Dynamics hw #6

# Dynamics hw #6 - 15-58. The 1-lb ball A is thrown so that...

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Unformatted text preview: 15-58. The 1-lb ball A is thrown so that when it strikes the ID-Ib block B it is traveling horizontally at 20 ft/s. If the coefficient of restitution between A and B is e = 0.6, and the coefﬁcient of kinetic friction between the plane and the block is {14; = 0.4, determine the time before block B stops sliding. Thus , + (a) Emlvi=£m1u2 ’0“ [ﬁﬂum +0 = [ﬁﬂm )2 {ﬁyw}: (Va )1 = ft/S —) (VA): = —9.091fth = 9.091st (—- BlockB : on: +10%): = 20 0,400) = 4-16 + _ 10/5 (‘3’) "WI +£Ith=mvz e = (Vsh "(VA )2 (VA)1“(V3)| [ 10) — (2.909)—4:=0 32.2 0.6 = (VB )2 "(VA )2 20-0 1:0.2265 Ans (“a )2 ‘(VA )2 = 12 15—65. The 1—1b ball is dropped from rest and falls a distance of 4 ft before striking the smooth plane at A. If e = 0.8, determine the distance d to where it again strikes the plane at B‘ T1+V1=T3+V2 o + o = %(m)(v.4)? — m (32.2)(4) (m). = ‘/2(32.2)(_4) = 16.05 [Us \+ (11,01: = %(16.05) = 9.63 NS 4 /+ (why 2 0.8 (16.05) = 10.27 flfs (1m): = ,/(9.63)2 + (10.27)2 = 14.08'1’Us 10.27 .. —' __ 9 ‘ [an ( 9.63 ) = 46.85° «b = 46.85" — tan“’ = 9.977” + (6) s = so + v0! (1 = 0 + 14.08 cos 9.977°(t) 16.05 EUR 4 I 10.27 st [4.08 ft}. <9. i 9.63 W5 1 (+ i) s = so + var + Eat-r2 l d = O -— 14.08 sin 9.977°(:) + 232.2)!2 I = 0.798 S d = 13.8 ft Ans *15-68. The 2-kg ball is thrown at the suspended 20-kg block with a velocity of 4 m/s. If the time of impact between the ball and the block is 0.005 5, determine the average normal force exerted on the block during this time. Take e = 0.8. System : 4' (a) 2m, v‘ = 2m; v2 (2X4) + 0 = (2)(VA )2 + (20)(Vs )2 (VA )2 + 100’s )2 = 4 e_ (VB)2"(VA)2 _ (VA )1 ' (Va )1 8 = (Va )2 ‘(VA )2 4-0 0. (V101 '(VA)1 = 3-2 Solving : (vA )2 = —2.545 m/s (V3 )2 = 0. mIS Block: mvl+2Jth=mvz 0 + F (0. 005) = 20(0. 6545) F=2618N=2.62kN Ans 20(7. JDN 15-71. The 0.2-lb ball bearing travels over the edge A with a velocity of 0,4 = 3 ft/s. Determine the speed at which it rebounds from the smooth inclined plane at B. Take e = 0.8. vA=3ﬂls . (a) S=SQ+V03 d=0+31 (+ ~L) s =so +vor+éacz2 d:0+0+%(32.2):2 4:0.559 ft r=0.1863s (s) v: v0 (ml, = 3 {Us —) (+ i) v = v0 + act (v, )l, = 0 + (32.2)(0. 1863) = 6 ft/s i (Vah = V (3)2 + (6)2 = 6.708 ftls 9=mn"(§)=63.43° ¢=63.43°—45°= 18.43“ (+\) (Va)2.- = (v, )I oosq) = 6.364st _ (VB)2y' "0 (+7) 8—0-(—6.708sin18.43°) (va)2,- = 1.697 ftls (Va )2 = 1/ (6.364)2 + (1.697)2 = 6.59 NS Ans "‘15-'76. The 5-"; box B is dropped from rest 5 ft from the top of the 10er plate P, which is supported by the spring 3 -—_I._ having a stiffness. of k = 30 Ib/ft. If e = 0.6 between the i box and plate, determine the maximum compression 5n imparted to the spring. Neglect the mass of the spring. P Box : k: 30 lbm Datum at P. _ E E+H=B+K 1 S 2 Initial compression of Spring: 5 S =— 4-— +0 “ ( ) 2(32.2)(VB)‘ 1510 = =—=—=0.333n (v5)1=17.94ﬂ1s F. let, x. k 30 I System: Datum at initial position of plate : (+ i) 2m. v1 = Em v2 5 5 m 13 + ‘4 = T2 + 15 [ﬁynw +0 = (ﬂaw): {gym} 1 10 2 1 2 1 .. 