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A square plate of mass m and side length ‘a’ is kept as shown in figure. In equilibrium two of its side are horizontal and two are vertical. What is its angular acceleration just after lower spring is cut.

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g/a

2g/a

3g/2a

4g/a

answer is C.

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Detailed Solution

Initially in equilibrium condition

$\begin{array}{l} F_{S_{1}} + F_{S_{2}} = m g . . . . . . (1) \\ F_{S_{1}} \cdot \frac{a}{2} = F_{S_{2}} \cdot \frac{a}{2} . . . . . . . (2) \end{array}$

$∴ F_{S_{1}} = F_{S_{2}} = \frac{m g}{2}$

When lower spring is cut, $F_{S_{2}}$ will immediately become zero.

$F_{S_{1}} . \frac{a}{2} = (\frac{m a^{2}}{6}) α$

$\Rightarrow α = \frac{3 g}{2 a}$

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