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Tag(#2) 8 m₂ x 想想 Tao | z = to - r = 0 (c) (a) (b) Figure (a) above shows an inverted slider-crank mechanism driven by a motor torque Q on crank 2. In Fig. (b), symbols my ms and ma represent the centres of mass of the crank, slider, and the swinging link, respectively. The geometric dimensions, masses (m), and mass moments of inertia (1) of the links are given as follows: · Ground (link 1); ry = 0.22 m • Crank (link 2): my = 0.5 kg. 12 = 1.6 kg-m² r2 = 0.12 m, rig = 0.06 m • Slider (link 3); my = 0.3 kg. 1y = 0.3 kg-m² • Swinging link (link 4); ma = 1.6 kg. 4 = 6.5 kg-m² rag = 0.28 m For carrying out the kinematic and dynamic analyses, a vector loop and a fixed coordinate system (X, Y) are established as shown in Fig. (c).
Now, let us consider the forward dynamics problem of the mechanism. Assume that the mechanism is working on a vertical plane, and the friction effect is neglected. Suppose that crank 2 is initially staying at rest at 8 = 52° (see Fig. (a)), and a motor torque Q = 490 N-m (CCW) is to be applied onto crank 2 at the instant. Referring the vector loop established in Fig. (c), we have been able to derive the kinematic coefficients of the slider (link 3) at the initial position as: h3 = 0.3372, f3= 0.0503 h₂0.0507, f= -0.0722 df₂ where h3 = 1, f3d, hd, and fd. Accordingly, by using the power-equation method and kinematic coefficients, we can = do derive the equation of motion at the initial position as follows: 490 A₂+ B+C = (Unit: N-m) with 8₂ (0) D (rad) and ₂(0) = E (rad/s) What are the values of A, B, C, D, and E? Note that the gravitational acceleration is 9.81 m/s² downward. Give your answer to a precision of 4 decimal digits. For example, if the answer is "0.548145", then input "0.5481". 1) What is the value of A? (30%) 2) What is the value of B? (30%) 3) What is the value of C? (30%) 4) What is the value of D? (5%) 5) What is the value of E? (5%)