Question

A door shown in the figure) undergoes rotational motions about the vertical axis. The governing equation of rotational motion is given by Jyö + C70 + k 0 = 0 where Jo is the moment of inertia of the door, Ct is the rotational viscous damping and kt is the rotational stiffness of the door hinge. Assume that the door is 0.8 m wide (L = 0.8 m) and has a mass m of 15 kg. The moment of inertia can be found from It is found that it takes a force of 15 N (applied at the outer edge of the door) to keep the door opened at an angle of 90°. Door Suppose that you do not know the damping ratio but you know that when the initial values are 0o =O and vo = 1 rad/s, the door crosses the 0 = 0 position at time t = 2.1 seconds. Find the corresponding value of damping ratio.

Answer 1

Jo ät Gto+kto=0 Jo ir moment of Inestia of the door Jo a mi? 15X(0-812 - 5x0.64 = 3.2 kg m2 5 Torsional spring constant, kt = I T= 15x0.8= 12 N-m and o= 1/2 so, kt = h = 764 Nm/radian Let O=Oo sincwat to) is the solution of Governing equation - at t=0 0=0 so, 9=0 do O. Wd cosCWdt+Q) at to do = 1 ralls and d=0 So, 1= 0o wd and 0-0 at t - 2-1 seconds 0=Oo sin ( Wd X 2-1) possible only of wd x2-1= 20 l: sinar=o so wd = m = 1.496 rad Is 2-1 and wd = windi-a? and wnaeth = 1.5 45 radis so 1-496 = 1-545NT-C2 Damping Ratio, 4 = 0-25]