2 . v _ v (1% W3): ﬂﬁyasm) + 5(2.0)(0333). : 0+ goon): +0333) — 10:: (W) “H; mm x’=0.974ft Solving : Thus. (VP)2 :9_570ﬁ-Js J, _ In,“ =0.97—4+0.333 = 1.31 ft Ans (th =1.196ft!sT 15-81. Two smooth billiard balls A and B each have a mass of 200 g. If A strikes B with a velocity (1)101 = 1.5 m [s as shown, determine their ﬁnal velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.85. Neglect the size of each ball. (VA, )l = “1.5 c0540” = -1.1491 mis (+ i) W (Va, )1 = "M (W): (Va, )l = —1.5 sin40" = —0.9642 mIs (VA, )2 = 0.9642 this (6) "IA (deh 4"“ (VB. )1 = "14 (V4, )2 +mg (vaxh PM 3 ; —0.2(1.1491)+0 = 0.202,“)z +0.20%;2 (+ T) "120231 = ms (V8,): * (VAx)2“(VB) (Ir' ) —(v ) (v ) :0 a] a: :2; 0.85: A12 5:2 H a) 1 ( (an )1 _(VA,)1 1.1491 cm‘ (Va )1 =(Va,)2 = 1.06 mls (- Ans (94 )2 = V (4.08613? + (0.9642)2 = 0.963 m/s 0.08618 9 =tzm"( j: . ° ( A)? 0-9 2 511 Ans Solving , (11A,); = “0.08618 mls (Max )2 = —L0629 m/s Ans 15-87. Two smooth disks A and B have the initial velocities shown just before they collide at 0. If they have masses m,‘ = 8kg and mg = 6kg, determine their speeds just after impact. The coefﬁcient of restitution is e = 0.5. +L/Emv1 = )"Jnv2 —6(3cosﬁ7.38°) + 306086733") = 6(VE),« + 80%))“ _ 0th "‘(VA )2 6/) — (V331 * (V3): 05 = (v3)1' "' (‘ML’ ' 7005 67.38“ + 3 (:03 67.38” Solving, (v3), = 2.14 mis (VA )1, : 0.220 mfs (my. = 35in6738° = 2.769 mis (vA)y. =-‘.'sin67.38° =~6.462 mfs v5 = {(2.14)2 + (23169)2 = 3.50 mls Ans VA = V(0.220)2 + (6.462 2 = 6.47 mfs Ans 15-99. The ball B has a mass of 10 kg and is attached MIN [0 the end of a rod whose mass may be neglected. If the MW rod is subjected to a torque M = (32‘2 +5! +2) N - m, where t is in seconds, determine the speed of the ball when t = 2 s. The ball has a speed v = 2 mls when t = O. Principle of 'Angular'lmpluse and Momentum: Applying 15—22, 1.5 m we have .2 M: 3:1 5 2 N- (H:)1+Zf Mammal r ( + '70398331»: ‘I 25 l.5(10)(2) + (3!!2 + 5! + 2)dt :2 15(10):) 1; 0 v = 3.47 m/s Ans 15—106. The 10-lb block is originally at rest on the smooth surface. It is acted upon by a radial force of 2 lb and a horizontal force of 7 lb. always directed at 300 from the tangent to the path as shown. Determine the time required to break the cord.which requires a tension T = 301b.What is the speed of the block when this occurs? Neglect the size of the block for the calculation. XE. = : ma" 1-“ low 10 v2 30 — 7sm30 — 2 = 372%?) 7 2» v = 17.764 ftls 3°‘ N (Hm + SIAM: = mm 0 + (7cos30°)(4)(t) = %(17.764)(4) t = 0.910 s Ans 15-109. A small particle having a mass m is placed inside the semicircular tube.Ihe particle is placed at the position shown and released. Apply the principle of angular momentum about point 0 (2M0 = Ho), and show that the motion of the particle is governed by the differential equation 9 + (g/R) sin 6 = O. (12% = 1%,- -ng sin9 = §‘(va) 0 g 8mg _ dv d7": ~ —— = —'—2 d: d: R 8 But, 5 = R0 Thus. gsin9 = —R0 or, b+(%)sme=o Q.E.D. ...
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## Dynamics hw #6 - 15-58. The 1-lb ball A is thrown so that...

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