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

Step 1: Understanding the Concept:
When the rod is released, gravity exerts a torque about the pivot point.
This torque results in an angular acceleration according to Newton's second law for rotation: \(\tau = I\alpha\).
Step 2: Key Formula or Approach:
Torque: \(\tau = r \times F = rF \sin \phi\), where \(r\) is distance from pivot to center of mass and \(\phi\) is angle between \(r\) and \(F\).
Moment of inertia of a rod about its end: \(I = \frac{1}{3}mL^2\).
Step 3: Detailed Explanation:
The weight of the rod acts at its center of mass, which is at a distance \(L/2\) from the pivot.
When the rod makes an angle \(\theta\) with the vertical, the angle between the position vector \(\vec{r}\) of the center of mass and the gravitational force \(\vec{mg}\) is \(\theta\).
The magnitude of the torque is: \[ \tau = (L/2) \cdot mg \cdot \sin \theta \] Applying \(\tau = I\alpha\): \[ \frac{1}{3}mL^2 \cdot \alpha = mg \cdot \frac{L}{2} \cdot \sin \theta \] Cancel \(m\) and one \(L\) from both sides: \[ \frac{1}{3}L \cdot \alpha = \frac{g}{2} \sin \theta \] Solve for angular acceleration \(\alpha\): \[ \alpha = \frac{3}{L} \cdot \frac{g \sin \theta}{2} = \frac{3g \sin \theta}{2L} \] Step 4: Final Answer:
The angular acceleration is \(\frac{3g \sin \theta}{2L}\).