Question

A uniform square plate of mass $m$ is supported with its plane vertical as shown in the figure. Assume the centre of mass is on a horizontal line passing through $A$ . If the cable at $B$ breaks then just after that the angular acceleration of the plate is

• A. $\frac{3 g}{2 \sqrt{2} b}$
• B. $\frac{3 g}{\sqrt{2} b}$
• C. $\frac{g}{\sqrt{2} b}$
• D. $\frac{2 g}{\sqrt{3} b}$

Answer 1

Answer: (A)

The plate will rotate about the hinged point ( point $A$ ) after breaking of the cable
$\therefore $ using $\tau=Iα$ for the plate
$Mg\frac{b}{\sqrt{2}}=\frac{2}{3}Mb^{2}\alpha $ ( Moment of inertia about point $A$ is $\left.\frac{ Mb ^{2}}{6}+ M \left(\frac{ b }{\sqrt{2}}\right)^{2}=\frac{2}{3} Mb ^{2}\right)$
$\therefore \alpha =\frac{3 g}{2 \sqrt{2} b}$