A thin uniform metal rod of mass ' $M$ ' and length ' $L$ ' is swinging about a horizontal axis passing…

A thin uniform metal rod of mass ' $M$ ' and length ' $L$ ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' $\omega$ '. Its centre of mass rises to a maximum height of ( $\mathrm{g}=$ Acceleration due to gravity)
  1. $\frac{\mathrm{L}^2 \omega^2}{3 \mathrm{~g}}$
  2. $\frac{L^2 \omega^2}{2 g}$
  3. $\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}$
  4. $\frac{\mathrm{L}^2 \omega^2}{4 \mathrm{~g}}$

Solution

By conservation of energy, $\frac{1}{2} \mathrm{I} \omega^2=\mathrm{Mgh}$ $\therefore \quad \mathrm{h}=\frac{\mathrm{I} \omega^2}{2 \mathrm{Mg}}=\mathrm{h}...(i)$
The M.I of a uniform rod about an axis passing through its centre is $\frac{\mathrm{ML}^2}{12}$. As the axis passing through the end, using parallel axis theorem, $\mathrm{I}=\frac{\mathrm{ML}^2}{12}+\mathrm{M}\left(\frac{\mathrm{~L}}{2}\right)^2=\frac{\mathrm{ML}^2}{3}...(ii)$
Putting (ii) into (i), $h=\frac{L^2 \omega^2}{6 g}$

Asked in: MHT CET 2024 (15 May Shift 1)

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