Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
Show Hint
Always include the moment of inertia of connecting rods and use the parallel axis theorem when rotation is not about the centre of mass.
Step 1: Understanding the Concept:
Angular acceleration is calculated using \(\alpha = \tau / I_{total}\). The total moment of inertia (\(I\)) is the sum of MOIs of the rod and the two discs about the central axis \(AB\). Step 2: Key Formula or Approach:
MOI of rod (mid-point): \(I_r = \frac{1}{12}ML^2\).
MOI of disc about diameter: \(I_d = \frac{1}{4}MR^2\).
Parallel Axis Theorem: \(I = I_{cm} + Md^2\). Step 3: Detailed Explanation: 1. MOI of Rod (\(I_{rod}\)): \(M = 600 \text{ g}, L = 30 \text{ cm}\).
\[ I_{rod} = \frac{1}{12} \times 600 \times 30^2 = 50 \times 900 = 45,000 \text{ g.cm}^2 \] 2. MOI of Discs (\(I_{discs}\)): The axis \(AB\) passes through the mid-point of the rod. Based on the diagram, the discs are \(10 \text{ cm}\) from the axis.
MOI of one disc about its diameter: \(I_{dia} = \frac{1}{4} \times 600 \times 10^2 = 15,000 \text{ g.cm}^2\).
Applying Parallel Axis Theorem (\(d = 10 \text{ cm}\)):
\[ I_{1\_disc} = 15,000 + 600(10^2) = 15,000 + 60,000 = 75,000 \text{ g.cm}^2 \]
Total MOI for two discs: \(2 \times 75,000 = 150,000 \text{ g.cm}^2\). 3. Total MOI (\(I_{total}\)):
\[ I_{total} = 45,000 + 150,000 = 195,000 \text{ g.cm}^2 \] 4. Angular Acceleration (\(\alpha\)):
\[ \alpha = \frac{\tau}{I_{total}} = \frac{43 \times 10^5}{1.95 \times 10^5} \approx 22.05 \text{ rad/s}^2 \] Step 4: Final Answer:
The angular acceleration is \(22 \text{ rad/s}^2\).
