Question

One end of a uniform rigid rod $OA$ of length $L$ and mass $m$ is attached to a frictionless hinge at $O$. The other end of the rod is connected to the roof at $B$ with a massless inextensible thread $AB$. Initially the rod is horizontal and at rest. The gravity is acting vertically downward as shown. Immediately after the thread $AB$ is cut, the reaction on the rod at $O$ is:

• A. $\dfrac{mg}{4}$ in the positive $y$-direction
• B. $\dfrac{mg}{2}$ in the negative $y$-direction
• C. $\dfrac{3mg}{4}$ in the negative $y$-direction
• D. $mg$ in the positive $y$-direction

Hint:

When a rod begins to rotate after a support is removed, find torque about the hinge, compute angular acceleration, then use COM acceleration to determine hinge reactions.

Answer 1

The Correct Option is A

Right after the thread is cut, only gravity acts on the rod. The rod begins to rotate about hinge $O$. The weight $mg$ acts downward at the center of mass, which is at a distance $\tfrac{L}{2}$ from $O$. The torque about $O$ immediately after cutting the thread is:
\[ \tau = mg \cdot \frac{L}{2} \] The moment of inertia of a uniform rod about one end is:
\[ I = \frac{1}{3} m L^2 \] Thus, the angular acceleration is:
\[ \alpha = \frac{\tau}{I} = \frac{mg(L/2)}{(1/3)mL^2} = \frac{3g}{2L} \] Now consider the acceleration of the center of mass. The linear acceleration of the COM is purely vertical at this instant and equals:
\[ a_{\text{COM}} = \alpha \cdot \frac{L}{2} = \frac{3g}{2L} \cdot \frac{L}{2} = \frac{3g}{4} \] Apply Newton’s second law in the vertical direction to the rod: \[ R_y - mg = m(-a_{\text{COM}}) \] Since the COM accelerates downward, $-a_{\text{COM}} = -\tfrac{3g}{4}$: \[ R_y - mg = -m\left(\frac{3g}{4}\right) \] Solving for $R_y$ gives:
\[ R_y = mg - \frac{3mg}{4} = \frac{mg}{4} \] This is a positive value, meaning the reaction is upward (positive $y$-direction). Thus, the reaction at hinge $O$ immediately after cutting the thread is:
\[ \frac{mg}{4} \text{ upward.} \]