Question
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
The correct answer is
$27$
To find the angular acceleration of the discs about the given axis \(AB\), we need to calculate the moment of inertia of the entire system (the rod and the two discs) and then apply the torque equation.
- Determine the Moment of Inertia:
- The moment of inertia of a disc about its center is given by: \(I_{\text{disc}} = \frac{1}{2} m r^2\)
- For each disc: \(I_{\text{disc}} = \frac{1}{2} \times 600 \, \text{gm} \times (10 \, \text{cm})^2 = 30000 \, \text{gm.cm}^2\)
- Total moment of inertia for two discs is: \(2 \times 30000 = 60000 \, \text{gm.cm}^2\)
- Moment of inertia of the rod about an axis through one end (using the parallel axis theorem): \(I_{\text{rod}} = \frac{1}{12} m L^2 + m \left( \frac{L}{2} \right)^2\)
- Substituting \(m = 600 \, \text{gm}\) and \(L = 30 \, \text{cm}\): \(I_{\text{rod}} = \frac{1}{12} \times 600 \times (30)^2 + 600 \times \left( \frac{30}{2} \right)^2 = 45000 \, \text{gm.cm}^2 + 67500 \, \text{gm.cm}^2\)
- Total moment of inertia of the system: \(I_{\text{total}} = 60000 + 112500 = 172500 \, \text{gm.cm}^2\)
- Apply Torque Equation:
- The torque about the axis is given as \(\tau = 43 \times 10^5 \, \text{dyne.cm}\)
- The angular acceleration \(\alpha\) is calculated using: \(\tau = I_{\text{total}} \cdot \alpha\)
- Solving for \(\alpha\): \(\alpha = \frac{\tau}{I_{\text{total}}} = \frac{43 \times 10^5}{172500} \, \text{rad/s}^2 = 24.93 \, \text{rad/s}^2 \approx 27 \, \text{rad/s}^2\) after rounding to the nearest whole number
Hence, the angular acceleration of the discs about the given axis \(AB\) is approximately \(27 \, \text{rad/s}^2\).
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