EN40: Dynamics and Vibrations. Final Examination Wed May : 2pm-5pm

Transcription

1 EN4: Dynacs and Vbratons Fnal Exanaton Wed May 1 17: p-5p School of Engneerng Brown Unversty NAME: General Instructons No collaboraton of any knd s pertted on ths exanaton. You ay brng double sded pages of reference notes. No other ateral ay be consulted Wrte all your solutons n the space provded. No sheets should be added to the exa. Make dagras and sketches as clear as possble, and show all your dervatons clearly. Incoplete solutons wll receve only partal credt, even f the answer s correct. If you fnd you are unable to coplete part of a queston, proceed to the next part. Please ntal the stateent below to show that you have read t `By affxng y nae to ths paper, I affr that I have executed the exanaton n accordance wth the Acadec Honor Code of Brown Unversty. PLEASE WRITE YOUR NAME ABOVE ALSO! 1- [41 ponts] 1 [1 POINTS] [1 POINTS] TOTAL [63 POINTS]

2 FOR PROBLEMS 1- WRITE YOUR ANSWER IN THE SPACE PROVIDED. ONLY THE ANSWER APPEARING IN THE SPACE PROVIDED WILL BE GRADED. ILLEGIBLE ANSWERS WILL NOT RECEIVE CREDIT. 1. The vehcle shown n the fgure starts fro rest and accelerates (wth constant acceleraton) to 1/s over a dstance of 5. Its acceleraton s j (a).5 /s (b) 1 /s (c) /s (d) None of the above Use the constant acceleraton straght lne oton forulas v v = a( x x) 1 = a 5 a = s Or alternatvely v v = at t = 1 / a x = vt + at 5 = a a = 1 / 5 = / s a ANSWER C ( POINTS). The vehcle n proble 1 has four wheel drve and each wheel s at the pont of slp (so T = µ N for each wheel). The coeffcent of frcton at the contact between the tres and road s approxately: (a).1 (b). (c).5 (d) None of the above A free body dagra and Newton s law shows that ( TA + TB) + ( N A + NB g) j= ax. The frcton law says TA = µ NA TB = µ NB so fro the coponent of Newtons law ax = ( N A + NB) and fro the j coponent N A + NB = g, so a x = µ g. The frcton coeffcent s therefore.. ANSWER B ( POINTS)

3 3. A prey partcle wth ass 3 gras travels at constant speed of 1/s around a crcular path wth radus 1. It s subjected to a vscous drag force wth agntude.4n and drecton opposte to the drecton of oton, as well as a propulsve force. The agntude of the propulsve force on the partcle s (a).3n (b).4n (c).5n (d).1n Newtons law for the partcle s Fp.4t =.3 1 / 1n F p = =.5N ANSWER C ( POINTS) R n V t 4. The fgure shows a force-extenson curve for a length of clbng rope. The energy (or work) requred to break a 1 length of rope s Break (a) 1.5 kj (b).5 kj (c) 5 kj (d) 1 kj, or cannot be deterned Force (kn) % change n length A 1 length of rope wll stretch by 1 pror to fracture. The area under a force-extenson curve for such a rope would be.5kj ANSWER B ( POINTS) 5. The center of the wheel oves wth (nstantaneous) velocty v O = V and the wheel rolls wthout slp. The rate of work done by the force P actng at pont D on the wheel s: (a) zero (b) PV/ (c) PV (d) PV R D O C P j The rollng wheel forula gves the angular velocty as ω = V / R. The velocty of pont D s vd = vo + ωk Rj= ( V ωr) = V. The force oves wth the sae velocty and the rate of work s F v ANSWER D ( POINTS) 3

