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JEE Main 2018 · Physics · Rotational Motion · easy

A uniform thin rod of length 1.2 m is pivoted at one end and released from rest in the horizontal position. When it becomes vertical, the speed of…

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A uniform thin rod of length 1.2 m1.2\ \mathrm{m} is pivoted at one end and released from rest in the horizontal position. When it becomes vertical, the speed of its free end is (take g=10 m s−2g=10\ \mathrm{m\,s^{-2}})
  1. A. 6 m s−16\ \mathrm{m\,s^{-1}}
  2. B. 8.5 m s−18.5\ \mathrm{m\,s^{-1}}
  3. C. 4.9 m s−14.9\ \mathrm{m\,s^{-1}}
  4. D. 5 m s−15\ \mathrm{m\,s^{-1}}
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Answer: (A) 6 m s−16\ \mathrm{m\,s^{-1}}

The moment of inertia about the pivot is I=ML23I=\dfrac{ML^2}{3}. The centre of mass falls by L/2L/2, so 12Iω2=MgL2\tfrac12 I\omega^2=Mg\dfrac L2, giving ω2=3gL=301.2=25\omega^2=\dfrac{3g}{L}=\dfrac{30}{1.2}=25, i.e. ω=5 rad s−1\omega=5\ \mathrm{rad\,s^{-1}}. The free end moves on a circle of radius LL, so its speed is v=ωL=5×1.2=6 m s−1v=\omega L=5\times1.2=6\ \mathrm{m\,s^{-1}}. The value 55 is the angular speed, not the linear speed of the end.
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