Question

A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two mass and m are attached to it at distances b and 3b respectively from the pivot. The rod is held horizo and then released. The angular acceleration of the rod at the instant it is released is :

• A. bg
• B. $\frac{g}{b}$
• C. gb + 5
• D. $\frac{5g}{1lb}$

Answer 1

Answer: (D)

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

Answer 2

The problem asks for the angular acceleration of a nearly massless rod, pivoted at one end, with two masses attached. The rod is initially held horizontal and then released.

1. Interpret the Masses and Distances: The question states: "Two mass and m are attached to it at distances b and 3b respectively from the pivot." This phrasing is ambiguous. However, comparing it to the similar question and its solution, which leads to one of the given options, we can infer the intended mass distribution. The similar question states: "Two masses $m$ and $2m$ are attached to it at distances $b$ and $3b$respectively from the pivot." The solution for the similar question calculates torque as $\tau = (2mg \times b + mg \times 3b)$ and moment of inertia as $I = 2mb^2 + m(3b)^2$. This implies that the mass $2m$ is at distance $b$, and the mass $m$ is at distance $3b$. Assuming the current question intends the same setup to yield one of the given options, we proceed with:

• Mass at distance $b$: $M_1 = 2m$
• Mass at distance $3b$: $M_2 = m$

2. Calculate the Total Torque ($\tau$): When the rod is horizontal, the gravitational force on each mass creates a torque about the pivot. The torque due to a force $F$ at a distance $r$ is $\tau = F \cdot r$ (since the force is perpendicular to the rod).

• Torque due to $M_1$: $\tau_1 = (M_1 g) \times b = (2m)g \times b = 2mgb$
• Torque due to $M_2$: $\tau_2 = (M_2 g) \times (3b) = (m)g \times (3b) = 3mgb$ Both torques act in the same direction (causing clockwise rotation). Total torque: $\tau = \tau_1 + \tau_2 = 2mgb + 3mgb = 5mgb$

3. Calculate the Total Moment of Inertia (I): For a point mass $M$ at a distance $r$ from the pivot, the moment of inertia is $I = Mr^2$.

• Moment of inertia due to $M_1$: $I_1 = M_1 b^2 = (2m)b^2 = 2mb^2$
• Moment of inertia due to $M_2$: $I_2 = M_2 (3b)^2 = m(9b^2) = 9mb^2$ Total moment of inertia: $I = I_1 + I_2 = 2mb^2 + 9mb^2 = 11mb^2$

4. Calculate the Angular Acceleration ($\alpha$): Using Newton's second law for rotational motion, $\tau = I \alpha$. Substitute the calculated values of $\tau$ and $I$: $5mgb = (11mb^2) \alpha$ Solve for $\alpha$: $\alpha = \frac{5mgb}{11mb^2}$ Cancel $m$ from the numerator and denominator, and cancel one $b$: $\alpha = \frac{5g}{11b}$

This result matches option (D).

The final answer is $\boxed{\text{5g/11b}}$.

Explanation of the solution: The angular acceleration is found by dividing the net torque by the total moment of inertia. The net torque is the sum of torques due to gravity on each mass, calculated as (mass × g × distance). The total moment of inertia is the sum of moments of inertia for each point mass, calculated as (mass × distance²). Assuming masses 2m at distance b and m at distance 3b, the total torque is $5mgb$ and the total moment of inertia is $11mb^2$. Dividing these yields the angular acceleration $\frac{5g}{11b}$.

Answer: The angular acceleration of the rod at the instant it is released is $\frac{5g}{11b}$. The correct option is (D).