hw10 - PROBLEM 17.5 Two uniform disks of the same material...

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Unformatted text preview: PROBLEM 17.5 Two uniform disks of the same material are attached to a shaft as shown. Disk A has a mass of 20 kg and a radius r = 150 mm. Disk B is twice as ' thick as disk A. Knowing that a couple M of magnitude 3O N-m is applied when the system is at rest, determine the radius nr of disk B if the angular velocity of the system is to 136-480 rpm after 5 revolutions. SOLUTION Moments of Inertia. 1 2_1 M ..._.-:M~- ‘ 'M-fl‘..- ‘a. w «MW.-. hm.<~m. 5. m -w--. ..«_W. m 1.. _ 4w..." .— — w ~mmm- DiskA: 1,, = EmArA E(20)(0.150)2 = 0.225 kg-m2 2 DiskB: m3 = mA [2)(511) = (20)(2)(n)2 = 40172 kg r U rA 1 1 13 = 3mg; = 3(40n2)(0.150n)2 = 0.45n4 kg-m2 1‘ :1 Total: 1 = [A + IE = (0.225 + 0.45n“)kg-m2 i Work. 192 —— 01 = 5 rev = 107r radian UH2 = M(612 — 01) = (30)(107r) = 3007: J Kinetic energy. 011 = 0 I; = 0 (02 = 480 rpm = 167: rad/s T2 = £16022 Principle of Work and Energy. -- I] + U1 _,2 = T2 Solving for 1, I = kg” = M = 0.74604 kg-m2 02 (1672') Radius of Disk B. Equating the two expressions for I, I 0.225 + 0.45n4 = 0.74604 n4 = 1.1579 n = 1.03732 r3 = nrA = (1.03732)(150 mm) 150 mm The double pulley shown has a mass of 14 kg and a centroidal radius of gyration of 165 mm. Cylinder A and block B are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block B and the surface is 0.25. Knowing that the system is released from rest in the position shown, determine (a) the velocity of cylinder A as it strikes the ground, (b) the total distance that block B moves before coming torest. 250 mm SOLUTION Let vA = speed of block A, VB = speed of block B, a) = angular speed of pulley. Kinematics. vA = rAa) = 0.25000 vB = rBa) = 0.150a) SA = rA0 = 0.2500 S3 = r349 = 0.15049 (a) CylinderA falls to ground. _ 0.150 s =0.900m s — A B 0.250 (0.900) = 0.540 m Work ofweightA: UH2 = mAgSA = (11.5)(9.81)(0.900) = 101.534 J Normal contact force acting on block B: N = mg 2 (9)(9.81) = 88.29 N Friction force on block B: Ff = ,ukN = (0.25)(88.29) = 22.0725 N Work offriction force: UH2 = —FfsB = —(22.0725)(0.540) = —11.919 J Total work: UHz =1'01.534 — 11.919 = 89.615 J Kinetic energy: ' T2 = émAvfi + —;-Ia)2 + é-va; 1 22 1 22 1 22 T=—mra) +—mka1 +—mra) 2 2AA 2C 233 = %[(11.5)(0.250)2 + (14)(0.165)2 + (9)(0.150)2]a)2 = 0.6512w2 Principle of Work and Energy. 1] + U1 _,2 = T: 0 + 89.615 = 0.6512a)2 a) = 11.7309 rad/s Velocity ofcylinderA: vA = (0.250)(11.7309) vA = 2.93 m/s l4 PROBLEM 17.1 1 CONTINUED (b) Block B comes to rest. For blockB and pulley C. T3 = élwz + évag; T4 = 0: / T3 = é—mckzwz + émBrng = %|:(l4)(0.165)2 + (9)(0.150)2](11-.7309)2 =40.159 J Work of friction force: U3_,4 = —Ffs}, = —22.07253§B Principle of Work and Energy. \ T3 + U3_,4 = T4: 40.159 — 22.07259}, = 0 s}; = 1.819 m Total distance for block B. ' d = SB + $33: d=0.540+l.819 d = 2.36 m 4 SOLUTION PROBLEM 17.15 A slender 6-kg rod can rotate in a vertical plane about a pivot at B. A spring of constant K = 600 N/m and an unstretched length of 225mm is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through 90°. Position 1. BG = 0.270 m CD = 0.5252 + 0.1802 = 0.555 m Spring: x1 2 CD — I0 = 0.555 — 0.225 = 0.330 m (Ve)1 = é—kxf = —2-(600)(o.330)2 = 32.67 J Gravity: h1 = BG = 0.270 m (Vg)l = Wh1 = mgh = (6)(9.81)(—0.270) = 715.892 J Kinetic energy: T1 = 0 Position 2. CD = 0.525 — 0.180 = 0.345 In Spring: x2 = CD —— 10 x2 = 0.345 — 0.225 = 0.120m (Ve)2 = %(600)(0.120)2 = 4.32 J Gravity: h2 = 0 )2 = 0 1m2 1 Kinetic energy: 2 E L =,I-2—(6)(0.900)2 = 0.405 kg-m2 172 = (13cm)2 = 0.270w2 T2 = v; + in); = %(6)(0.270a)2)2 + %(0.405)a)22 = 0.4212w22 Conservation of Energy. 