A thin uniform rod of mass and length is pivoted at a height as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is [2026]

Question

A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $(\frac{L}{3})$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is __________.

(g = gravitational acceleration) [2026]

Smart Achievers · Accepted Answer

(4)

$m g \frac{ℓ}{6} = \frac{1}{2} I ω^{2}$

$Here I = \frac{m ℓ^{2}}{12} + \frac{m ℓ^{2}}{36} = \frac{m ℓ^{2}}{9}$

$m g \frac{ℓ}{6} = \frac{m ℓ^{2}}{18} ω^{2} \Rightarrow ω^{2} = \frac{3 g}{ℓ}$

$ω = \sqrt{\frac{3 g}{ℓ}}$

Solution

1 $\sqrt{\frac{3}{2} \frac{g}{L}}$

2 $\frac{3}{\sqrt{2}} \sqrt{\frac{g}{L}}$

3 $\frac{1}{\sqrt{2}} \sqrt{\frac{g}{L}}$

4 $\sqrt{\frac{3 g}{L}}$

Ans.
(4)

$m g \frac{ℓ}{6} = \frac{1}{2} I ω^{2}$

$Here I = \frac{m ℓ^{2}}{12} + \frac{m ℓ^{2}}{36} = \frac{m ℓ^{2}}{9}$

$m g \frac{ℓ}{6} = \frac{m ℓ^{2}}{18} ω^{2} \Rightarrow ω^{2} = \frac{3 g}{ℓ}$

$ω = \sqrt{\frac{3 g}{ℓ}}$

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Solution

1 32gL

2 32gL

3 12gL

4 3gL