Question

The uniform rod of mass $20kg$ and length $1.6m$ 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 as shown in figure is

• A. $\frac{15g}{16}$
• B. $\frac{17g}{16}$
• C. $\frac{16g}{15}$
• D. $\frac{g}{16}$

Answer 1

The correct option is A $\frac{15g}{16}$

Given: The uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings freely in the vertical plane.

To find the angular acceleration of rod just after the rod is released from rest in the horizontal position

Solution:

We know,

The torque of the weight, $ma=I\alpha$

$⟹mg\times \frac{l}{2}=\frac{m{l}^2}{3}\alpha ⟹\alpha =\frac{3g}{2l}$

Now substituting the values we get

$\alpha =\frac{3g}{2\times 1.6}⟹\alpha =\frac{30g}{2\times 16}⟹\alpha =\frac{15g}{16}$

is the angular acceleration of rod just after the rod is released from rest in the horizontal position