4 6. A frctonless collson takes place between two bllard balls. { nt,} denote unt vectors parallel and perpendcular to the lne connectng the centers of the spheres at the nstant of collson. Please crcle the answers to ndcate whether the followng stateents are true or false: t n R (a) Moentu of the entre syste n the n drecton s conserved durng the collson (b) Moentu of the entre syste n the t drecton s conserved durng the collson (c) Moentu of the black ball n the n drecton s conserved durng the collson (d) Moentu of the whte ball n the n drecton s conserved durng the collson (e) Moentu of the black ball n the t drecton s conserved durng the collson (f) Moentu of the whte ball n the t drecton s conserved durng the collson TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE (All these are solved usng the pulse-oentu forula I= p1 p for ether the full syste or a partcle. Moentu s conserved n a drecton f there s no pulse n that drecton. There s no external pulse on a syste consstng of both balls so oentu s conserved n both n and t. For the ndvdual balls there s a large pulse n the n drecton but no pulse n the t drecton because the contact s frctonless. (3 POINTS) 7. The two asses n the fgure both have asses 1 kg. At te t= sphere A has velocty 1 /s and B s statonary. They collde wth a resttuton coeffcent e=. The pulses exerted on the two partcles durng the collson are: A R 1 /s B R A B R R Collson (a) IA = 1 IB = 1 Ns (b) IA = 5 IB = 1 Ns (c) IA = 5 IB = 5 Ns (d) IA = IB = Ns Total oentu s conserved and the two partcles ove at the sae speed after collson (fro the resttuton forula). Ther veloctes are therefore 5/s after collson. The pulse on each partcle s equal to ts change n oentu so I = v = (5 1) I = v = (5 ) A A B B ANSWER C ( POINTS) 4

5 8. The fgure shows a vbraton easureent fro an acceleroeter. The apltude of the dsplaceent s (a) 1.7 (b) 4 (c) 1.7 (d) 41 The apltude of the acceleraton s 4 /s and the perod s.s. The angular frequency s therefore ω = π /. = 1π. The dsplaceent apltude follows as / 4 / (1 ) 41 X = A ω = π ANSWER D ( POINTS) 9. The fgure shows a datoc olecule. In statc equlbru, the dstance between the atos s r=a. When the atos vbrate, the dstance between the changes by a sall aount x, so that rt () = a+ xt (). The equaton of oton for x s d x a F a = dt a+ x a+ x The natural frequency of sall apltude vbratons of the olecule s F (a) ω 8 n = a F (b) ω 4 n = 1 (c) ω n = 4 1 (d) ω n = 8 F a F r=a+x(t) Lnearze the EOM by takng a Taylor expanson of the rght hand sde for sall x a a a a F 16F + 16F x= 4 16 x a+ x a+ x 6 1 ( a x) ( a x) + + a The lnearzed equaton of oton s therefore coeffcent of the acceleraton ter s 1/ n a 4 16F ω so ω n = 8 F /( a). x= d x + x = ; a Case I vbraton EOM. The dt ANSWER A ( POINTS) 5

6 1. The fgure shows a D dealzaton of a vbraton solaton syste. There s no slp at any of the contacts. The syste has (a) degrees of freedo and vbraton odes (b) 4 degrees of freedo and 4 vbraton odes (c) 6 degrees of freedo and 6 vbraton odes (d) 6 degrees of freedo and 4 vbraton odes Horzontal base oton Vertcal base oton There are two ovng asses n the syste, each of whch s a rgd body that can translate freely n both horzontal and vertcal drectons, and can rotate about the k axs. Ths s x3=6 DOF. There are no rgd body odes so 6 vbraton odes. ANSWER C ( POINTS) 11. Syste A n the fgure s crtcally daped. Syste B s (a) Overdaped (b) Crtcally daped (c) Underdaped (d) Favorably dry c k A c k B c k The dapng factor for A s ζ = c / k = 1. Snce the sprngs and dashpots are n parallel the dapng factor for B s ζ = c / k = > 1 so the syste s overdaped. ANSWER A ( POINTS) 1. For the ost effectve base solaton, a vbraton solaton syste should have (a) A hgh natural frequency and hgh dapng (b) A low natural frequency and low dapng (c) A low natural frequency and hgh dapng (d) A hgh natural frequency and low dapng ANSWER B ( POINTS) 6

