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46962_HW4 - I HOMEVVORK II TA PO-WEN CHEN LIU YEN-FU l{a...

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Unformatted text preview: I. HOMEVVORK II TA: PO-WEN CHEN; LIU YEN-FU l.{a) The potential and kinetic energies of this system can be expressed in term of polar coordinates and velocities as)? y : y5 ——lsin6 a: = *lcosfl y : 315 "icosfié j: : zsineé (1) 1 .. 1 .2 .2 T : iMy5+§m(2: +y) V : imglcosfl L : T—V. (2) Substituting these transformations into Eq.(1) yields 1 1 L : 5114:1751 §m($2+§rz)+mglcosfl 1 .. 1 . - 2 , .2 7 EM 3J5" Em (y5+lcos66) +(ls1n66) +mglcos3 1 .. l .2 - 2 . - 1 E11/131515111 y5+ (l3) +2(y5zecose) +mglcosf9 (3) ('3) (BL (1815 , 0 86 dtas _ mglsin0+ml é+my51cose : 0 (4) (c) 6 << 1, sine 2 6,cosfi 2 1, é+67mw2y0 coswt 0 mw2 (905) : 1 2 yo cos wt — w 2. “5' use Figure 2 to describe the system. We have , however, for the Lagrangian function 1 11 - -2 T : —M '2 '2 ——MR2cb 2 (:0 +1; )+ 22 , 1 . . '1 2 . .- 2 11 _2-2 +§m (xibsmejocp) +(y+bc0scqu) +§§mr9, V : 0, L : TiV. (5) \Ve find the equation of motion by using above Eq.(l) 3?: 3L daL _ 0 5‘3: dt 51$ 7 d . 01E(Ma§+mi~1mbsin¢¢) : 0 Mi+mi—mbsin¢$ : 0 (6) ____ : 0 (93; dt 33,; d . 0—E(1Mg+mg+mbcos¢b¢) : 0 Mfl+mfl+mbcosqfi¢ : 0 {7) 6: 62,36; _ 0 66 dtag ’ 9 : 0 {8) (I): 111,36; _ 0 ago dt a¢ i 7 , , . 2, egg éMRngimibsinngrmVsinqbq'b _ mmbcos ¢¢+mb smqbcos gm alt ( +mgbcosqfi+mbg 005W?) 7 0, imibcos¢¢+mb2sin¢cosqbgb2 , EM R2¢+mibsin¢i mbzsin m *mgjbcosgbimbz cos (15¢ : 0. (9) 7-15. m b = unextended length of spring 6 = variable length of spring T:lm(zz+ezez) 2 1 2 1 2 ‘ 11251413719) +mgy:ik(£ib) imgfcosfi L:T U:1m(t21C292) 1(E b)2+mg€c0s6 2 2 Taking Lagrange's equations for 6 and Bgives d . . 6:7 f : wtk 67b 6 dim] m ( )+mgcos t9: iLmCfléJ: 7mg 6 sin 19 This reduces to ZiC€2+£(£ib)igcos6:0 m 5+3 w+§sin .970 C C 7-34- The coordinates of the wedge and the particle are xM:x xm:rcost9+x . (1) yM : 0 ym : —r sm (9 The Lagrangian is then L: M; m x2 +E( '2 + 7/292 +25cf’ cos 8—2atr9si11 6)+ mgr sin 6 (2) r Note that we do not take 1’ to be constant since we want the reaction of the wedge 0n the particle. The constraint equation is f (x, 6, r) : r — R : 0 . a) Right now, however, we may take 1' = R and r : i‘ : 0 to get the equations of motion for x and 6. Using Lagrange’s equations, 56 :aR(i§i sin 6+ 92 cos 6) (3) é_xsm6;gcos6 (4) where :1 E m/(M + m) . b) We can get the reaction of the wedge from the Lagrange equation for r xlzmjé cos 6—mR92 —mg sin 6 (5) We can use equations (3) and (4) to express it in terms of Band 9 , and substitute the resulting expression into (5) to obtain a — 1 - . To get an expression for 9 , let us use the conservation of energy H : M + m 12 + §(R292 , 25th9 sin 19) , ng sin 6 : ing sin 60 (7) where 60 is defined by the initial position of the particle, and —ng sin (90 is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain 92: : aRt? sin 19 , and substitute this into the energy equation to obtain an expression for (9 92 : 23(sin 6—sin (90) R(1—a sin2 6) (8) Finally, we can solve for the reaction in terms of only Band 60 mMg(3 sin 6—3 sin3 (9—2 sin 60) (M+ m)(1 , a sin2 (9)2 ,1 : (9) If we take angles 49 and (b as our generalized coordinates, the kinetic energy and the potential energy of the system are 1 . 