A uniform rod of length ℓ and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is (m ℓ2/3) ):

Question

A uniform rod of length ℓ and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is (m ℓ2/3) ): (A) (3 g/2 ℓ) (B) (2 ℓ/3 g ) (C) (3 g /2 ℓ2) (D) mg (ℓ/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1f2c9207735f35f1215c9e0e6e1a60bf-.png" /> Here τ= operatornameI α ⇒ ( mg )((ℓ/2))=(( m ℓ2/3))(α) ⇒ α=(3 g /2 ℓ)

Q. A uniform rod of length $ℓ$ and mass $m$ is free to rotate in a vertical plane about $A$ . The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about $A$ is $\frac{m ℓ ^{2}}{3}$ ):

$\frac{3 g}{2 ℓ}$

$\frac{2 ℓ}{3 g}$

$\frac{3 g}{2 ℓ ^{2}}$

$m g \frac{ℓ}{2}$

Here $τ = I α$
$\Rightarrow (m g) (\frac{ℓ}{2}) = (\frac{m ℓ ^{2}}{3}) (α)$
$\Rightarrow α = \frac{3 g}{2 ℓ}$

Q. A uniform rod of length ℓ and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is 3mℓ2​ ):

2ℓ3g​

3g2ℓ​

2ℓ23g​

mg2ℓ​