analytic mechanics 7E ch5

analytic mechanics 7E ch5 - CHAPTER 5 NONINERTIAL REFERENCE...

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Unformatted text preview: CHAPTER 5 NONINERTIAL REFERENCE SYSTEMS 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force acting on him is zero. The scale exerts an upward force, N . whose value is equal to the scale reading --- the “weight,” W”. ofthe observer in the accelerated frame. Thus N+m§—m;lo =0 * N—mg—on=N—mg—m§=N—3mg=0 Ac 4 4 ’ W' = N = 2mg = 2W (b) 4 ‘ 4 W' = 1501/). (b) The acceleration is downward. in the same direction as g If, N—mg+m[§)=0 ll“"=ll"—L:3W 4 4 4 5.2 (a) F, =-m(Dx((Dxr') L s”, For (7) .L 7". Run = mafr'ér. a) = 500 s" =1000m‘I Eu," 2 10‘“ x (10007:)2 x 5 é, : 571'2 dynes outward 2 2 -' 1000;: 5 (b) = ""0 ’ = (—)— = 5.04x104 Fx mg 980 5.3 mg + T‘ —m}1n = 0 (See Figure 5.1.2) —nzgj+TCOSOj+TSin(9i—m(%)i=0 'l'cost9=mg, and Tsinf):fl l0 tangzi. 0:571: 10 '1‘: mg = I.005mg c056 5.4 The non-inertial observer thinks that g' points downward in the direction of the hanging plumb bob. .. Thus -, _. “ ‘. g 2 z -A = ——‘I g g o g] 10 For small oscillations of a simple pendulum: 4—— ‘30 box. so f— non = mii’ ((7 is the acceleration of the box relative to the truck. See Equation 5.1.4b.) Now. the only real force acting horizontally, so the acceration relative to the road is f #mg g (b) a=—=— =-/1g=-— m m 3 (For + in the direction of the moving truck, the — indicates that friction opposes the forward sliding of the box.) An = —§ (The truck is decelerating.) 2 from above. ma — mA0 2 ma’ so (a) (I'=a—A°=_§+£:__g_ 3 2 6 5.6 (a) F=i(xo+RcosQI)+j'Rsith " —lQRsith+jQRcosQI ." ' 1r: =S22R2 v = QR circular motion of radius R V l ‘1-l (b) I" = r' —(D>< i" where r’ = ix' + jy’ = —iQR sin Qt+j§2R cos Qt —a)/ix(i:t'+]y') = —iQR sin Q! + j'QR cos Qt — jcox' + fwy' = (uy' —QRsin Q! 'l y = —a)x' + QR cos Q! (c) Let u' = x' + iy' here i = \/-—1 l 12' = .ic' + ij" = coy' — QR sin Qt — iwx' + iQR cos QI iam' = = —(oy' +imx' zi' + iam' = —QR sin Q! + iQR cos Qt = iQ Re'n’ Try a solution of the form 11’ = Ae""" + Be’n’ ti' = —ia)Ae"“" + iQBe'Q' IQ! [5011' = iwAe"“" + ine QR zi'+ia)u' =i(a)+Q)Be"‘” so B: (0+ Q Also at t = 0 the coordinate systems coincide so 11': A + B = x'(0)+iy'(0) = + R QR R .‘.A=.\;+R—B=.\;+R— so, A=xfi+ w w+Q w+Q R QR Thus. 14': x0 + a) 6"” + c’n’ w+Q 0+0 5.7 The x, y frame of reference is attached to the Earth. but the x-axis always points away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y frame of reference is fixed inertial frame attached to the Sun. (a) In the x, y rotating frame of reference x(t) = Rcos(Q-w)t — R y(t)=—Rsin(Q—(o)t where R is the radius of the asteroid‘s orbit and RE is the radius ofthe Earth’s orbit. Q is the angular frequency ofthe Earth‘s revolution about the Sun and (n is the angular frequency ofthe asteroid‘s orbit. (b) .f(!)