The uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings freely in the vertical plane. Angular acceleration of rod just after the rod is released from rest in the horizontal position as shown in figure is

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$\frac{15 g}{16}$

$\frac{17 g}{16}$

$\frac{16 g}{15}$

$\frac{g}{15}$

answer is A.

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Detailed Solution

Initial velocity of each point on the rod is zero so angular velocity of rod is zero. Torque about 0 –

$\begin{array}{l} τ = I α \\ \Rightarrow 20 g (0.8) = \frac{m l^{2}}{3} α \\ \Rightarrow 20 g (0.8) = \frac{20 {(1.6)}^{2}}{3} α \\ \Rightarrow \frac{3 g}{3.2} = α = a n g u l a r a c c e l e r a t i o n \\ \Rightarrow α = \frac{15 g}{16} \end{array}$

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