Question

A rod of length $50\ \text{cm}$ is pivoted at one end. It is raised such that it makes an angle of $30^\circ$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$^{-1}$) will be

• A. $\sqrt{30}$
• B. $\dfrac{\sqrt{20}}{3}$
• C. $\dfrac{\sqrt{30}}{2}$
• D. $\dfrac{\sqrt{40}}{2}$

Hint:

For rotating rigid bodies, always use conservation of energy and the correct moment of inertia about the axis of rotation.

Answer 1

The Correct Option is A

Step 1: Let the length of the rod be \[ L = 50\ \text{cm} = 0.5\ \text{m} \]
Step 2: The centre of mass of the rod is at a distance $\dfrac{L}{2}$ from the pivot. The vertical fall of the centre of mass when the rod moves from $30^\circ$ to the horizontal is: \[ h = \frac{L}{2}\sin 30^\circ = \frac{L}{4} \]
Step 3: Loss of potential energy: \[ \text{P.E.} = mg\frac{L}{4} \]
Step 4: Gain in rotational kinetic energy: \[ \text{K.E.} = \frac{1}{2}I\omega^2 \] For a rod about one end, \[ I = \frac{1}{3}mL^2 \]
Step 5: Apply conservation of energy: \[ mg\frac{L}{4} = \frac{1}{2}\cdot\frac{1}{3}mL^2\omega^2 \]
Step 6: Simplify: \[ \frac{mgL}{4} = \frac{mL^2\omega^2}{6} \] \[ \omega^2 = \frac{3g}{2L} \]
Step 7: Substitute $L=0.5\ \text{m}$: \[ \omega^2 = \frac{3g}{1} = 3g \] Taking $g \approx 10\ \text{m s}^{-2}$, \[ \omega = \sqrt{30}\ \text{rad s}^{-1} \]