First slide

Rotational motion

Question

A uniform thin rod of length l and mass m is hinged at a distance l/4 from one of the end and released from horizontal position as shown in the figure. The angular
velocity of the rod as it passes the vertical position is

Easy

A

$2 \sqrt{\frac{5 g}{7 l}}$
B

$2 \sqrt{\frac{6 g}{7 l}}$
C

$\sqrt{\frac{3 g}{7 l}}$
D

$2 \sqrt{\frac{g}{l}}$

Solution

Loss in potential energy = Gain in Kinetic energy
$\begin{array}{l} \Rightarrow m g (\frac{l}{4}) = \frac{1}{2} [\frac{m l^{2}}{12} + m {(\frac{l}{4})}^{2}] ω^{2} \\ \Rightarrow ω = \sqrt{\frac{24 g}{7 l}} = 2 \sqrt{\frac{6 g}{7 l}} \end{array}$

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