EN40: Dynamcs and Vbratons Homewor 7: Rgd Body Knematcs, Inertal propertes of rgd bodes Due Frday Aprl 20, 2018 School of Engneerng Brown Unversty 1. The rgd body shown n the fgure s at rest at tme t=0, and rotates counterclocwse wth constant angular acceleraton vector 2α Fnd 1.1 The angular velocty vector as a functon of tme 2L r B -r A B θ L We can ust ntegrate the angular acceleraton ω= 2α t A 1.2 The spn tensor W (as a 2x2 matrx, also a functon of tme) Usng the formula 0 2α t W = 2α t 0 1.3 The rotaton tensor (a 2x2 matrx for a 2D problem) R that rotates the rectangle from ts ntal to ts fnal poston shown n the fgure. Chec your answer by computng dr R T dt R cosαt snαt snαt = cosαt dr T 2αtsnαt 2αtcosαt cosαt snαt R dt = 2α t cosα t 2α t snα t snα t cosα t 0 2α t = 2α t 0 Ths reduces to W as expected. 1.4 Hence, express the rotated vector rb r A n (,) components. r r can be found usng the mappng r r = Rp ( p ) B A B A B A cosαt snαt 2L 2Lcosαt Lsnαt snαt cosαt L = 2 Lsnαt + Lcosαt
2. The rectangular prsm shown n the fgure s subected to two sequental rotatons: (1) A 45 degree rotaton about the axs (2) A -90 degree rotaton about the axs 2.1 Wrte down the rotaton tensor (matrx) for each rotaton We can use the general formula 2 cos θ + (1 cos θ) n (1 cos θ) nn sn θn (1 cos θ) nn + snθn 2 R = (1 cos θ) nn + snθn cos θ + (1 cos θ) n (1 cos θ) nn snθn 2 (1 cos θ) nn x z sn θny (1 cos θ) nn y z + snθnx cos θ + (1 cos θ) n z For the frst rotaton θ = 45, n = n = 0 n = 1 R (1) x x y z x z y x y z y y z x x z y cos 45 0 sn 45 1/ 2 0 1/ 2 = 0 cos 45 (1 cos 45) 0 + = 0 1 0 sn 45 0 cos 45 1/ 2 0 1/ 2 For the second rotaton θ = 90 0 1 0 (2) R = 1 0 0 0 0 1 [3 POINTS] (2) (1) 2.2 Fnd the rotaton matrx R = R R that descrbes the combned effects of both rotatons (you can use matlab to do the matrx multplcaton, but there s no need to submt the code or scrpt). Matlab gves 0 1 0 R = 1/ 2 0 1/ 2 1/ 2 0 1/ 2
2.3 Fnd the axs n and rotaton angle θ that wll complete the rotaton R drectly. We can use the formula 1+ 2cosθ = R + R + R xx yy zz ( Rzy Ryz ) ( Rxz Rzx ) ( Ryx Rxy ) n 1 = + + 2snθ 1 2 Ths gves cosθ = 1 θ = 1.71777rad 2 2 1 1 1 1 n= (1 ) 0.3574 0.3574 0.8628 2sn(1.71777) + + + = + 2 [3 POINTS] Optonal: You can chec your answer by downloadng a matlab scrpt from the homewor page of the course webste that wll anmate a rgd rotaton through an angle θ about an axs parallel to a unt vector n. You can use the code by navgatng to the drectory storng the fle n the Matlab command wndow, and then typng Anmate_rotaton(angle,[ nx, ny, n z ]) Where angle s your soluton for the rotaton angle θ (n radans), and nx, ny, n z are the components of your soluton for the unt vector n. 3. The fgure shows a four-bar chan mechansm that appears nsde both natural and artfcal nee onts. Member OB rotates counterclocwse at constant angular speed ω. Calculate the angular veloctes and angular acceleratons of members CB and AC. At the nstant shown, calculate the angular velocty and acceleraton of lns AB and BC. The rgd body formulas for the three members gve vb vo = ω ( rb ro) = ω ( L+ L) = Lω( + ) vc vb = ωbc ( rc rb ) = ωbc ( L) = LωBC va vc = ωac ( ra rc ) = ωac L( ) = LωAC ( + ) ω C O L B A L We now that vo = vc = 0 so f we add all the equatons lsted above we get Lω( + ) Lω + Lω ( + ) = 0 BC + = The component of ths equaton gves ω ω AC 0 The component gves ω ω + ω = 0 BC AC AC Solve these to get ω = ω ω = 2ω AC BC [3 POINTS]
