Rotational Motion Part 3

Concept:Use the relation $\tau = I_{\text{total}} \alpha$ to find angular acceleration.Calculate total moment of inertia of the system about axis $AB$ using the parallel axis theorem.Explanation:Given: each disc mass $M = 600\ \text{g}$, radius $R = 10\ \text{cm}$; rod mass $M_{\text{rod}} = 600\ \text{g}$, length $L = 30\ \text{cm}$.Torque $\tau = 43 \times 10^5\ \text{dyne} \cdot \text{cm}$.First, find moment of inertia of the rod about axis $AB$.Centre of mass of rod is at $15\ \text{cm}$ from either end. Distance of CM from axis $AB$ = $15 - 10 = 5\ \text{cm}$.Using parallel axis theorem: $I_{\text{rod}} = \frac{M_{\text{rod}} L^2}{12} + M_{\text{rod}} d^2 = \frac{600 \times 30^2}{12} + 600 \times 5^2 = (50 \times 900) + (600 \times 25) = 45000 + 15000 = 60000\ \text{g} \cdot \text{cm}^2$.Now for each disc: axis $AB$ is parallel to its diameter.Moment of inertia of a disc about its diameter = $\frac{1}{4} M R^2$.Left disc: its centre is $10\ \text{cm}$ from axis $AB$.$I_1 = \frac{1}{4} M R^2 + M r_1^2 = \frac{1}{4} (600)(10)^2 + 600(10)^2 = 15000 + 60000 = 75000\ \text{g} \cdot \text{cm}^2$.Right disc: its centre is $20\ \text{cm}$ from axis $AB$.$I_2 = \frac{1}{4} M R^2 + M r_2^2 = \frac{1}{4} (600)(10)^2 + 600(20)^2 = 15000 + 240000 = 255000\ \text{g} \cdot \text{cm}^2$.Total moment of inertia:$I_{\text{total}} = I_{\text{rod}} + I_1 + I_2 = 60000 + 75000 + 255000 = 390000\ \text{g} \cdot \text{cm}^2 = 3.9 \times 10^5\ \text{g} \cdot \text{cm}^2$.Angular acceleration:$\alpha = \frac{\tau}{I_{\text{total}}} = \frac{43 \times 10^5}{3.9 \times 10^5} \approx 11.025\ \text{rad/s}^2$.The nearest integer is $11$.Answer:$\alpha = 11\ \text{rad/s}^2$ (Option D).