Question

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

A. $3g/2l$
B. $2l/3g$
C. $3g/2l^2$
D. $l/2$

Answer 1

Hint:to find the angular acceleration, use the formula that, Torque = Moment of inertia * Angular acceleration, substitute the values and solve the question.
Formula used - $T = I \times \alpha $ , $T = mg\dfrac{l}{2}$ , $I = \dfrac{{m{l^2}}}{3}$

Complete step by step answer:
Given a uniform rod of length l and mass m which is free to rotate in a vertical plane about point A.
Also, moment of inertia of rod about A is $\dfrac{{m{l^2}}}{3}$
Now, we know that Torque = Moment of inertia * Angular acceleration
Which implies, $T = I \times \alpha - (1)$ , here T is the torque, I is the moment of inertia and $\alpha $ is angular acceleration.
Now we also know that, Torque = Force * perpendicular distance.
Torque is acting at the centre of gravity of the rod so we can say-
$T = mg\dfrac{l}{2} - (2)$
here mg is the force due to gravity and l is the length of the rod.
Now equating equation (1) and (2), we get-
$I \times \alpha = mg\dfrac{l}{2}$
now, putting $I = \dfrac{{m{l^2}}}{3}$ we get-
$\dfrac{{m{l^2}}}{3} \times \alpha = mg\dfrac{l}{2}$
solving further we get-
$
  \dfrac{{m{l^2}}}{3} \times \alpha = mg\dfrac{l}{2} \\
   \Rightarrow \dfrac{l}{3} \times \alpha = g \times \dfrac{1}{2} \\
   \Rightarrow \alpha = \dfrac{{3g}}{{2l}} \\
$
Therefore, the angular acceleration of the rod is 3g/2l.
Hence, the correct option is A.

Note – While solving such types of questions always first write down the information provided in the question then to find the angular acceleration use the formula as mentioned in the solution. Also, the inertia of the rod is given in the question as $I = \dfrac{{m{l^2}}}{3}$ and we know the torque is also given by force * perpendicular distance, so equating the two equations $I \times \alpha = mg\dfrac{l}{2}$ and then solving further we obtained the value of initial angular acceleration.