HW9_solutions_ - Problem 18.38 The mass of the disk is 45...

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Unformatted text preview: Problem 18.38 The mass of the disk is 45 kg and its radius is R = 0.3 m. The spring constant is k =f3-IIIL'IN/m. The disk is rolled to the left until the spring is com— pressed 05 m and released from rest. (a) If you assume that the disk rolls, what is its angular acceleration at the instant it is released? (b) What is the minimum coefficient of static friction for which the disk will not slip when it is released? Solution: x0 = 70.5 k : 600 N/m m : 45 kg R : 03 m 10 : 5M3 : 2.025 N—mzt F3. : kx ZEY: _ FA _ : 1715104 ER: N — mg = Rolling implies am : —Roz We have, at x : 70.5 m * kx * f : mam N 7 mg 2 U — Rf : 100: not = 7R0: Four eqns, four unknowns (£101, or, N, f) (4) Solving f = I00 N, N = 441.5 N at : —l4.81 rad/s2 (clockwise) um- : 4.44 m/s2 (b) for impending slip, : Mr N m = f/N = 100/4415 m = 0.227 Problem 18.40 A 42-lb sphere with radius R = 4 in is placed on a horizontal surface with initial angular velocity coo = 40 rad/s. The coefficient of kinetic fric- tion between the sphere and the surface is uk 2 0.06. What maximum velocity will the center of the sphere attain, and how long does it take to reach that velocity? Strategy: The friction force exerted on the spinning sphere by the surface will cause the sphere to acceler- ate to the right. The friction force will also cause the sphere’s angular velocity to decrease. The center of the sphere will accelerate until the sphere is rolling on the surface instead of slipping relative to it. Use the relation between the velocity of the center and the angular veloc— ity of the sphere when it is rolling to determine when the sphere begins rolling. Solution: Given W = 42 1b, g = 32.2 ft/sz, m = W/g, R = 4/12 a, 11k = 0.06 We have ZFX : ,ukN 2 ma ZEviN—ngO 2 9 ZMG tut/NR = ngYt Solving we find _ SUM: a = 14.49 rad/$2, a = 111. g = 1.932 ft/s2 From kinematics we learn that a! = 14.49 rad/32, a) = (14.49 rad/32): — (40 rad/s) a = 1.932 ft/sz, v = (1.932 mg): when we reach a steady motion we have u = —Ra) => (1.932 131/52): = —(0.33 ft)[(l4.49 rad/52y — (40 rad/5)] Solving for the time we find r=l.97s => v=3.81ft/s Problem 18.45 The l8-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. At the instant of release, determine (a) the angular acceleration of the ladder and (b) the normal force exerted on the ladder by the floor. (See Active Example 18.3.) Solution: The vector location of the center of mass is rG = (L/2) sin 30°i + (L/2) cos 300j : li + 1.732j (m). Denote the normal forces at the top and bottom of the ladder by P and N. The vector locations of A and B are rA = L sin 30°i = 2i (In). r3 = L cos 30°j = The VfiClOKS I'A/G = I'A — l‘G = ll— I'B/G = r3 7 1‘0 = 7 li + 1.732j (m). The moment about the center of mass is M=l'B/G XP-i-l‘A/G XN. 1 J k 1 J k M: —l 1732 0 + l —l.732 0 P 0 0 O N 0 : (4.73219 + N)k (N-m). From the equation of angular motion: (1) —l.732 P + N : Ia. From Newton‘s second law: (2) P = mrrr, (3) N — mg = max. where ax, a". are the accelerations of the center of mass. From kinematics: ac; : aA + 05 x rG/A — wer/A. The angular veloc- ity is zero since the system was released from rest, 1 j k ac = rrAi—l— 0 O a = (mi—1.732ai—aj fl 1.732 0 : (aA — l.732n!)i — Otj (m/s2). from which a, = 70:. Similarly, I j k ao=ag+o£><rG/g.a(;=33+ 0 0 a l —l.732 0 = (sz + l.732ai + otj. from which at- : l.7326¥. Substitute into (1), (2) and (3) to obtain three equations in three unknowns: 71.732P + N = Id. P = m(l.732)cr, N — mg 2 —ma. Solve: (a) Ct = 1.84 rad/52, P = 57.3 N, (b) N = 143.47 N 30° mg 4m Problem 18.48 The masses of the bar and disk are 14 kg and 9 kg, respectively. The system is released from rest with the bar horizontal. Determine the bar’s angular acceleration at that instant if (a) the bar and disk are welded together at A, (b) the bar and disk are connected by a smooth pin at A. Strategy: In part (b), draw individual free-body diagrams of the bar and disk. Solution: L=l.2m R=O.3m mg:l4 kg 171;):9kg 0 is a fixed point For the bar 1 7. l w 7 ‘ i -7 -7 _ 7 - [0—1 mBL — 12(14)“ ) —l68Nm L 2 [OH = [C +1713 3 103 z 6.72 N—m2 For the disk: :A : §mDR3 = grows)? = 0.405 N_m2 [00 21,; + mng 213.37 N—m2 The total moment of inertia of the welded disk and bar about 0 is ['1‘ = 103 + 10;) = N-m2 ZR: 0X = 0 = mam 2F}: 0"- — mgg — mpg : (1713 + 1711))(16‘. L 2M0: — mgg — meg : [Tu We can solve the last equation for or without finding