A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is (ml2/3) the initial angular acceleration of the rod will be

Question

A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is (ml2/3) the initial angular acceleration of the rod will be <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f036775e6ed1bc2ae95e542f090ddfe8-.png" /> (A) (2g/3l) (B) (mgl/2) (C) (3gl/2) (D) (3g/2l)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/848a4a9bf1e24c4ac9d37e935d7cff7d-.jpeg" /> Torque about A τ=mg (l/2) also, τ=I α Angular acceleration α=(τ/I)=(mgl /2/ml2/ 3)=(3g/2l) text(Given I=(ml2/3))

Q. A uniform rod $A B$ of length $l$ and mass $m$ is free to rotate about point $A$ . The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $\frac{m l ^{2}}{3}$ the initial angular acceleration of the rod will be

$\frac{2 g}{3 l}$

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

$\frac{3 g l}{2}$

$\frac{3 g}{2 l}$

Torque about $A$
$τ = m g \frac{l}{2}$
also, $τ = I α$
Angular acceleration $α = \frac{τ}{I} = \frac{m g l /2}{m l ^{2} /3} = \frac{3 g}{2 l}$
$(Given I = \frac{m l ^{2}}{3})$

Q. A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is 3ml2​ the initial angular acceleration of the rod will be

3l2g​

2mgl​

23gl​

2l3g​