A uniform rod \(A B\) of mass \(2 \mathrm{~kg}\) is hinged at one end \(A\). The rod is kept in the…
A uniform rod \(A B\) of mass \(2 \mathrm{~kg}\) is hinged at one end \(A\). The rod is kept in the horizontal position by a massless string tied to point \(B\). Find the reaction of the hinge $(\text{in } N)$ on end $A$ of the rod at the instant when string is cut. $(g=10 \, m/s^{2})$
Solution
The rod will rotate about the end A.
Let $a$ be the linear acceleration of $\mathrm{COM}$ of rod and $\alpha$ be the angular acceleration.
$\begin{aligned}
a &= \frac{\tau}{I} = \frac{mg \frac{1}{2}}{\frac{ml^{2}}{3}} = \frac{3g}{2l}, \\
a &= \frac{1}{2} \alpha = \frac{1}{2} \frac{3g}{2l} = \frac{3g}{4l}.
\end{aligned}$
From force equation:
$\begin{aligned}
mg - F &= ma, \\
F &= mg - m \frac{3g}{4l} = m \frac{g}{4l} = \frac{20}{4} = 5.
\end{aligned}$
Asked in: JEE Mains - Rotational Motion - Chapter Test
