A square plate \(A B C D\) of mass \(m\) and side \(l\) is suspended with the help of two ideal strings P…
A square plate \(A B C D\) of mass \(m\) and side \(l\) is suspended with the help of two ideal strings P and Q as shown. Determine the acceleration \(\left(\mathrm{in~} \mathrm{m} / \mathrm{s}^{2}\right)\) of corner \(A\) of the square just at the moment the string \(Q\) is cut. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
Solution
\(\begin{aligned}
& T \frac{l}{2}=\frac{m l^2}{6} \times \alpha, m g-T=m a \\
& \alpha=\frac{3 T}{m l}=\frac{3}{m l}(m g-m a)
\end{aligned}\)
Constraint relation: Acceleration of \(A\) in the vertical direction should be zero.
\(\begin{aligned} & \frac{l}{2} \alpha=a \\ & \Rightarrow \alpha=\frac{2 a}{l}=\frac{3(g-a)}{l} \\ & 5 a=3 g>a=3 \times \frac{g}{5}=6 \mathrm{~m} / \mathrm{s}^2\end{aligned}\)
Asked in: JEE Mains - Rotational Motion - Chapter Test
\(\begin{aligned} & \frac{l}{2} \alpha=a \\ & \Rightarrow \alpha=\frac{2 a}{l}=\frac{3(g-a)}{l} \\ & 5 a=3 g>a=3 \times \frac{g}{5}=6 \mathrm{~m} / \mathrm{s}^2\end{aligned}\)