The uniform rod of mass 20 kg and length of 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

Question

The uniform rod of mass 20 kg and length of 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 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7376b296c4d0a32c3c8945526cc60ca4-.png" /> (A) (15g/16) (B) (17g/16) (C) (16g/15) (D) (g/15)

Tardigrade Admin · Accepted Answer

Initial velocity of each point on the rod is zero so angular velocity of rod is zero

Torque about O

T = I α

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 = angular acceleration

Q. The uniform rod of mass $20 k g$ and length of $1.6 m$ 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

$\frac{15 g}{16}$

$\frac{17 g}{16}$

$\frac{16 g}{15}$

$\frac{g}{15}$

Initial velocity of each point on the rod is zero so angular velocity of rod is zero
$Torque about O$ $T = I α$
$20 g (0.8) = \frac{m ℓ ^{2}}{3} α$
$\Rightarrow 20 g (0.8) = \frac{20 ( 1.6 ) ^{2}}{3} α$
$\Rightarrow \frac{3 g}{3.2} = α = angular acceleration$

Q. The uniform rod of mass 20kg and length of 1.6m 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

1615g​

1617g​

1516g​

15g​