Question

A uniform rod of length \(60\ \text{cm}\) is placed with one end in contact with the horizontal table and is then inclined at an angle \(30^\circ\) to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]

• A. \(9\ \text{rad s}^{-1}\)
• B. \(6\ \text{rad s}^{-1}\)
• C. \(5\ \text{rad s}^{-1}\)
• D. \(8\ \text{rad s}^{-1}\)

Hint:

For a uniform rod rotating about one end, \[ I=\frac{1}{3}mL^2. \] When a rod falls under gravity, use conservation of energy: \[ \text{Loss in P.E.}=\text{Gain in rotational K.E.} \] to find the angular velocity.

Answer 1

The Correct Option is C

Step 1: Use energy conservation for falling rod.
Centre of mass falls \( h = \frac{L}{2}\sin 30^\circ = \frac{0.6}{2}\times 0.5 = 0.15\text{ m} \).
\( mgh = \frac{1}{2}I\omega^2 \), where \( I = \frac{mL^2}{3} \).

Step 2: Solve for \( \omega \).
\( mg(0.15) = \frac{1}{2}\cdot\frac{m(0.6)^2}{3}\cdot\omega^2 \)
\( 10\times0.15 = \frac{1}{2}\times0.12\times\omega^2 \Rightarrow \omega^2 = \frac{1.5}{0.06} = 25 \Rightarrow \omega = 5\text{ rad/s} \)

\[ \boxed{\omega = 5\text{ rad/s}} \]