For acceleratons, use αb αo = ω ( rb ro) = Lω ( + ) αc αb = αbc ( rc rb ) ωbc ( rc rb ) = LαBC + L4ω αa αc = αac ( ra rc ) ωac ( ra rc ) = αac L( + ) + Lω ( + ) At the statonary ponts ao = ac = 0, and add everythng: 2 Lω ( + ) LαBC + L4 ω + LαAC ( + ) + Lω ( + ) = 0 2 The component gves LαAC + 2Lω = 0. The component gves αbc + αac = 0 ThusαAC = 2ω αbc = 2ω [3 POINTS] 4. Ths publcaton descrbes a varable speed transmsson for a wnd turbne 1. The transmsson has 6 possble gear ratos, whch are acheved by connectng one of 6 possble (blue) gears to the output shaft (connected to the generator). The numbers of teeth on each gear or pnon are shown n the table (n each case the red gear drves a blue pnon. Calculate the 6 possble transmsson ratos ω / ω generator gearbox Let NA, N B be the numbers of teeth on the man gearbox and gearset 1. Let NC, N D be the numbers of teeth on Gearset 2 and the generator gear. Then use the gear formula 1 In practce most modern turbnes are drect drve: t s very dffcult to desgn a gearbox capable of transmttng the necessary power for 20 years wthout falure.
N / A NC generator gearbox NB ND ω ω = The resultng numbers (from matlab) are lsted below
5. The fgure (from ths publcaton) shows a schematc of the splt-power transmsson system for the second generaton Chevy Volt. The transmsson contans two epcyclc gears: the wheels are always connected to the planet carrers through the fnal drve gearng. The sun gears are drven by two electrc tracton motor, whle the rng gear can ether be loced (close clutch C1 and open C2), to mae the car an electrc vehcle, or connected to the nternal combuston engne (open C1 and close C2,C3) to mae the car a hybrd. We can fnd the followng specfcatons for the vehcle: Tre dameter 16.5 nches The gear rato between the planet carrers and wheels s ωpc / ω W = 2.64 Planetary gear 1 has 60 teeth on the sun and 112 teeth on the rng gear Planetary gear 2 has 52 teeth on the sun and 108 on the rng gear. 5.1 In EV mode the car drves wth the nternal combuston engne off, and rng gears of both planetary gearsets statonary (clutch 2 s closed). For a vehcle drvng at 30mph n EV mode, calculate the angular speeds of the two sun gears. The angular speed of the wheel s ω = v / r = 13.4 / 0.20955 = 63.95 rad / s The planet carrers spn at ω PC = 2.64 63.95 = 168.8 rad / s
The general expresson relatng angular speeds n an epcyclc gear s ωr ωpc N = S ω ω N S PC R The rng gear s loced n EV mode so ω R = 0 gvng N 1 R ωpc + = ωs NS Substtutng numbers gves ω = 484 rad / s (4622rpm) for PG1 ω = 519 rad / s (4956rpm) for PG2 [3 POINTS] 5.2 Another mode used at hgher speeds s Fxed Rato Extended Range mode, n whch the IC engne s runnng; both clutches are closed (so the sun gear on PG1 s statonary and the rng gear n PG2 s statonary). For a vehcle drvng at 35mph n ths mode, calculate the angular speeds of MGB and the nternal combuston engne. The angular speed of the wheel s ω = v / r = 15.65 / 0.20955 = 74.68 rad / s The planet carrers spn at ω PC = 2.64 74.68 = 197.17 rad / s Wth the clutches closed the sun gear of PG1 s statonary, the rng gear of PG2 s statonary. For PG1 the general formula gves N 1 