homework6 - Problem 16.11 (2011] Problem 16.“), use the...

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Unformatted text preview: Problem 16.11 (2011] Problem 16.“), use the princi— ple of impulse and momentum to determine the collar’s velocity as a function 0|" time. (b) Use the result of (a) to determine the time at which the collar reaches the right hand end of the bar. Solution: ’2 (it) Thu impulse uciing on the collar is j ZFdr : Huh-2 — my”: 'l f I a If — a} d. d L L. — ‘7; I1 )l- _r- r _ )_! Ur __. ‘- n W1 1 I, 7 K K H .1 ‘ r g l tn- 312 4 * 322 [h] Start with m r A : t", Integrate] (M :f r 11!. n'! ()0 u U (10 32.2 r4 and evaluate 4 : ( ) («n chcc r L 2.34 s_ ()0 4 Problem 16.33 The "IT-kg skier is traveling at IO m/s at position I, from which he goes to position ] to 2 in 0.7 s. (a) Neglecting friction and aerodynamic drag, what is the time average of the tangential component of force exerted on the skier as he moves from posi— tion ] to position 2? (b) If his actual velocity is measured at position 2 and determined to be 13.] m/s, what is the time average of the tangential component of force exerted on the skier as he moves from position 'I to position 2? Solution: (21) The impulse is !2 f th = Fawm — Ii) = mm 7 mm. 11 If friction and aerodynamic drag are neglected, the velocity at 2 is determined from energy considerations: 1 l Emu; = Emu? + High. front which u; 2 Jr)? +2g : J10? + 23(4) : 13.35 ms. The time average of the tangential force is (b) [f v; : I3‘I this, the average tangential force is Problem 16.45 Suppose that the railroad track in Prob- lem 16.44 has a constant slope of 0.2 degrees upward toward the right. If the cars are 6 ft apart at the instant shown, what is the velocity of their common center of mass immediately after the impact? Solution: Time to couple (both accelerate at the same rate) is t = 6 ft 6 s. (2ft/s—lftfs) : Impulse — momentum is now 12 l . ( 0,000 2b)(2 flI,S)+(T’00001I2))(1 ms) 32.2 fth 32.2 ftfs . D [90,000 lb — (190,000 lb) sm 0.2 (6 s) = “centururmuss “centernl'mass = 0-957 W5 l mfs Problem 16.61 Two cars with energy-absorbing bum- pers collide head-on. Their speeds are 1),; :3 mi/h and v3 = 2 rni/h in the directions shown. The weights of the cars are WA 2 2800 lb and W3 =4400 lb. If the coefficient of restitution of the impact is e = 0.2, what are the velocities of the cars immediately after the collision? Solution: (2800b )(3 h)+( 4400 lb )( 2m h) m fl _ 32.2 his? p 32.2 M;2 p (2800lb)u,+(4400|b) , = ———“ U 32.2 «152 A 32.2 Ms2 B 0.2(3 — [—2]) mph : vg’ —- UA’ Solving we find of = —0.667 mph. vg’ : 0.333 mph M Problem 16.80 The one gives the cue ball A a velocity 3; parallel to the y axis. The cue ball hits the eight ball B and knocks it straight into the corner pocket. If the magnitude of the velocity of the cue ball just before the impact is 2 mls and the coefficient of restitution is e = 1, what are the velocity vectors of the two balls just after the impact? (The halls are of equal mass.) Solution: Denote the line from the S-ball to the corner pocket by 3?. This is an oblique central impact about HP. Resolve the cue ball velocity into components parallel and normal to HP. For a 45G angle, the unit vector parallel to BP is 83}: = fihi +j). and the unit vector normal to HP is egpn : fa +j), Resolve the cue ball velocity before impact into components: VA = UAPEEP + 'uApnegpn. The magnitudes 0,”: and 1),; p” are determined from Vlvnpenplz + ivnpnespnlz = I‘m = 2 W5 and the condition of equality imposed by the 45° angle, from which 11,“: : UAPn : x/i rn/s. The cue ball velocity after impact is v14 2 v’Apegp + UApnegpn. (since the component of VA that is at right angles to HP will be unchanged by the impact). The velocity of the B-ball after impact is v’BP 2 Upreap. The unknowns are the magni“ tudes u’BP and vi”, These are determined from the conservation of linear momentum along BP and the coefficient of restitution. mitt)“; = mAu’AP +msv’BP. and I n‘ 1: "BF _ “Ar 1MP For mA 2 ms, these have the solution 1):“, = 0, 1);”, : 1MP, from which V’A : UAPneBPn = i +.i (W5) and VIE : 11,4983}: 2 —i+j Problem 16.89 The bar rotates in the horizontal plane about a smooth pin at the origin. The 2-kg sleeve A slides on the smooth bar, and the mass of the bar is negligible in comparison to the mass of the sleeve. The spring constant k :40 N/m, and the spring is unstretched when r = 0. At t: 0, the radial position of the sleeve is r 2 0.2 m and the angular velocity of the bar is we : 6 rad/s. What is the angular velocity of the bar when r = 0.25 m? Solution: Since the spring force is radial, it does not affect the angular momentum which is constant. (0.2 m)(2 kg)[(0.2 m){6 rad/5)] = (0.25 m)(2 kg)[(0.25 m)wI (u r 3.84 rad/s Problem 16.93 A l-kg disk slides on a smooth horizontal table and is attached to a string that passes through a hole in the table. (a) If the mass moves in a circular path of constant radius r = 1 m with a velocity of 2 m/s, what is the tension T? (b) Starting from the initial condition described in part (a), the tension T is increased in such a way that the string is pulled through the hole at a constant rate until r z 0.5 m. Determine the value of T as a function ofr while this is taking place. Solution: (a) Circular motion T : intvz/rer |T|={l)(2)2/1= 4 N (b) By conservation of angular momentum, I'llvll mmun I mrur. UT : n— r and from Newton's second law mm: 2 lTl:mv%/r =m( ) /r i" 2 T=(l)(<11(2))/r24lrva ...
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This note was uploaded on 03/20/2008 for the course EM 311 M taught by Professor N/a during the Spring '08 term at University of Texas at Austin.

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homework6 - Problem 16.11 (2011] Problem 16.“), use the...

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