Two circular discs, each of radius 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in the figure. If the mass of each disc is 600 gm and applied torque between the two discs is $$43\times 10^{5} dyne.cm$$. The angular acceleration of the discs about the given axis AB is_______$$rad/s^{2}$$.

Moment of Inertia for Rod:

$$I_{rod} = \frac{1}{12} m_r L^2 + m_r d_r^2$$

$$I_{rod} = \frac{1}{12} (600)(30)^2 + (600)(5)^2$$

$$I_{rod} = 45000 + 15000 = 60,000 \text{ g cm}^2$$

Moment of inertia for left disc:

$$I_1 = \frac{1}{4} m_d r^2 + m_d d_1^2$$

$$I_1 = \frac{1}{4} (600)(10)^2 + (600)(10)^2$$

$$I_1 = 15000 + 60000 = 75,000 \text{ g cm}^2$$

Moment of inertia for right disc:

$$I_2 = \frac{1}{4} m_d r^2 + m_d d_2^2$$

$$I_2 = 15000 + (600)(20)^2$$

$$I_2 = 15000 + 240,000 = 255,000 \text{ g cm}^2$$

Total Moment of Inertia:

$$I_{total} = 60,000 + 75,000 + 255,000 = 390,000 \text{ g cm}^2$$

$$\tau = I_{total} \alpha$$

$$\alpha = \frac{43 \times 10^5}{3.9 \times 10^5}$$

$$\alpha = \frac{43}{3.9} \approx 11.025 \text{ rad/s}^2$$