1 . T:§m[[R—p]6]2+§l’¢2 (1) U=[R—(R—p)cos 19ng (2) where m is the mass of the sphere and Where U = 0 at the lowest position of the sphere. I is the moment of inertia of sphere with respect to any diameter. Since I : (2/5) rm):1 , the Lagrangian becomes 1 L—T U—2m(R p)292:1 511%;)ng2 [R (R p)cos€]mg (3) When the sphere is at its lowest position, the points A and B coincide. The condition A0 = BO gives the equation of constraint: Hflfl=R—fl9flm=0 (Q Therefore, we have two Lagrange’s equations with one undetermined multiplier: a é_i[§]+fl_f0 as dt as as (5) §i£ 67L +Ai=0 ea: dt 3g» 5gb After substituting (3) and (if/66': Rip and 6f/6¢= 7,0 into (5}, we find 7(Rip)mg sin 67m[Rip)29+fi(Rip)=0 (6) 2 i _E mPZ¢—3p=0 (7) From (7) we find it: 2 __ fi:——mp¢ (8) 5 or, if we use (4), we have 1=—§m(R—p)5 (9) Substituting (9) into (6), we find the equation of motion with respect to (9: {9:702 sine (10) Where a) is the frequency of small oscillations, defined by ,f 53 a), 7(R—p) (11) Let us choose 5,8 as our generalized coordinates. The x,y coordinates of the center of the hoop are expressed by (1) x:§+Scosa+rsina y:rcoscx+(£—S)sina Therefore, the kinetic energy of the hoop is 1"th : :12 m (5:2 Hf) +2121¢2 _%m[(é‘+3cosa)2+(—Ssin a)2]+%1g§§2 (2) Using I : mr2 and S : m, (2) becomes T :mPSZJrg-KZJng-EScosa] (3) hoop _ In order to find the total kinetic energy, we need to add the kinetic energy of the translational motion of the plane along the x-axis which is Tpm gm: (4) Therefore, the total kinetic energy becomes T—mS2+:(m+M)§2+m§S'cosa (5) The potential energy is Uzmgyzmg[rcos a+(f—S) sin a] (6) Hence, the Lagrangian is l: 14132 +%(m + M)? + méS- cos a— mg[r cos a+ (f— 8) sin a] (7) from which the Lagrange equations for g and S are easily found to be 2m§+mgcosa—mgsma:0 (8) (m+M)%+m§cosa:0 (9) or, if we rewrite these equations in the form of uncoupled equations by substituting for g: and S , we have 2 [2_Wls_gm=o m+ M (10) _ mg sin a cos a _ 2(m+M)imcos2 05 Now, we can rewrite (9) as %[(m+M)§+mScosa]—O (11) where we can interpret (m + MM as the x component of the linear momentum of the total system and m3 cos a as the x component of the linear momentum of the hoop with respect to the plane. Therefore, (11) means that the x component of the total linear momentum is a constant of motion. This is the expected result because no external force is applied along the x—axis. 7-37. Let’s choose the coordinate system as shown: The Lagrangian of the system is 2i with the constraints Jc1+y:l1 and xz—y+x3—y:I2 dzx dzx dzx 1 + 2 + 3 :0 1 dt2 dtz dtz U which imply 2x1 + x2 + x3 7(211+12): U :> 2 The motion equations (with Lagrange multiplier 2.) are dzx mlgimlfi+2120 (2) ng—m2%——l—O (3) mBg—mS‘ifnl—O (4) Combining (1)—(4) we find ’1 4 _ig 1 ...
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