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

Answer 1

Answer: (A)

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