the location and acceleration of the center of mass. G. Solving, at = —9.38 rad/s2 (clockwise) (b) In this case, only the moment of inertia changes. Since the disk is on a smooth pin, it does not rotate. It acts only as a point mass at a distance L from point 0. In this case, [gm 2 mle and I} = 103 + [gm 2 19.68 N-m2 We now have , I" I / ZMU. — (7)1ngg—meg=frcr Solving oz’ : 79.57 rade2 (clockwise) Problem 18.65 Bars 0Q and PQ each weigh 6 lb. The weight of the collar P and friction between the collar and the horizontal bar are negligible. If the system is released from rest with 6 = 45°, what are the angular accelerations of the two bars? Solution: Let (mg and OIPQ be the clockwise angular acceleration of bar 0Q and the counterclockwise angular acceleration of bar PQ. The acceleration of Q is i k HQ =ag+a0Q XFQm= 0 0 —0£0Q 2 cos 45° 2 sin 45° 0 : ZUQQ sin 450i — 20t0Q COS 45Dj. The acceleration of P is 2110 = aQ + (3po X l‘p/Q i k (“xi = 20100 sin45°i * ZOEOQ C05450j + 0 0 org 2 cos 45° 2 sin 45° 0 Equatiflg i and j components, (1]) = 20mg sin 45° — Zapg sin 45° (1) O = —2(xOQ cos 45° + 20!}3Q c0545° (2). The acceleration of the center of mass of bar PQ is ac = 3Q +£2po X l'g/Q = 200g sin 450i i j k —2achos45°j+ 0 0 OtPQ cos 45° — sin 45° 0 Hence, ((0‘- : ZOIOQ Sin 4-5D -|- DIPQ Sin 450 (3): no). : 72mm) cos 45° + apQ cos 45° (4). From the diagrams: The equation of angular motion of bar 0Q is 2 M0 = 100mg: Q,(2sin 450) — Q_v(2cos45°) + 6cos45° : §(6/32.2)(2)2a09 The equations of motion of bar PQ are 2 Fir : — Qt : (6/32.2)aci (6) 2 F)» : N — Q3» — 6 : (6/32.2)ac_v (7) 1 7 2M = (N + Qy + Q,.r)(cos45°) = fi(6/32.2)(2)'CEPQ (8). Solving Equations (l)—(8), we obtain 0mg : apQ : 6‘83 radJ’s2 (5). 18.124 should be solved for the information to solve 18.125 Problem 18.124 Each bar is l m in length and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what are the angular accelerations of the bars at that instant? 30° Solution: For convenience, denote H = 45°. r: = 303 and L = 1 m. The acceleration of point A is i k at = don >< Parr) = 0 0 arm Leosr‘) LsinH U at = 0,0,, {—iL sin!) + jL cos in (M53). ThwlrflanJl-nf'n' . . . e e . . 6 race 6 1m) 0 IS 1 so 3|“: b} The equations of motion tor the bars: lor the pin supported lelt hill‘: 3.41 = 31; + urib’ X Page L (5) AIVL cost! — Axis siné! — mg (;) C056 = {0,1961A. i j k 8,11: :13 + 0' 0 (My . “IL: 4 a —LcosH Lsinr) 0 where 10,; = ( 3 )= E kg-m‘. n - . 3 3A = 38 — 10M 8 L “"9 ‘JQ’A-‘llL €059 ("VS—l The equations of motion for the right bar: From the constraint on the motion at B‘ Equate the expressions for . . . . — AI- — 3 sin = mang -, the acceleration ol A to obtain the two equations: ‘ ‘6 ' . . (7 —/l‘—HI +Bcos =mu;, .. (l) — 00,; L 5111“ = in; (305 ff — 0,13]. .N'ln lot ) " g fl ( 1B" I. i'. . I. . and (’2) arm.me = (if; sinfi — «mi. e039. (8) A). (g) cos-(J + A; (g) $me + B (;)51n{}CUNfi The acceleration of the center of mass of AB is L . — B (3039 smfi = f(-‘.\mx’,13, “(L-t3 =1l.-t + «an X roam-i l l w i j k where 10,“; = (l—j) oth = (g) kg-m: = 31 + 0 0 “at; _ h ’ L cos o L sin ti . . . . . . a — A} These eight equations In eight unknowns are solved by iteration: Ar = ' ' —|9.27 N, A}. = 1.15 N. mm = 0.425 radisz. (1,”; = —|.59 radi’sg, a = 45.43 N, mm]. = —0.8610 111153, am ,1,"- = —0.2001 m3 LCM}; _. . €3sz _ . 3 ' 30,1, 3 = 31.11 + $111 0| + $035 U" [mils }. from which . Loco; . 1 (3)130131 = —o'0,iL smb’ + sInH (mIS'l. [Xle (4) “0,11”- : 09,; L cost‘i + cos 6. Problem 18.125 Each bar is l m in length and has a mass of 4 kg. The inclined surface is smooth. If the system is released from rest in the position shown, what is the magnitude of the force exerted on bar 0A by the support at O at that instant? Solution: The acceleration of the center of mass of the bar 0A is i j k 0 0 0'0“ a;.=ot, I‘;',=a‘. ‘ (01 “1X (’1 ‘+ LeosH Lsth 2 2 L sin 9 _ L cos 6' _ 1 flow = — 7 (Xe-11+ a (mu (mist). The equations 0f mono“: Use the solution to Problem 18.140: 9 = 45°. and = 0.425 radt'sl. A.- = —l9.27 N, m = 4 kg, from which [3 = 18.6? N. F)- = Fx + Ax = Mann Fr + A." - me = mam-- 33.00 N. from which |F| = VH1? + F3 = 42.00 N ...
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This note was uploaded on 03/29/2009 for the course MECHENG 240 taught by Professor Perkins during the Spring '09 term at University of Michigan.

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HW9_solutions_ - Problem 18.38 The mass of the disk is 45...

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