If the door is opened 90˚ and released, how long it will take the door to be within 1˚ of closing? Assume the return spring of the door to be critically damped.

$x = A_1 e^{-w_n t} + A_2 + e^{-w_n t}$

$\frac{dx}{dt} = x = A_1 (-w_n) e^{-w_n t} + A_2 [e^{-w_n t} + t (-w_n t)e^{-w_n t}$

$0 = A_1 (-w_n) - A_2$

$\frac{11}{2} (-w_n) + A_2 = 0$

$A_2 = \frac{w_n n}{\sum}$

For critically damped system, equation of motion is given as.

$x = (A_1 + A_2 t)e^{-w_n t}$

Given that [ t = 0 , X = $\pi/2$ ; Also, t = 0, so $\dot{x}$ = 0 ]

We get,

$\frac{\pi}{2} = A_1$. . . . by 1st B.C.S

$\frac{w \pi}{2} = A_2$. . . .by 2nd B.C.S

$\therefore$ In general,

$x = [ \frac{\pi }{2} + \frac{2n \pi }{2} t] e^{-w_n t}$

$x = \frac{\pi }{2} [ 1+ w_n t ] e^{-w-n t}$

Also : $\therefore$ $\frac{\pi}{180} = \frac{\pi}{2} [ 1 + w_n t ] e^{-w_n t}$

$e^{w_n t} = 90 [ 1+ w_n t ]$ - - - (1)

On solving above equation (1)

$W_n t = 6.516$

M.I of door about axis $X-X_1$

$I_{xx} = \frac{1}{2} m (a^2 + b^2) + m (\frac{9}{2})^2$

$= \frac{1}{2} ma^2 + \frac{1}{2} mb^2 + \frac{ma^2}{4}$

$I_{xx} = \frac{3}{4} ma^2 + \frac{1}{2} mb^2$

Given : a = 75 cm

B = 4 cm

M = 35 kg

$\therefore$ $I_{xx} = 147936 \ kg-cm^2$

$\therefore$ $w_n = \sqrt{ \frac{kt}{I_{xx}}} = \sqrt{ \frac{0.1}{14.79}}$

NOTE : $k_t$ = 1kg.cm/rad = 981 kg $cm^2$ / rad. $sec^2$ = 0.1 N.m / rad

$I_{xx}$ = 147936 kg – $cm^2$ = 14.79 $kg-m^2$

$\therefore$ we know that,

$W_n t = 6.516$

$\therefore$ $t = \frac{6.516}{0.08} = 81.45 \ sec$