Rotational Motion

JEE Main level, numerical answer, medium

171 of 280

Question 171 of 280

One end of a straight uniform 1m1m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30o{30}^{o} from the horizontal (see figure). Its angular speed when it hits the table is given as nrads−1\sqrt{n}rad{s}^{-1} , where nn is an integer. The value of nn is ____________

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Answer: 15

according to energy conservation
mgl2sin⁡⁡30∘=12ml23ω2mg\dfrac{l}{2}\sin ⁡30^{\circ}=\dfrac{1}{2}\dfrac{m{l}^{2}}{3}{\omega }^{2}
Solving
ω2=15{\omega }^{2}=15
ω=15\omega =\sqrt{15}

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