1. Given data

• Length of rod: $L = 50\text{ cm} = 0.5\text{ m}$
• Pivoted at one end
• Initial angle with horizontal: $30^\circ$
• Released from rest
• Need angular speed when rod becomes horizontal
• $g = 10\text{ m s}^{-2}$

Assume the rod is uniform.

2. Use conservation of mechanical energy

As the rod falls from $30^\circ$ above horizontal to horizontal, its center of mass loses gravitational potential energy.

For a uniform rod pivoted at one end, the center of mass is at distance

$\frac{L}{2}$

from the pivot.

Initial height of center of mass above the horizontal position:

$h = \frac{L}{2}\sin 30^\circ$

So,

$h = \frac{0.5}{2} \cdot \frac{1}{2} = 0.125\text{ m}$

Hence loss in potential energy is

$\Delta U = mgh = mg\left(\frac{L}{2}\sin 30^\circ\right)$

$\Delta U = mg\left(\frac{0.5}{2}\cdot \frac{1}{2}\right)$

$\Delta U = mg\left(\frac{1}{8}\right)$

3. Rotational kinetic energy at horizontal position

Moment of inertia of a uniform rod about one end:

$I = \frac{1}{3}mL^2$

If angular speed is $\omega$, rotational kinetic energy is

$K = \frac{1}{2}I\omega^2 = \frac{1}{2}\cdot \frac{1}{3}mL^2\omega^2 = \frac{1}{6}mL^2\omega^2$

4. Equate loss in PE to gain in KE

$mg\left(\frac{L}{2}\sin 30^\circ\right) = \frac{1}{6}mL^2\omega^2$

Cancel $m$:

$g\left(\frac{L}{2}\cdot \frac{1}{2}\right) = \frac{1}{6}L^2\omega^2$

$g\frac{L}{4} = \frac{1}{6}L^2\omega^2$

Multiply by 6:

$\frac{6gL}{4} = L^2\omega^2$

$\frac{3gL}{2} = L^2\omega^2$

$\omega^2 = \frac{3g}{2L}$

Substitute $g=10$ and $L=0.5$:

$\omega^2 = \frac{3\cdot 10}{2\cdot 0.5} = \frac{30}{1} = 30$

Therefore,

$\omega = \sqrt{30}\text{ rad s}^{-1}$

5. Check options

• A: $\sqrt{\frac{30}{7}}$ ❌
• B: $\sqrt{30}$ ✅
• C: $\frac{\sqrt{20}}{3}$ ❌
• D: $\frac{\sqrt{30}}{2}$ ❌

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.