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Chap. 18 Planer Kinetics of a Rigid Body: Work and Energy APPLICATIONS The work of the torque (or moment) developed by the driving gears on the two motors on the concrete mixer is transformed into the rotational kinetic energy of the mixing drum. If the motor gear characteristics are known, could the velocity of the mixing drum be found? 18-2 18.1 Kinetic Energy 1 2 T = ∫m dmvi 2 vi = v P + v i / P = (v P ) x i + (v P ) y j + ω k × ( xi + y j ) = [(vP ) x - ωy ]i + [(vP ) y + ωx] j 18-3 ⇒ v i ⋅ v i = vi = [( v P ) x − ω y ]2 + [( v P ) y + ω x ]2 2 = (v P ) x − 2(v P ) x ω y + ω 2 y 2 + (v P ) y + 2(v P ) y ω x + ω 2 x 2 2 2 = v P − 2(v P ) x ω y + 2(v P ) y ω x + ω 2 r 2 2 1 1 2 2 ∴ T = ( ∫ dm ) v P − (v P ) x ω ( ∫ ydm ) + ( v P ) y ω ( ∫ xdm ) + ω ( ∫ r 2 dm ) m m m 2 m 2 1 1 2 = m v P − ( v P ) x ω y m + ( v P ) y ω x m + I Pω 2 2 2 若 P 為質心 G , 則 x = y = 0 1 1 2 T = mv G + I G ω 2 2 2 (平移) (旋轉) 18-4 Translation Qω = 0 0 1 1 2 ∴T = mvG + I Gω 2 2 2 1 2 T = mvG 2 18-5 Rotation about a Fixed Axis 1 1 2 T = mvG + I Gω 2 2 2 1 1 2 = m ( rGω ) + I Gω 2 2 2 1 2 2 = ( mrG + I G )ω 2 1 2 T = I Oω 2 18-6 General Plane Motion 1 1 2 T = mv G + I G ω 2 2 2 18-7 18.2 The Work of a Force Work of a Variable Force U F = ∫ F cos θds s 18-8 Work of a Constant Force U Fc = ∫ FC cos θds =( FC cos θ ) s 18-9 Work of a Weight 若物體向上移動 , 即 Δy > 0 則 U W = −WΔy 為負 若物體向下移動 , 即 Δy < 0 則 U W > 0 為正 18-10 Work of a Spring Force Fs = -ks 伸長時(往外),彈簧力向內 壓縮時(往內),彈簧力向外 即位移和力相反 1 2 1 2 Us = - ( ks2 - ks1 ) 2 2 1 2 1 2 = ks1 - ks2 2 2 18-11 Forces That Do No Work (1)外力作用於固定點 (2)外力與位移的方向垂直 18-12 18.3 The Work of a Couple r r dU M = F ( dθ ) + F ( dθ ) 2 2 = ( Fr)dθ = Mdθ 18-13 θ2 U M = ∫ Mdθ θ1 18-14 若M為constant U M = M (θ 2 − θ1 ) 18-15 18-16 18-17 18.4 Principle of Work and Energy T1 + ΣU 1− 2 = T2 Procedure for analysis: 1. Kinetic energy (Kinematic diagram) translation, rotation 2. Work (Free-body diagram) 3. Principle of Work and Energy 18-18 18-19 18-20 18-21 18-22 18-23 18-24 p. 470, 18-2 The double pulley consists of two parts that are attached to one another. It has a mass of 25 kg and a radius of gyration about its center of kO = 0.24 m. If it rotates with an angular velocity of 20 rad/s clockwise, determine the kinetic energy of the system. Assume that neither cable slips on the pulley. 18-25 p. 471, 18-8 The drum has a mass of 50 kg and a radius of gyration about the pin at O of kO = 0.23 m. If the 15-kg block is moving downward at 3 m/s, and a force of P = 100 N is applied to the brake arm, determine how far the block descends from the instant the brake is applied until it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is μK = 0.5. 18-26 18-27 p. 474, 18-20 The15-kg ladder is placed against the wall at an angle of θ = 45° as shown. If it is released from rest, determine its angular velocity at the instant just before θ = 0° . Neglect friction and assume the ladder is a uniform slender rod. 18-28 1.2 18-29 18-30 p. 476, 18-32 The assembly consists of two 7.5-kg slender rods and a 10-kg disks. If the spring is unstretched when θ = 45° and the assembly is released from rest at this position, determine the angular velocity of rod AB at the instant θ = 0° .The disk rolls without slipping. 18-31 18-32 Conservation of Energy (能量守 恆) 總位能 : V = Vg + Ve 保守力所作的功 : (ΣU1→ 2 ) cons = V1 − V2 又T1 + (ΣU1→ 2 ) cons + (ΣU1→ 2 ) noncons = T2 ⇒ T1 + (V1 − V2 ) + (ΣU1→ 2 ) noncons = T2 ⇒ T1 + V1 + (ΣU1→ 2 ) noncons = T2 + V2 若 (ΣU1→ 2 ) noncons = 0 ⇒ T1 + V1 = T2 + V2 ( 如摩擦力 ) 稱為能量守恆 18-33 EXAMPLE 18.6 Given:The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when θ = 0°. Neglect the mass of the pistons. Find: The angular velocity of rod AB at θ = 0° if the rod is released from rest when θ = 30°. Plan: Use the energy conservation equation since all forces are conservative and distance is a parameter (represented here by θ). The potential energy and kinetic energy of the rod at states 1 and 2 will have to be determined. 18-34 Solution: Initial Position EXAMPLE (continued) Final Position Potential Energy: Let’s put the datum in line with the rod when θ = 0°. Then, the gravitational potential energy and the elastic potential energy will be zero at position 2. => V2 = 0 Gravitational potential energy at 1: - (10)( 9.81) ½ (0.4 sin 30) Elastic potential energy at 1: ½ (800) (0.4 sin 30)2 So V1 = - 9.81 J + 16.0 J = 6.19 J 18-35 Initial Position EXAMPLE (continued) Final Position Kinetic Energy: The rod is released from rest from position 1 (so vG1 = 0, ω1 = 0). Therefore, T1 = 0. At position 2, the angular velocity is ω2 and the velocity at the center of mass is vG2 . 18-36 EXAMPLE (continued) Therefore, T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(ω2)2 At position 2, point A is the instantaneous center of rotation. Hence, vG2 = r ω = 0.2 ω2 . Then, T2 = 0.2 ω22 + 0.067 ω22 = 0.267 ω22 Now apply the conservation of energy equation and solve for the unknown angular velocity, ω2. T1 + V1 = T2 + V2 0 + 6.19 = 0.267ω22 + 0 => ω2 = 4.82 rad/s 18-37 p. 484, 18-40 At the instant shown, the 25-kg bar rotates clockwise at 2 rad/s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 0.6 m and a stiffness of k = 100 N/m, determine the angular velocity of the bar the instant it has rotated 30° clockwise. 18-38 18-39 p. 486, 18-52 The 25-kg square plate is pinned at corner A and attached to a spring having a stiffness of k = 300 N/m. If the plate is released from rest when θ = 0°, determine its angular velocity when θ = 90°. The spring is unstretched when θ = 0°. 18-40 18-41 18-42 p. 489, 18-64 The 12.5-kg slender rod AB is attached to spring BC which has an unstretched length of 1.2 m. If the rod is released from rest when θ 30°, determine its angular velocity at the instant θ = 90°. 18-43 18-44