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JEE Main 2019 (9 Jan, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2019 (9 Jan, Shift 2)

Q57JEE Main 2019 · 9 JAN · SHIFT 2Medium

Rotational Kinematics & Dynamics

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

Figure

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Answer: (b)

  1. Pivot at one end, I=mL23I = \frac{mL^2}{3}; centre of mass falls L2sin⁡30∘=L4\frac{L}{2}\sin30^\circ = \frac{L}{4}.
  2. mgL4=12⋅mL23ω2⇒ω2=3g2Lmg\frac{L}{4} = \frac{1}{2}\cdot\frac{mL^2}{3}\omega^2 \Rightarrow \omega^2 = \frac{3g}{2L}
  3. ω2=3×102×0.5=30\omega^2 = \frac{3\times10}{2\times0.5} = 30
  4. So the answer is (b) 30\sqrt{30} rad s−1^{-1}.
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