Rotational Motion Part 3

Concept:Use $\tau = I_{\text{total}}\,\alpha$ about the fixed axis $AB$.The axis $AB$ is perpendicular to the rod, so it is parallel to a diameter of each disc, and the parallel-axis theorem applies.Given:Mass of each disc, $M = 600\ \text{g}$; radius, $R = 10\ \text{cm}$.Mass of the rod, $M_{\text{rod}} = 600\ \text{g}$; length of the rod, $L = 30\ \text{cm}$.Applied torque, $\tau = 43 \times 10^5\ \text{dyne}\cdot\text{cm}$.From the figure, $AB$ is $20\ \text{cm}$ from the centre of the right disc, hence $30 - 20 = 10\ \text{cm}$ from the centre of the left disc.Moment of inertia of the rod about $AB$:The rod's centre of mass is at its midpoint, $15\ \text{cm}$ from either end, i.e. $15 - 10 = 5\ \text{cm}$ from $AB$.$I_{\text{rod}} = \frac{M_{\text{rod}}L^2}{12} + M_{\text{rod}}d^2 = \frac{600 \times 30^2}{12} + 600 \times 5^2 = 45000 + 15000 = 60000\ \text{g}\cdot\text{cm}^2$.Moment of inertia of the discs about $AB$:Moment of inertia of a disc about its diameter is $\frac{1}{4}MR^2$.Left disc, whose centre is at $d_1 = 10\ \text{cm}$ from $AB$:$I_1 = \frac{1}{4}MR^2 + Md_1^2 = \frac{1}{4}(600)(10)^2 + 600(10)^2 = 15000 + 60000 = 75000\ \text{g}\cdot\text{cm}^2$.Right disc, whose centre is at $d_2 = 20\ \text{cm}$ from $AB$:$I_2 = \frac{1}{4}(600)(10)^2 + 600(20)^2 = 15000 + 240000 = 255000\ \text{g}\cdot\text{cm}^2$.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:$\alpha = \frac{\tau}{I_{\text{total}}} = \frac{43 \times 10^5}{3.9 \times 10^5} = 11.03\ \text{rad/s}^2 \approx 11\ \text{rad/s}^2$.Answer:$\alpha \approx 11\ \text{rad/s}^2$ (Option D).