Question

a massless rod is pivoted at one end and two masses of 2m and M are attached at distance b and 3b respectively find the angular acceleration when rod is released from horizont​

Answer 1

The angular acceleration at the instant the rod pivoted at one end is released is:

$\frac{5g}{11b}$

The Complete Question is:

A light rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m and m are attached to it at distance b and 3b respectively from the pivot. The rod is released from the horizontal. The angular acceleration at the instant it is released is:

Step-by-step explanation:

At the instant the rod is released, the torque will be

$\tau=2mg\times b+mg\times 3b$

$\implies \tau=2mbg+3mbg=5mbg$

The moment of inertia of the masses about the pivoted end

$I=2mb^2+m(3b)^2$

$\implies I=2mb^2+9mb^2=11mb^2$

Therefore, the angular acceleration

$\alpha=\frac{\tau}{I}$

$\implies \alpha=\frac{5mg\times b}{11mb^2}$

$\implies \alpha=\frac{5g}{11b}$

Hope this helps.

Similar questions:

Q: A light rod of length 1 m is pivoted at its center and two masses of 5 kg and 2 kg are hung from the ends as shown in the figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning?

Check here: https://brainly.in/question/4775391

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:

Check here: https://brainly.in/question/6536135