Question

The motion of the uniform 40 kg door as shown in Figure 5 is guided at its ends by the track. If it is released from rest at θ-00, determine the door's angular velocity at the instant θ-300 The spring is originally stretched 0.3 m when the door is held open, θ-00. Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the two sides of the door. k= 150N/m 2.4 m 2.4 m 1 Figure 5

Answer 1

Calculating potential energy at theta = 0 degree V_spring = 2 times [1/2 times K times s_1^2] = 2 times [1/2 times 150 times 0.362] = 13.5 Nm Total potential energy v-1 = v-spring + v_door V_1 = 13.5 = 0 v-1 = 13.5 N/m Door is initially at rest, So, Kinetic energyis zero, I_1 = 0 distance moved by pulley, 2l_0 = 2l-1 + 9distance a moves downward) 2l_0 = 2l_1 + 2.4 sin 30 degree l_0 - l_1 = 1.2 m total stretch of spring in final position,S_2 = a_3(l_0 - l_1) = 13 + 1.2 m = 1.5 m Total potential energy of both springs at theta = 30 degree V_spring = 2 times (1/2 times k times s_2^2) = 2 times 1/2 times 150 times (1.5)^2 = 337.5 Nm Total potential energy at theta = 30 degree, V_2 = V_spring + V-door = 337.5 - W times (2.4.2) sin 30 degree = 337.5 - 40 times 1.2 times 1/2 = 313.5 Nm Instantaneous centre (o) for rotation