7] + Vl = T2 + V2: 0 + 32.67 —15.892 = 0.421203 + o + 4.32 wz = 5.44 (02 = 5.44 rad/s) 4 Wt . ' a.— J: , 1" W- ,4" .7" .11 SOLUTION Moment of inertia about B. Position 2. PROBLEM 17.18 Two identical slender rods AB and BC are welded together to form an L-shapedassembly.~ The assembly is pressed against a spring at D and released from the position shown. Knowing that the maximum angle of rotation of the assembly in its subsequent motion is 90° counterclockwise, determine the magnitude of the angular velocity of the assembly as it passes through the position where rod AB forms an angle of 30° with the horizontal. 0=30° l . o 1 0 Position 3. Conservation of Energy. “’2 My 6g 2 _ 3g WAB (1 — sin30°) + WBC c0530.o 2 6g 9=90° l V3=WABE T3=0 e+n=n+w a); + WABésin30° — WBcécos3O° = 0 + WAR—[2— : .3;§[;1;_ Sin'3.0.° + c0530°l WAB + WBC 2 l 602 = 6.29 rad/s { PROBLEM 17.24 The 1.5-kg uniform slender bar AB is connected to the 3-kg gear B which meshes with the stationary outer gear C. The centroidal radius of gyration of gear B is 30 m. Knowing that the system is released from rest in the position shown, determine (a) the angular velocity of the bar as it passes through the vertical position, (b) the corresponding angular velocity of gear B. _ 58 mm SOLUTION l ' ' . _ _ _ AB Kinematics. v3 — [ABwAB — rBwB wB — r wAB B __ 1 1 VAB = 3V3 = ElABwAB 1 Moments of inertia. TAB = Em A3133, TB = mBk2 Kinetic energy. T = émABVjB + éTABwiB + é—vag + £7360; T-l—l-m +—1—m +m +m1€i 2 4 AB 12 AB B Br; 1 1 1 0.030 2 2 = E[20.5) + 1—2-(1.5)+ 3 + 3EO.050;2](0.12) wfiB = 0.032976ij Position 1. As shown in the drawing above. VI = 0 Position 2. Point B is directly below A. V2 = —WAB% — WBZAB = —(1.5)(9.81)(0.060) — (3)(9.81)(0.120) = —4.4145 J Conservation of energy. Tl + VI = T2 + V2: 0 + 0 = 0.032976(wAB): — 4.4145 (am): = 133.87 (a) Angular velocity of the bar. (DAB = 11.57 rad/s )1 4 wB = M0157) (03 = 27.8 rad/s) 4 0.050 PROBLEM 17.35 The uniform rods AB and BC are of mass 2.4 kg and 4 kg, respectively, . and the small'wheel at C is of negligible mass. If the wheel is moved slightlyto the right and then released, determine, the velocity of pin B after rod AB has rotated through 90°. SOLUTION Moments of inertia. BarAB: IA = émABLflB = %(2.4)(0.360)2 = 0.10368 kg-m2 Bar BC: 1? = TIEmBCLgc = T15(4)(0.600)2 = 0.1200 kg-m2 Position 1 . As shown with bar AB vertical. Point G is the midpoint of BC. Vl = m ,1th” + mBCghBC = (2.4)(9.81)(0.180) + (.4)(9.81)(0.180) = 11.3011 J Bar BC is at rest. Kinematics. 1 2 1’ -2 1— 2 T =—Ia) +—m v +—Ia) 2 2 A AB 2 BC 2 BC 1 v 2 1 1 2 1 v 2 = ——(0.10368)[ 3 j + —(4)[—ij + —(0.1200)[ B j 2 0.360 2 2 2 0.600 =1.06667v§ Conservation of energy. 1] + = T2 + V2: 0 + 11.3011 = 1.06667v12, v3 = 3.25 ...
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hw10 - PROBLEM 17.5 Two uniform disks of the same material...

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