=—(Q—a))Rsin(Q-a))t—>0 at t=0 y(t)=—(Q—(0)Rcos(Q—a))t—>—(Q—a))R at t=0 (C) (7:3‘1-Ac-QXF—2éxlé—QXQXI" Where (7 is the acceleration of the asteroid in the x. y frame of reference, 1:1, A are the accelerations of the asteroid and the Earth in the fixed. inertial frame of reference. Thus: I =(Quw2)re_2oxc Therefore: (ix+jy)=(g22 —a)2)[iRcos(Q—(u)r—szin(£2—(u)t] ~2jm+2i§2y Thus: .‘r‘=(QZ—a)2)Rcos(Q-a))/+29y j}=—(Qz—a)2)Rsin(Q—a))I—2§lt Let .¥=(Q—(0)y and j?=(§2—a)).i” Then. we have y=(Q—(0)Rcos(Q—w)t+ 5» which reduces to y=—(§2—a))Rcos(Q—w)l Integrating... y=—Rsin(Q—w)t —>0 at (=0 Also. —.€(Q—a)) = —(§22 —(02)R sin (Q —(u)t — ZQWC’ or :‘c=(Q+w)Rsin(Q—w)t+2ch X=—(Q—(0)Rsin(Q—w)l Integrating x = Rcos(Q—a))l +c0nst x=Reos(Q—w)t—R€ —>R—RE at (=0 5.8 Relative to a reference frame fixed to the turntable the cockroach travels at a constant speed v ’ in a circle. Thus I? _, v . a = ——e,.. b Since the center of the turntable is fixed. x' i2: = 0 The angular velocity. (u. of the tumtable is constant, so (D = (uk'. with (D = 0 - A —- —< a ‘I A r’ = be,., so a) x (60x 1") = —ba)’e,, f” = v'é,,., so (I) x 17' = —a)v’ér. From cqn 5.2.14, (3 = (7' + 2(Dx 17' +E)x((7)x f") and putting in terms from above vi? a, = —— — Zwv' — ba)2 1) For no slipping S King , so lfil S yfig Since v' was defined positive, the +square root is used. if" = —(u/) + , [bpfig (b) r" = —v'é,,. (I) x F = +(uv'é, :2 1' 2 (1,. = —/— + 2021" — ba) 2 I7 % — 2wv' + ba)2 S yfig 13'" = (ob+ by‘g / . —- " . 5.9 As in Example 5.2.2. (B = I“ j' and An = V" i’ P ,0 For the point at the front of the wheel: V2 A " and 17'=—Vl€’ b J a 15! r : U! 5.10 (See Example 5.3.3) mwzx' = mi” x'(1) = Acm' + 136"" i" (t) = (uAew' — (oBe‘m' Boundary Conditions: -. A = i = i 4 4 , 1 ., l . (a) v (I) = ~2-cosh a)! x (I) = (03.9111th (b) x'(T) = g +% = écosh (0T when the bead reaches the end of the rod costh=2 or T=icosl1"2= 1317 a) a) (c) = wésinh (0T 2 misinh [0031]" 2] = (01(1 .732) = 0.866(1)] 2 2 I or (uL[(:osh2 (0T —1]3 = = 0.866(01 2 2 SJ] §u=400jnnph=58657j'n-s‘ 6::727x10*(cos4rj'+sn14réj s4 (5x52:—(727x10e)(58567)(an4r)? fi-s4 FM _ l—2ma‘3x t7" Fmv mg __2(127x10*)(586b7)(gn4rj n gw=amn gruv The Coriolis force is in the —(Z)>< 6' direction i.e.. +i' or east. 5.12 (See Figure 5.4.3) é=qf+QW A —/ _ 7t -r V — val + Vy._] —‘I 03 x r = —a)z.vy.i' + 601.13,] — myvxk' -- -v yI 7/ “a (a) x v )lmm _ —a)_,.t yvz + mpvxr/ l l — 7, __ 2,2 225: (2 3);: ,r |( X} )Imn: —((o,ry. +(o:.v»‘..) (0:. v", +ty. cola Fm = —2m(Z) x 17' K in.) = 2m|((?)x \7’)hm = 2mwfv', independent of the direction of F' . i hnn: 3- ‘ 5.13 From Example 5.4.1 l 3 5 x,',=la)[%] cos/l and y,2=0. 3 g l 3 3 5 x;,=l(7.27x10‘5.s") c0541” 3 32ft.s‘ x,” = 0.404 fl to the east. 5.14 From Example 5.4.2: wH 2 Va was struck due East at Yankee Stadium at latitude A = 41'” N (problem 5.13). 