i-iw=lt~ PROBLEM13.61

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A ID i-iw=lt~ PROBLEM13.61 A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 40 N/m and undeformed length equal to the arc of circle AB. A 00-g collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released ftom rest when fj = 30, determine (a) the maximum height above point B reached by the collar, (b) the maximum velocity of the collar. A (a) Maximum height Above B is reached when the velocity at E is.zero Tc = 0 1t fj = 30 = - rad 6 Point C R = 0.3m (VC)g = WR(I- cosfj) = (0. kg x 9.81m/s)(0.3m)(l- cos300) (VE)e=0 (spring is unattached) (VE)g = WH = (0. x 9.81)(H) = 1.96H (1) o + 0.4935 + 0.07886 =0 + 0 + 1.96H H = 0.9m ~

1-1IN =It'3 PROBLEM 13.61 CONTINUED (b) The maximum velocity is at B where the potential energy is zero, vb = vrnax Tc = 0 Vc = 0.4935 + 0.07886 = 0.574 J 1 1 = - ( 0. kg ) vrnax TB = -mvb TB = 0.1vrnax VB = 0 0+ 0.574 =(O.I)v~ax Vrnax=.39 m/s... ---

- ------ A ID /fw:tt3 PROBLEM 13.6 A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 40 N/m and undeformed length equal to the arc of circle AB. A 00-g collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest at an angle 0 with respect to the vertical, determine (a) the smallest value of 0 for which the collar will pass through D and reach point A, (b) the velocity of the collar as it reaches point A. (a) Smallest angle 0 occurs when the velocity at D is close to zero Vc = 0 Tc = 0 TD = 0 Point C R = 0.3m MBC = (0.3m)O = 0.30 m W = (0.kg) x (9.81 m/s) = 1.96 N (VCt =~k( MBC) (VCt = 1.80 (VC)g = WR(1 - coso) (VC)g = (1.96N)(O.3m)(l- coso) Point D Vc = (Vct + (VC)g = 1.80 + 0.5886(1 - coso) (VD)e = 0 (spring is unattached) (VD)g = W(R) = ()(1.96 N)(O.3m) = 1.177J 0+ 1.80 + 0.5586(1- coso) = 1.177 J By trial (1.8)0 - (0.5886)cosO = 0.5886 o = 0.75 rad

HW#-3 PROBLEM 13.6 CONTINUED (b) Velocity at A Point D VD = 0 TD = 0 VD = 1.177J (see Part (a») Point A 1 1 TA=- mva = - ( 0.kg ) VA VA = (VA)g = W(R) = (1.96N)(O.3m) = 0.5886J O.lv~ + 0.5886 = 0 + 1.177 VA =.43 m/s ~ - - -- -- -----

--- -- -- -- 1+w~?::' PROBLEM 13.63 A 6-lb collar can slide without friction on a vertical rod and is resting in equilibrium on a spring. It is pushed down, compressing the spring 6 in., and released. Knowing that the spring constant is k = 15 lib/in., determine (a) the maximum height h reached by the collar abbve its equilibrium position, (b) the maximum velocity of the collar. -- ~=o (a) Maximum height when Vz = 0 :. 11 = Tz= 0 Position CD Xl = 6 lb + 6 in. = 0.4 + 6 = 6.4 in. 15 lb/in. (~)l =!kxf =!(15Ib/in.)( 6.4 in.)z = 307. lb.in. = 5.6Ib.ft Position~ (Vg) = mgc~ + h) = 6(0.5 + h) (Ve)z= 0 11 + VI = Tz+ Vz: (Vg)l+ (~)l = (Vg)z+ (~)z 5.6 = 6(0.5 + h) h = 3.767 ft h = 45. in.... (b) Maximum velocity occurs when acceleration is 0, equilibrium position 1 z 1 6 3.) z z 13= -mv3 = - - v3 = 0.093167v3 ( ~ = (Vg)3 + (Ve)3 = 6(6) + ~k(xi - 6)z = 36 + 7.5( 6.4-6)z = 37.lb.in. = 3.llb.ft 11+ VI = T3+ ~: 5.6 = 0.093167vi + 3.1 vrnax= 15.54 ft/s...

PROBLEM 13.69 CONTINUED (b) Force of rod on collar AC Fz = 0 (no friction) F = Fxi + Fyj 8 = tan-]~ = 14.04 300 Fe =(kmac)(cos8i + sin8k) Fe = (30)(0.1093)(cosI4.04 i + sini4.04 k) Fe =33.909i + 8.4797k (N) LF =(Fx + 33.909~i of:(fy -4~905~j +8.4797k Fx + 33.909N = 0 F =4.905N + (0.5) y Fx = -33.909 N Fy = 33.309 N = ~; j +. mgk ( 8.51 m/s ) 0.15 m F = -33.9 N i + 33.3 N j ~ - - -- --

- - --- HW:#-?;> PROBLEM 13.70 A A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant k = 40 N/m and undeformed lengthequalto the arc of circle AB. A 00-g collar C is unattached to the spring and can slide without friction along the rod. Knowing that the collar is released from rest when () = 30, determine (a) the velocity of the collar as it passes through point B, (b) the force exerted by the rod on the collar as is passes through B. (a) Ve = 0, Te = 0 Iv 1 TB = -mvb TB =.!.(0. kg)v~ TB = O.lv~ arc BC = MBe = R(} MBe = (0.3m)(300 ) (Jr) 180 MBe = 0.15708m (Vet = ~k(mbe) = ~(40 N/m)(0.15708 m) = 0.49348J (Ve)g = WR(I- cos(}) = (0. kg)(9.81 m/s)(0.3m)(l- cos300) (Ve)g = 0.078857J Ve = (Ve t + (Ve)g = 0.49348 J + 0.078857 J = 0.5734 J VB = (VBt + (VB)g = 0 + 0 = 0 Te + Ve = TB+ VB; 0 + 0.5734 = O.lv~ v~ = 5.734 m/s vb =.39 m/s ~ (b) +t"f,f =FR - W = mv~ R FR = 1.96 N + (0. kg) (5.734 m/s) (0.3 m) FR = 1.96 N + 3.8156 N = 5.7776 N FR = 5.78N ~