7 13. The sprng-ass syste shown n the fgure s subjected to a haronc force. The apltude of the force s 1 kn and the frequency s equal to the undaped natural frequency of the syste. The apltude of vbraton s 1. If one dashpot s reoved, and the syste s subjected to the sae force (.e the apltude and frequency of the force s the sae), the steady-state apltude of vbraton wll be: (a) (b) 1 (c).5 (d).5 c c F(t) k The forula for apltude s X = KM ( ω / ωn, ζ) F. When ω = ωn the agnfcaton M = 1/ ( ζ ) where ζ = c / ( k). Reovng a dashpot does not change the undaped natural frequency and halves ζ. The apltude wll double. ANSWER A ( POINTS) 14. The rotaton tensor represents 1/ 1/ 1/ R = 1/ 1/ 1/ 1/ 1/ (a) A 9 degree rotaton about an axs parallel to n= ( + j )/ (b) A 9 degree rotaton about an axs parallel to n= ( j )/ (c) A 45 degree rotaton about an axs parallel to k (d) A 45 degree rotaton about an axs parallel to. The forula gves 1+ cosθ = R + R + R = 1 cosθ = θ = 9 o xx yy zz 1+ cosθ = Rxx + Ryy + Rzz = 1 cosθ = θ = 9 o n= [( R R ) + ( R R ) j+ ( R R ) k]/sn θ = / + j / zy yz xz zx yx xy ANSWER A ( POINTS) 7

8 15 Pont C on the robotc actuator shown n the fgure oves vertcally wth velocty v C = 1 j /s. The angular veloctes of ebers AB and BC are (a) ωab =, ωbc = 1 rad/s (b) ωab = 1, ωbc = 1 rad/s (c) ωab = 1, ωbc = 1 rad/s (d) ωab = 1, ωbc = rad/s A 45 B j 1 C V Cy = 1/s a C = Rgd body kneatcs forulas vb va = ωabk ( + j) = ωab ( + j ) vc vb = ωbck = ωbc j. Add: vc va = 1 j= ωab ( + j) + ωbc j ωab = ωbc = 1 ANSWER A ( POINTS) 16 Pont C on the robotc actuator shown n the fgure oves vertcally wth velocty v C = 1 j /s and zero acceleraton. The angular acceleratons of ebers AB and BC are (a) αab =, αbc = 1 rad/s j (b) αab = 1, αbc = 1 rad/s 45 A (c) αab = 1, αbc = 1 rad/s (d) αab = 1, αbc = rad/s Rgd body kneatcs forulas αb αa = αabk ( + j), αc αb = αbck 1. Hence, addng αc αa = αab ( + j) + αbc j =. Hence αab = 1 αbc = 1 rad/s ANSWER B ( POINTS) B 1 C V Cy = 1/s a C = 8

9 17. Wth the rng gear fxed, the planet carrer and sun gears n the epcyclc gear syste shown have a rato ωzs / ω zpc = 3. If the sun gear has teeth, the rng and planet gears have (a) NR = 4, NP = 1 teeth (b) NR = 4, NP = teeth (c) NR = 8, NP = teeth (d) NR = 8, NP = 3 teeth ω zs Sun gear R S r R R Planet Carrer ω zpc ω zp Planet gear ω zr The general forula relatng angular veloctes n a planetary ω ω / ω ω = ( N / N ) gear syste s ( ) ( ) zr zpc zs zpc S R Substtutng nubers gven ω / 3ω ω = / N N = 4. Fnally N = N N N = 1 ( ) zpc zpc zpc R R P R S P Rng gear ANSWER A ( POINTS) 18. The two gears A and B n the fgure have rad R and R, and ass oents of nerta and R, respectvely. Ther centers are statonary. If gear A rotates at angular speed total knetc energy of the two gears s / R ω A, the (a) (b) 4R ω A 3R ω A ω Α (c) R ω A R ω A / (d) The total KE s ( IAzzωA + IBzzωB )/; we know ω = ω R / R = ω / and so the total KE s B A A B A R ωa + R ωa = R ωa ( ) ( /) /4 / / A R R B ANSWER D ( POINTS) 9

10 19. Three bars wth ass and length L are connected to for an equlateral trangle as shown n the fgure. The poston of the center of ass of the trangle wth respect to the orgn at O s (a) L+ Lj 1 (b) L+ Lj 3 1 (c) L+ Lj 3 (d) L+ Lj 3 Use the COM forula 1 1 ( 1 3 ) ( 3 3 ) 1 L L L L L M r = = + 3 j j 3 j. O L j L L ANSWER C ( POINTS). Three bars wth ass and length L are connected to for an equlateral trangle as shown n the fgure. The (D) ass oent of nerta of the assebly about the center of ass s (a) (b) (c) IGzz IGzz IGzz = L /4 = L = L (d) IGzz = L / L G j L L The ass oent of nerta of one bar about ts own COM s ( L) /1 L / 3 =. The COM of each bar s a dstance L / 3 fro the COM of the assebly. Therefore usng the parallel axs I = 3 L /3 + L /3 = L theore Gzz ( ) ANSWER C ( POINTS) 1

