Question 179 of 280

A rod of length $50cm$ is pivoted at one end. It is raised such that it makes an angle of ${30}^{o}$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad{s}^{-1}$ ) will be $\left(g=10m{s}^{-2}\right)$

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

Show answer and solution▾

Answer: (B) $\sqrt{30}$

From conservation of mechanical energy


Loss in gravitational $PE=$ gain in $KE$

$mg\dfrac{l}{2}\sin ⁡{30}^{o}=\dfrac{1}{2}I{\omega }^{2}$

$mg\dfrac{l}{2}\sin ⁡{30}^{o}=\dfrac{1}{2}\dfrac{m{l}^{2}}{3}{\omega }^{2}$

$\omega =\sqrt{\dfrac{3g}{2l}}=\sqrt{\dfrac{3\times 10}{2\times 50\times {10}^{-2}}}=\sqrt{30}rad/s$