Uniform rod \(A B\) is hinged at end \(A\) in horizontal position as shown in the figure. The other end is…
Uniform rod \(A B\) is hinged at end \(A\) in horizontal position as shown in the figure. The other end is connected to a block through a massless string as shown. The pulley is smooth and massless. Masses of block and rod is same and is equal to \(m\). Then acceleration of block just after release from this position is
\(6 g / 13\)
\(g / 4\)
\(3 \mathrm{~g} / 8\)
None
Solution
$\begin{aligned}
& \tau_{ext}=la \\
& \tau_T+\tau_{mg}=la \\
& \Rightarrow T L-\frac{mg L}{2}=\left(\frac{mL^2}{3}\right) a
\end{aligned}$
Substituting from Eq (i),
$\begin{aligned}
& TL-\frac{mgL}{2}=\frac{mL^2}{3} \times \frac{a}{L} \\
& \Rightarrow 2 T-mg=\frac{2 ma}{3} \cdots
\end{aligned}$
Multiplying Eq (ii) with 2 & adding with Eq (iii),
$\begin{aligned}
& 2 mg-mg=\frac{8 ma}{3} \\
& mg=\frac{8 ma}{3} \\
& \therefore a=\frac{3 g}{8}
\end{aligned}$
Asked in: JEE Mains - Rotational Motion - Chapter Test
