GATE ME 2017 Set 2 — Question 38

Let the coordinates of P be $(0, y_P)$ and Q be $(x_Q, 0)$.
At the instant shown, the rod makes an angle of $45^\circ$ with the horizontal.
So, $x_Q = L \cos(45^\circ) = L/\sqrt{2}$ and $y_P = L \sin(45^\circ) = L/\sqrt{2}$.The center of mass G of the rod is at $(x_G, y_G)$.
$x_G = (L/2) \cos(45^\circ) = L/(2\sqrt{2})$
$y_G = (L/2) \sin(45^\circ) = L/(2\sqrt{2})$Since the ends slide along frictionless faces, the instantaneous center of rotation (IC) is the intersection of the perpendiculars from P and Q to the axes. This point is $(x_Q, y_P) = (L/\sqrt{2}, L/\sqrt{2})$.The distance from the center of mass G to the instantaneous center of rotation IC is $r_{IC}$.
$r_{IC} = \sqrt{(x_Q - x_G)^2 + (y_P - y_G)^2}$
$r_{IC} = \sqrt{(L/\sqrt{2} - L/(2\sqrt{2}))^2 + (L/\sqrt{2} - L/(2\sqrt{2}))^2}$
$r_{IC} = \sqrt{(L/(2\sqrt{2}))^2 + (L/(2\sqrt{2}))^2} = \sqrt{2 \left(\frac{L}{2\sqrt{2}}\right)^2} = \sqrt{2 \frac{L^2}{8}} = \sqrt{\frac{L^2}{4}} = L/2$.The moment of inertia of the rod about the instantaneous center of rotation $I_{IC}$ can be found using the parallel axis theorem:
$I_{IC} = I_G + M r_{IC}^2$
Given $I_G = ML^2/12$.
$I_{IC} = ML^2/12 + M(L/2)^2 = ML^2/12 + ML^2/4 = ML^2/12 + 3ML^2/12 = 4ML^2/12 = ML^2/3$.The torque about the instantaneous center of rotation is due to the gravitational force $Mg$ acting at the center of mass G.
The perpendicular distance from the IC to the line of action of gravity (a vertical line through G) is $d = x_Q - x_G = L/\sqrt{2} - L/(2\sqrt{2}) = L/(2\sqrt{2})$.Torque $\tau = Mg \cdot d = Mg \frac{L}{2\sqrt{2}}$.Now, apply Newton's second law for rotation: $\tau = I_{IC} \alpha$.
$Mg \frac{L}{2\sqrt{2}} = \left(\frac{ML^2}{3}\right) \alpha$Solving for $\alpha$:
$\alpha = \frac{MgL/(2\sqrt{2})}{ML^2/3} = \frac{3g}{2\sqrt{2}L}$Substitute the given values:
$L = \sqrt{2}$ m
$g = 10$ m/s$^2$$\alpha = \frac{3 \times 10}{2\sqrt{2} \times \sqrt{2}} = \frac{30}{2 \times 2} = \frac{30}{4} = 7.5$ radian/s$^2$.The magnitude of angular acceleration is $7.5$ radian/s$^2$.