S ωr = + ωpc = 302.79 rad / s NR (2891 rpm) For PG2 we have ω PC N 1 R + = ωs = 606.6 rad / s NS (5793 rpm) 5.3 At hgh speeds the car drves n Hgh Extended Range Mode wth clutch 1 closed, clutch 2 open, and the IC engne runnng. In ths mode the sun gear of PG1 has the same speed as the rng gear of PG2, and MGB s run slowly or bacwards. For a car runnng at 53 mph, calculate the speeds of the IC engne and MGA n ths mode, assumng MGB s statonary. The angular speed of the wheel s ω = v / r = 23.69 / 0.20955 = 113.1 rad / s The planet carrers spn at ω PC = 2.64 113.1 = 298.5 rad / s The sun gear of PG2 s statonary, so ts rng gear has speed N 1 S ωr = + ωpc = 442.16 rad / s NR
The sun on PG1 spns at the same speed. The general epcyclc gear formula can be rearranged to gve Substtutng numbers gves ω N (1 S N ) S R = ω PC + S NR N ω R ω R = 221.47 rad / s 6. The fgure shows three partcles wth equal mass m connected by rgd massless lns. 6.1 Calculate the poston of the center of mass of the assembly 1 1 L rg = m r ( ml ml ) ( ) M = 3m + = 3 + 6.2 Calculate the 2D mass moment of nerta of the system about the center of mass ( ) IGzz = m dx + dy where = dx + dy = G [3 POINTS] d r r s the poston vector of the th partcle wth respect to the center of mass. L L L 2L 4 2 I m d d m 2m = + = + + + = ml 3 3 3 3 3 Use the formula Gzz ( x y ) 6.3 Suppose that the assembly rotates about ts center of mass wth angular velocty ω (the center of mass s statonary). What are the speeds of the partcles A,B and C? Use the crcular moton formula. Partcles B and C are dstances 5 L /3 from the COM, and partcle A s a dstance 2 L /3 from the COM. Therefore VB = VC = 5 Lω /3 VA = 2 Lω /3 6.4 Calculate the total netc energy of the system (a) usng your answer to 6.2; and (b) usng your answer to 6.3 1 2 (a) The netc energy formula n terms of mass moment of nerta s T = I ω = 2 ml ω /3 2 Gzz z (b) Summng the netc energes of the masses drectly gves 1 ( ) 2 5 /3 ( 2 /3) 3 T = m Lω + m L = ml ω /3 2 C A m m L m B L
7. The fgure shows a pyramd wth heght h and base axa and unform mass densty ρ Usng a Matlab Lve Scrpt, calculate h θ O a a a/2 a/2 7.1 The total mass M h a(1 zh / ) a(1 zh / )/2 1 2 M = ρ COM = ρdxdydz = ρa h /3 M (see below for Lve Scrpt) 0 0 a(1 zh / )/2 7.2 The poston vector of the center of mass (wth respect to the orgn shown n the fgure) h a(1 zh / ) a(1 zh / )/2 1 ρcom = ( x y z ) dxdydz M ρ + + 0 0 a(1 zh / )/2 3a h = + 8h 4 (see below for Lve Scrpt) 7.3 The nerta tensor (matrx) about the center of mass, n the bass shown The nerta matrx s (see below for Lve Scrpt) I dy + dz dxdy dxdz h a(1 zh / ) a(1 zh / )/2 ρ x y x z y z x y z 0 0 a(1 zh / )/2 dxdz dydz dx + d y = d d d + d d d dxdydz d = x d = y 3 a /8 d = z h /4 19 2 3 2 a + h 0 0 320 80 3 2 3 = M 0 h ah 80 160 3 7 2 0 ah a 160 64 [ 3 POINTS] Graders equvalent solutons that nclude the densty should get credt too
8.4 Usng the parallel axs theorem, calculate the mass moment of nerta about the tp O. The general formula s dy + dz dxdy dxd z IO = I G + M dxdy dx + dz dyd z dxdz dydz dx + d y In the problem here d = d = 0 d = 3 h/4 so x y z 19 2 3 2 a + h 0 0 320 5 3 2 3 = M 0 h ah 5 160 3 7 2 0 ah a 160 64 Equvalent solutons that nclude the densty should get credt (see below for numbers) The Lve Scrpt for ths problem s shown.