Question

Q.1. For the system shown in Figure 1, the spring constant = 200 N/m. a) Write the complete Energy Equation for the solution to the problem (all terms must be included) b) If the system is initially at rest and the weight and inertia of the pulley are neglected, determine the angular velocity of A after Block B has dropped 600 mm. c) Calculate the maximum displacement of block B when it is lowered slowly. (Use Work, Power, Energy Method). Diameter of A = 1200 mm Mass A = 150 kg K= 200 N/m Frictionless Pulley 400 mm 50 kg Figure 1

Answer 1

a) Given me 50 kg, M = 150 kg, p= 0.6 m, R=0.4 m k = 200 N/m ܠܥܥܥܥܠܥܠܥ — 0 Total energy (€;) = total energy (E) initially finally as initially the system is in rest to so Kinetic energy of the system U; = = mgh now when Block 'B' lowered a height down then it will gain some kinetic energy as well as Block 'n' will also move linearly as well as it will rotate also due to the spring force in this condition the force of spring depends on the displacement of the center of the block A hence (KE) If = mo²+ mw2 + 6 2 w² y = -mg Ch+x) += kx2

now from equation is -mgh = 2 mw² + 2 mv² + 1 W² - mg (h+x) + Rx 2 | 2 (M + m) o² + & 2 w² - mg x + 2 kx² = 0 .-11) here I a moment of Inertia of 1 block A I = J MR² w = angular frequency * displacement in block B (b) it x = 600 mm = 0.6 m then w= ? consider Bock A 3 due to the Torque a which brise by spring force F will rotate the block so Usew now from o 3 (+m) * * + ( ne) - mg + 2 ex^ = 0 if a=0.6 and place all known values then **(150+50) (0.4)*62 + 4x150 x(0.6)?w?- 50x10x0.6 + *** 2000.6

16w?+ 13.5W2 - 300 + 60 = 0 29.562 = 240 w = 8.135 16 = 2.85 rad /sec / if g = 10 m/s2 or I w = 2.0 rad/sec / ifg=9.8 m/s2 (C) Now calculate maximum displacement in block B when the system is in equilibrium. by for block @ FBD T-mg for Block A FOD To A fa'+014) Te klx) -

so from ② f ® ka = mg x = mg x 50 X 9.8 200 = 1x=2.45 mtr 7