11 1 An unbalanced wnd-turbne s dealzed as a rotor-excted sprngass syste as shown n the fgure. The ass represents the tower, and represents the cobned ass of the three rotor blades. The sprng and daper represent the stffness and energy dsspaton n the tower. The rotor s unbalanced because ts center of ass s a sall dstance Y away fro the axle. The total ass ( + ) of the syste s 5kg. c k ωt Y The fgure shows the results of a free vbraton experent on the turbne. 1.1 Use the data provded to deterne the followng quanttes: (a) The vbraton perod The perod s.5 sec [1 POINT] (b) The log decreent The log decreent s 1 log.5 = 1.6. [1 POINT] 11

12 (c) The undaped natural frequency 4π + d The undaped natural frequency s ωn = = 1.8 rad / s T (d) The dapng factor The dapng factor s ζ = δ 4π + δ =. (e) The sprng stffness ωn = k k = ωn = 41 kn / [1 POINT] [1 POINT] [1 POINT] (f) The dashpot coeffcent c ζ = c = ζ k = 18 Ns / k [1 POINT] 1

13 8 6 4 Dsplaceent () Te (s) 1. The fgure shows the easured dsplaceent of the syste durng operaton. The blades have a radus of 4, and the total ass of the syste ( + ) s 5kg. Assung that the rotor can be balanced by addng ass to the tp of one blade, estate the ass that ust be added to balance the rotor. We know that the vbraton apltude of the unbalanced rotor s X = Y ω / ωn n n + (1 ω / ω ) + ( ζω / ω ) We can use the vbraton easureent to estate the product Y : Fro the graph, we see that the perod s about 6 sec so ω 1 rad/s, so usng the nubers fro 1.1 1/1.8 X= Y =.61Y 5 (1 1/1.8 ) + (. /1.8) Snce the easured apltude s about 6, we conclude that Y 1kg. We want to ove the COM back to the center of the rotor recall that the COM s (1 / M) r so the requred ass at the blade tp s 1/4 kg. [4 POINTS] 13

14 . The fgure shows a spool (e.g. a yo-yo) wth outer radus R, ass and (D) ass oent of nerta IGzz = R / restng on a table. The hub has radus r. A constant vertcal force P s appled to the yo-yo strng. The goal of ths proble s to () fnd a forula for the (horzontal) acceleraton of the spool, and () fnd a forula for the crtcal value of P that wll cause slp at the contact between the spool and the table j R r O P.1 Draw a free body dagra showng the forces actng on the spool. Assue that the spool reans n contact wth the surface, and that no slp occurs at the contact. (The frcton force can go n ether drecton snce there s no slp) [3 POINTS] R r g N O P T. Wrte down the equatons of lnear (Newton s law) and rotatonal (the oent-angular acceleraton relaton) oton. Your equaton should nclude forces fro.1, and the lnear and angular acceleraton of the spool. Please state whch pont you are takng oents about for the oent equaton. F=a T+ ( N + P g) j= agx Rotatonal oton (takng oents about the contact pont) 1 1 rpk = Rj agx+ R azk rp = RaGx + R az [ POINTS].3 Wrte down a relatonshp between the angular acceleraton α z and lnear acceleraton a G of the center of ass of the spool The rollng wheel forula gves αg = Rα z [1 POINT] 14

15 .4 Use. and.3 to fnd forulas for (a) the angular acceleraton and (b) the lnear acceleraton of the spool n ters of P, and other relevant varables. Cobnng results fro. and.3: 1 rp rp = R αz + R αz αz = 3 R The acceleraton s therefore a G rp = 3 R [ POINTS].5 Fnd forulas for the reacton forces at the contact, n ters of P,, g, R and r F= a gves rp T = 3 R N = g P [ POINTS].6 The contact has a frcton coeffcent µ. Fnd a forula for the crtcal value of P at the pont where the contact begns to slp r g T = N P = ( g P) P = 3 R + r / (3 R) [ POINTS] 15