B PROBLEM 13.75 (1) () An 8-oz package is projected upward with a velocity v0 by a spring at A; it moves around a frictionless loop and is deposited at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach C, (b) the corresponding force exerted by the package on the loop just before the package leaves the loop at C. Loop 1 ~ B (a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero. P-> tt=- 0 ~ tma d I mv +,'LF=O+mg=-1L r v~ = rg = 1.5 ft (3. ft/s), y.. ",..~."".";,~: v~ = 48.30 AtA 1 TA = -m Vo VA =0 80Z = 0.5 lb ~ = 0.5 = 0.01553 ( 3. ) AtB 1 1 TB = -mvb = -m ( 48.30 ) =4.15 m VB = mg(7.5 + 1.5) = 9mg = 9(0.5) = 4.5Ib.ft TA+ VA= TB+ VB:!(0.01553)v~ = 4.15(0.01553) + 4.5. AtC v~ = 67.8 Vo= 5.056 Vo = 5.1 ft/s ~ Tc = -mvc 1 = 0.007765vc V: = 7.5mg = 7.5(0.5) = 3.75 TA + VA = Tc + Vc: 0.007765v~ = 0.007765v~ + 3.75 0.007765(5.056) - 3.75 = 0.007765v~ v~ = 144.87 --- ---- - ---

-- - PROBLEM 13.75 CONTINUED (b) Loop B (a) r~~' 'mq, N = 0.01553(144.87) 1.5 N = 1.49989 At B, tube supports the package so, {Package in tube} N c = 1.500 lb -... VB ~ 0 VB = 0, TB = 0 VB = mg(7.5 + 1.5) = 4.5 lb.ft ()!(0.01553)v~ = 4.5 => va = 4.073 VA = 4.1 ft/s... (b) At C Tc = 0.007765v~, Vc = 7.5mg = 3.75 TA+ VA= Tc + Vc: 0.007765(4.073) = 0.007765v~ + 3.75 V~ = 96.573 N~-CtJ = 0.5 Nc = 0.01553 ( 96.573 ) = 0.99985 1.5 {Package on tube} Nc = 1.000 lb...

/-Iv #~ PROBLEM 13.83 Knowing that the velocity of an experimental space probe fired from the earth has a magnitude VA =3.5 Mm/h at point A, determine the velocity of the probe as it passes through point B. ra = ha + R = 4.3Mm + 6.37Mm ra = 10.67Mm rb = hb + R = 7.7Mm + 6.37Mm, \:) \ rb = 19.07Mm AJB_ AtA, VA = 3.5 Mm/h = 908m/s TA =.!.m(908 m/s) =40.75 x 106 m ra = 10.67 Mm = 10.67 x 106 m R = 6370km = 637 x lq6m (9.81 m/s)( 6.37 x 106 mt 6 m = -37.306 x 10 m 6 10.67x 10 m ( ) AtB rb = 19.07 Mm =19.07 X 106m " - (9.81 m/s)( 6.37 X 106m) m- 6 = -0.874 x 10 m 6 19.07 x 10 m.. ( ) 6' 6 1. 6 TA + VA = TB + VB; 40.75 x 10 m - 37.306 x 10 m = -mvb. t-. '. v~.= [40.75 x 106-37.306 x 106 + 0.874 x 106.J -,0.874 x 10 m VB = 6.974 x 103m/s =5.107 Mm/h VB = 5.1 Mm/h...

I+'V~~ PROBLEM 13.116 A spacecraft of mass m describes a circular orbit of radius 1}around the earth. (a) Show that the additional energy M which must be impartedto the spacecraft to transfer it to a circular orbit of larger radius r is M = GMm(r - r.) 1}r where M is the mass of the earth. (b) Further show that if the transfer from one circular orbit to the other circular orbit is executed by placing the spacecraft on a transitional semielliptic path AB, the amounts of energy M A and M B which must be imparted at A and Bare, respectively; proportional to 1}and r: (a) For a circular orbit of radius r V _ GM r E = T + V =!mv _ GMm _ 1 GMm r - -~ (1) Thus M required to pass from circular orbit of radius 1}to circular orbit of radius r is M = GMm(r - 1}) 1}r () (Q.E.D.) (b) For an elliptic orbit we recall Equation (3) derived in Problem 13.113 (with vp =VI) B At point A: Initially spacecraft is in a circular orbit of radius 1} V. clrc -_ -GM 1} ~irc =!mv. 1 GM clrc - -m- Yi I ---- - ---

-- ~ -- --- PROBLEM 13.116 CONTINUED After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall VI =. GM r.- ('i + r) rl And T. _ 1 _ 1 GMr I - - mvi - - m rl (rl + r) At point A, the increase in energy is 1 GMr 1 GM M A = 11 - I::irc = - m ( ) - - m- 'i 'i + r rl Recall Equation (): (Q.E.D) A similar derivation at point B yields, (Q.E.D)

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