1:0 is the initial speed of the baseball whose range is H. From eqn 4.3.1 8b, without air resistance in an inertial reference frame, the horizontal range is A: |sin 1| is the deflection of the baseball towards the south since it v3 sin 20: g . H Solv1ng for v0 v0 =( .g ) Sin 2a J" =113fl-s" H: hal— r _ 32 fis‘lxzoofr ° sin30° ~ (7.27x10'53'I )(2003fi2) A ~ . l 13ft - s" A deflection of 0.2 inches should not cause the outfielder any difficulty. sin 41° = 0.0169fl=0.2 in 5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference. Thus 2': [d5 J (dc?) _ a r = — = —— + cox (1 dt fiwa dt rm c7=F'+(DXF'+2(DXF'+(DX((DXF') dz; '3: 'Z _. .1 .L I. .2. —4 l: [— =r'+a)xr'+a)xr'+2a)xr+2wxr' dt ml +(2')x(c?)xf')+(Dx<{bx7')+é)x(d)xi") wxfz=err'+(Dx((i)xf')+2(Dx((Dx?’)+€)x[i¢3x(a§x7')] Now ((Dxf’) is _L to (k) and 7". Let this define a direction f1: (DxF'=|a3xF'|Ii Since {Dir} , (z)x((DxF') is in the plane defined by (I) and 7" and |(Dx ((DX F')| = |a3 x fiHcDx F'] = (0|(Dx 7|. Since (DJ.0")><(("XF') l(Dx[(Dx(TXF'):”=(02|(DXF'| And (72x[a)x(E x?’)] is in the direction —fz Thus a3x[(f)x(a3xf")]=—(02((Dxf') (Bxa'=(Dxfi'+d’)x((§xf’)+2wx((fix;’)—w2(cf)xf') r- =7'+(Exf'+3szfi'+35;xfi'+(bx((bxi=') +2rax(5xf')+3(5x(a3xfi')—mz((bxf') .; . 5.16 With x: = y: = z; = x: = y: = O. and 2: = v: Equations 5.4.1521 — 5.4.15c become: x'(!) = Stag!3 cos/1 —a)!’vn cos/1 y'(]):0 l 7 2’: =—— t’+v't ( ) ° When the bullet hits the ground z'(l) = 0, so 2v: [: g 1 8 13 4,12 r =—wg cosA-(o i; t cos/1 3 g g , 4am:3 .x =— , cos/t 3g' x' is negative and therefore is the distance the bullet lands to the west. 5.17 With x; = y; = z: = 0 and if = we cosa y: = 0 2": = vfisina we can solve equation 5.4.15c to find the time it takes the projectile to strike the ground , 1 . 7 z (r)=——g12+v'lsma+(0vt cosacosl=0 2 ° ° 2v; sin at 2V; sin a or I = I— z — g—2wvocosacosl g We have ignored the second term in the denominator—since v: would have to be impossibly large for the value of that term to approach the magnitude g For example, for ,1 = 41" and a = 45“ g -— 2(01': cosa cos 1 z g — a) v: , km or i; z 5 z 144— 2 a) 5 Substituting t into equation 5.4.15b to find the lateral deflection gives . . 7 4(1) 1":3 . . 7 y(1) = —[(m-; cosasmlh' : — 2 Sin/ism“ acosa g 5.18 (7; = acceleration ofobjeet relative to Earth (I): = (ask = its angular speed x A = acceleration of satellite (7) = (air = its angular speed . 5° =Z+2a3xF+F x((7)xF)+ A“ (Equation 5.2.14) : ‘: c7=c7fl—Ao—2(Z)x§—d)xé)xf : As in problem 5.7 Evaluate the term Au=§5—AO—(DX(DXF=(DQXID:XR(—[bxd)x1§—(Dxa3x(R¢—R A” =(a13 —a)2 )RQ given the condition that (1)3163 = (1)21?3 but R.i{=(1§r~). F)=R2+r2+2chos(9 Letting x = 0036? RD '11 = R2 +)'2 +2Rx : R2 [l for small r R 3 _ 3 r: and szl€3[l+%) or .R_=(1+£) 1 ml»; 3 .. , — 2. ‘3 1 ‘ ~ A = —(0‘R0 l—[l +%) z —3a)‘x R” z —3(02xi for small r u Hence: (7 = Au — 20x 17 = —3a)2xi —2(ul€ x (133+ j'y) a = + j}; = —3a)2xi + 2wy‘i — 2mg} So — 2a)}? — 3(02.