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# homeworl11 - Problem 19.13 The 4-kg bar is released from...

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Unformatted text preview: Problem 19.13 The 4-kg bar is released from rest in the horizontal position 1 and falls to position 2. The unstretched length of the spring is 0.6 m and the spring constant is k = 20 N/m. What is the magnitude of the bar’s angular velocity when it is in position 2? Solution: In position 2 the spring is stretched a distance d =,‘\/(1.6 m —— 1.0 mcos60")2 + (1.0 msin60")2 — 0.6 m = 0.8 m The energies are l T]: 0,v1= 0, T2 = % (3M kg][l m12)w2 V2 2 _(4 kg)(9.81 m/s2)(0.5 m) sin 60° + £00 N/m)(0.8 m)2 = —10.6 N-m Work energy gives T] + V] = T2 + V2. => (1) = 3.99 rad/s Problem 19.22 The IOO-kg homogenous cylindrical disk is at rest when the force F = 500 N is applied to a cord wrapped around it, causing the disk to roll. Use the principle of, work and energy to determine the" angular velocity of the disk when it has turned 1 revolution; Solution: From the principle of work and energy: U = T2 — T., where T] = 0 since the disk is at rest initially. The distance traveled in one revolution by the center of the disk is r = 27rR .—; 0.67r m. As the cord unwinds, the force, F acts through a distance of 2 s. The work done is 23 I , U =/ Fds = 2 F(0.67() = 1884.96 N-m. 0 The kinetic energy is 1 2 1 2 T2— (2)1(0 :t-(2)mv, where I = émRZ, and v = Ra), from which T2 = i-mszz = 6.75102. U = T2, 1884.96 = 6.75012, from which a) = ~~16.7 rad/s (clockwise). Problem 19.39 The mass and length of the bar are m = 4 kg and l = 1.2 m. The spring constant is k = 180 N/m. If the bar is released from rest in the position 9 = 10°, what is its angular velocity when it has fallen to 6 = 20°? Solution: If the spring is unstretched when 0 = 0, the stretch of the spring is S = [(1 — c080). The total potential energy is v = mg(1/2)cos9 + §k12(1— cos a)? From the solution of Problem 19.37, the bar’s kinetic energy is 1 T = gmlzwz. We apply conservation of energy. T] + V] 2 T2 + V2: 0 + mg(l/2) cos 10° + gum — cos 10°)2 1 2 6mm; + mg(l/2) cos 20° + gum — cos 20°)? Solving for (02, we obtain (02 = 0.804 rad/s. Problem 19.71 In Problem 19.70, what is the angular velocity of the bar after the impact if the coefﬁcient of restitution is e = 0.8? Solution: Angular momentum about 0 is conserved: l ImAuA = lmAuﬁ‘ + Emmi}, + law}. (1) The coefﬁcient of restitution relates v; and v1”: VB e:——v\$P_v:‘. (2) .L HA 2 (”:3 We also have the relations 11’ ‘ Lw’ (3) .9. B “ 2 3' VI; . V33? v’ = lw’ . (4) . . up a . Before impact After impact Solving Eqs. (l)—(4), we obtain (of, = 3.93 rad/s. Problem 19.85 The 20-kg homogenous rectangular plate is released from rest (Fig. a) and falls 200 mm before coming to the end of the string attached at the corner A (Fig. b). Assuming that the vertical component of the velocity of A is zero just after the plate reaches the end of the string, determine the angular velocity of the plate and the magnitude of the velocity of the comer B at that instant. Solution: We use work and energy to determine the plate‘s down- ward velocity just before the string becomes taut. mgtU.2) : %m U2. Solving. at = 1.98 mls. The plate‘s moment of inertia is I 2 2 2 I : E(mutual) + (0.5) 1: 0.567 kg—m . Angular momentum about A is conserved: 0.25tmu) : 025(mu') + her. (I) The velocity of A just after is VIA :V'G [email protected]' x l'A/G t j k : ﬁu'j + 0 0 —w’ . 70.25 0.15 U Thej component of v‘A is zero: —u' + 0.25m’ = 0. (2) Solving Eqs. {1) and (2). we obtain u’ = I36 m/s and a)” : 5.45 rad/s. The velocity of B is “B = V's +69, X l'B/G i j k 0 0 ed 0.25 —0.]5 .0 : —vfj + v; = 70.8l8i — 2.726] (In/s). EWmmﬁ-l (a) 61V Just before Just after Problem 1934* The Apollo CSM (A) approaches the Soyuz Space Station (B). The mass of the Apollo is m,‘ = 18 Mg, and the moment of inertia about the axis through the center of mass parallel to the z axis is 14 = “4 Mg—mz. The mass of the Soyuz is my 2 6.6 Mg, and the moment of inertia about the axis through its center of mass parallel to the z axis is 13 = 70 Mg-mz. The Soyuz is stationary relative to the reference frame shown and the CSM approaches with velocity VA = 0.2i + 0.05j (m/s) and no angular velocity. What is the angular velocity of the attached spacecraft after docking“? Solution: The docking port is at the origin on the Soyuz. and the conﬁguration the instant after contact is that the centers of mass of both spacecraft are aligned with the .r axis. Denote the docking point of contact by P. (P is a point on each spacecraft. and by assumption. lies on the .r-axis.) The linear momentum is conserved: t l JPIAVUA =11”th + ”HEW”;- t’rom which mAtOVZt = nutt'm. + "'11 ”haw and (ii mMO‘OS} : my, "EM;- + mgn’cw‘, Denote the vectors from P to the centers of mass by l'pyGA = —7.3i (m). and l'pyﬁg = +4.3i (in). The angular momentum about the origin is conserved: r r rPfGA X mAVGA : rP/GA X ”I‘VGA + 1,101.4 + I'pyﬁg x "with” Denote the vector distance from the center of mass of the Apollo to the center of mass of the Soyuz by IBM 2 H.6i (m). From kinematics: the instant after contact: Vim :Vim +w' x r3”. Reduce: , _ r Von—Von + 2—6" OCH- car = “hm” “’iim + llﬁw’li» - - F _ r trom which am“ 7 no“. (2) vii-9‘ = “imi- + lwa' — l‘P/GA x mAvG,‘ i j k = 77.3 U 0 = (—0.365mﬂk. 0.21:1“ 0.05m,‘ 0 (Al i j k fit/on X "MV'UA = ‘7-3 0 0 "lava M "Hub,“ : t—lJmA ”in; )k. i j k rpm" xmpv:1m : 4.3 0 0 "'Bl’imt rrtAtv'GM + I Loaf) {l = 4.3"l3Ui—A‘. + I Ibro’lk. Collect ternts and substitute into conservation of angular momentum expression. and reduce: —0.365m4 = [+7.3mA + 4:3ntglv'5m + (49.9rn3 + [A +15Jw'. From (I) and t2). 005nm 7 I Lootgm’ mg. + my ' " 7 “o m- — These two equations in two unknowns can be further reduced by substitution and algebraic reduction. but here they have been solved by iteration: “ism- : 0.04704 mils. from which “hm : U'GM + l Lbro’ : 0.00807 this ...
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