Question

Four holes of radius \(5\) cm are cut from a thin square plate of \(20\) cm side and mass \(1\) kg. The moment of inertia of the remaining portion about the \(Z\)-axis is: 
 

• A. \(15 { kg} \cdot {m}^2\)
• B. \(0.37 { kg} \cdot {m}^2\)
• C. \(0.0017 { kg} \cdot {m}^2\)
• D. \(0.08 { kg} \cdot {m}^2\)

Hint:

- The moment of inertia of a solid square plate is \( I = \frac{1}{6} ML^2 \). - When holes are cut, subtract the moment of inertia contribution of each hole using \( I_{{hole}} = M_{{hole}} (R^2 + d^2) \). - Ensure all distances and masses are converted to standard SI units.

Answer 1

The Correct Option is C

Area mass density: \[ \sigma = \frac{M}{16R^2} \quad (\because {Area} = 4R \times 4R = 16R^2) \] 
Mass of each hole: \[ m_1 = \sigma \pi R^2 = \frac{M}{16R^2} \pi R^2 = \frac{\pi M}{16} \] 
Distance between center of plate and center of hole: \[ x = \frac{\sqrt{(2R)^2 + (2R)^2}}{2} = \frac{2\sqrt{2}R}{2} = \sqrt{2} R \] 
 

Moment of inertia of one hole about the Z-axis: \[ I_1 = \frac{1}{2} m_1 R^2 + m_1 x^2 = \frac{5\pi}{32} M R^2 \] 
Moment of inertia of whole plate about the Z-axis: \[ I = \frac{M (4R)^2}{6} = \frac{8}{3} M R^2 \] 
Required moment of inertia: \[ I_0 = I - 4 I_1 = \left[ \frac{8}{3} - 4 \left( \frac{5\pi}{32} \right) \right] M R^2 \] \[ I_0 = \left[ \frac{8}{3} - \frac{5\pi}{8} \right] M R^2 \] 
Given: \[ R = 5 { cm}, \quad M = 1 { kg} \] 
Final Calculation: \[ I_0 = \left[ \frac{8}{3} - \frac{5\pi}{8} \right] \times 1 \times 25 \times 10^{-4} \] \[ I_0 = 0.0017 { kg} \cdot {m}^2 \]