Question

A thin uniform rod of mass \(m\) and length \(L\) is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle \(\theta\) with the vertical is (consider negligible friction at the pivot) (\(g =\) acceleration due to gravity)

• A. \(\frac{3g\sin\theta}{2L}\)
• B. \(\frac{3g\cos\theta}{2L}\)
• C. \(\frac{2g\sin\theta}{3L}\)
• D. \(\frac{2g\cos\theta}{3L}\)

Hint:

Always take torque about pivot and use centre of mass at \(L/2\) for uniform rod.

Answer 1

The Correct Option is A

Concept:
Angular acceleration is given by: \[ \alpha = \frac{\tau}{I} \] where \(\tau\) is torque and \(I\) is moment of inertia. Step 1: Position of centre of mass. For a uniform rod pivoted at one end: \[ \text{Centre of mass is at } \frac{L}{2} \]
Step 2: Torque due to gravity. Force \(mg\) acts downward at centre of mass. Perpendicular distance: \[ = \frac{L}{2} \sin\theta \] \[ \tau = mg \cdot \frac{L}{2} \sin\theta \]
Step 3: Moment of inertia. For rod about one end: \[ I = \frac{1}{3}mL^2 \]
Step 4: Angular acceleration. \[ \alpha = \frac{\tau}{I} = \frac{mg \frac{L}{2}\sin\theta}{\frac{1}{3}mL^2} \] \[ \alpha = \frac{mgL\sin\theta}{2} \times \frac{3}{mL^2} \] \[ \alpha = \frac{3g\sin\theta}{2L} \]
Step 5: Conclusion. \[ \alpha = \frac{3g\sin\theta}{2L} \]