ers of sprng sffness k and x. k,l 0 k,l 0 k,l 0 x k,l 0 Two sprngs, usng he sprng PE forula PE = kx [1 PINT] 6. Usng kneacs forulas, fnd a forula for he velocy of he cener of ass of he rollers, as well as her angular velocy, n ers of dx / d and oher relevan varables. We can use he rollng wheel forulas: 1 dx 1 dx v = = d d 6.3 Hence, fnd a forula for he oal knec energy of he syse n ers of dx / d and oher relevan varables. Add he KE of he hree asses, nong ha IG = / for he rollers, and reeberng ha here are wo rollers (he nd and 3 rd ers n he KE below are ranslaonal and roaonal KE for he wo rollers): 1 dx 1 1 dx 1 dx 1 1 dx 1 7 dx KE = + v + = + + = d d d d 4 d 6.4 Use energy conservaon o derve an equaon of oon for x, and hence deerne a forula for he naural frequency. d dx 7 dx d x ( PE + KE) = 0 kx + = 0 d d 4 d d 7d x + x = 0 8k d Hence n = 8 k / (7 ).