\' 2 0 j} + 2(wé = 0 5.19 mi: =qE+q(i‘xI§) Equation 5.2.14 +Qx7+203xi7+d3x(a3xf') Equation 5.2.13 ii = 17' +(DxF' (b=—il§ so 65:0 2121 lizii’—q(§xfi')—1§x(ibxf')=qE+q[(fi'+a3xf’)x1}] 2 tlzii'+q(F'x§)+%(d)xF')xE:q§+q(fi'xl§)+q(cfixf’)x§ ((DXF')X B to la =%[;’T’: (r’)(sint9)(B)och r s ' . ' 2 '1" __ Neglecting terms in B . m; —qE 5.20 For x’ = 3; cos aft + ysin w’! y' = —x sin (0'! + y cos (0'! i' = cos w't — xa)’ sin (0'! + y sin (0'! + ya)’ cos (0'! y’ = —.i: sin w't — xw' cos (0’! + ycos w't — yw' sin (0'! i' = 3': cos (0'! + ysin (11'! + (u'y' y’ = —i' sin w't + y cos (0'! — (u'x' It" : cos co't — J'ca)’ sin w’t + j) sin (0’1 + ya)’ cos w'! + (u'y' = —Zr'sin (u't — i'w' cos (0'! + )5 cos w't — ya)’ sin (0'! — (032' it" = cos (0’1 + )7 sin a)'1+ 2(0'3‘1' + (o'zx' y' = —5c' sin w't + cos 0'! — Zw’J’c’ + w’zy’ Substituting into Eqns 5.6.3: cos 60'! + sin (0'! + 2(u'jl' + w'zx' = —£ xcos (o'r — %y sin (u't + 2(0’j2' l .. . , .. I I" I2 I xsmcoz+ycoswl—2a)x +0) y = +£ xsin (0’! — § y cos w’t — Zw'ic' 1 10 J] -..21 5.22 Collecting terms and neglecting terms in (0’2 : (.f+£x)cos (0’1 +[j}+%y]sin (0’! = 0 l .. Y . , .. g ,_ [x+7x)sm(ot—[y+7y]coswl—0 T: .24 hours 5m}. 24 T: I =73.7 hours sml9° Choose a coordinate system with the origin at the center of the wheel, the x' and y' axes pointing toward fixed points on the rim ofthe wheel. and the z' axis pointing toward the center of curvature of the track. Take the initial position of the x’ axis to be horizontal in the 47; direction. so the initial position ol‘the y' axis is vertical. . . . . . V . . The bicycle wheel is rotating With angular veloc1ty about its aXis, so _ A, V a), = k ‘ I) A unit vector in the vertical direction is: . 1*, . 2, n=l sun—+1 cos— b b At the instant a point on the rim of the wheel reaches its highest point: A ~ . ’ ~ V f" = bit: I) i'sm II“, +j'cos—J] b b Since the coordinate system is moving with the wheel, every point on the rim is fixed in that coordinate system. r;' = 0 and F' = 0 The x'y’z' coordinate system also rotates as the bicycle wheel completes a circle around the track: a A V, If 9, . VJ :, VJ w2=n =— ism +_]COS— p p b b The total rotation of the coordinate axes is represented by: « a « Viv. VJ ~., Var] ~14 (0=(U.+a),= zsm +1 008— +k_r ’ p b b b .: V02(:, V0! ’2, . VJ) ( =— 1008 —/su1 pl) b ‘ b /'2 .. ‘ 72 A (2.)>()—:':I—”[k'COSZ£‘1_+k'sin3Vat]=l/a kl p b p 11 A _, Kb -, . V1 V0! A, . VJ VJ , 2, . VJ ;, V I (oxr = k Sin—cos —k s1n cos Hg .1 sm——1 cos ° b b b b p b b 4 w A, V2 A . Vt ~, .VI V2 ;, . VI 1, V! (ox(a)xr)=—°— k’sm2 ° +k cos‘ ° +; —1sm—”—./ cos “ p b b h b b a b. “I A, [2 A V2 (UX((UXI‘)=k;—I ‘ p b Since the origin ofthe coordinate system is traveling in a circle of radius p : _ .IV3 / =k _° ,0 f=IE'—+IE'——n—°—+k' ° p ,0 b p . Xv? . /‘_7 i=3L—k'—l° fl ,0 b With appropriate change in coordinate notation. this is the same result as obtained in Example 5.2.2. 12 ...
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analytic mechanics 7E ch5 - CHAPTER 5 NONINERTIAL REFERENCE...

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