Question

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through \( O \) (the center of mass) and \( O' \) (corner point) is: \begin{center} \begin{tabular}{c} \includegraphics[width=0.3\textwidth]{32.png} % Replace with your figure environment or manual reference \end{tabular} \end{center}

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{8} \)
• D. \( \frac{1}{2} \)

Hint:

To compute the moment of inertia about an axis parallel to the center of mass axis, apply the parallel axis theorem: \( I = I_{\text{center}} + M \cdot d^2 \). Ensure correct calculations for \( d \), the perpendicular distance.

Answer 1

The Correct Option is B

The moment of inertia (\( I_O \)) of a rectangular sheet about an axis passing through its center of mass (\( O \)) is: \[ I_O = \frac{M}{12} \left( a^2 + b^2 \right), \] where \( a = 80 \, \text{cm} \) and \( b = 60 \, \text{cm} \) are the dimensions of the rectangle, and \( M \) is the mass of the sheet. Step 1: Calculate \( I_O \) Substitute the given values: \[ I_O = \frac{M}{12} \left( 80^2 + 60^2 \right). \] Simplify: \[ I_O = \frac{M}{12} \left( 6400 + 3600 \right) = \frac{M}{12} \cdot 10000 = \frac{10000M}{12}. \] Step 2: Use the Parallel Axis Theorem to Find \( I_{O'} \) The moment of inertia about \( O' \) is given by: \[ I_{O'} = I_O + M \cdot d^2, \] where \( d \) is the perpendicular distance between \( O \) and \( O' \). Calculate \( d \): \[ d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \sqrt{40^2 + 30^2} = 50 \, \text{cm}. \] Substitute \( d = 50 \, \text{cm} \): \[ I_{O'} = \frac{10000M}{12} + M \cdot 50^2. \] Simplify: \[ I_{O'} = \frac{10000M}{12} + 2500M = \frac{10000M}{12} + \frac{30000M}{12} = \frac{40000M}{12}. \] Step 3: Find the Ratio of \( I_O \) to \( I_{O'} \) \[ \frac{I_O}{I_{O'}} = \frac{\frac{10000M}{12}}{\frac{40000M}{12}} = \frac{10000}{40000} = \frac{1}{4}. \] Final Answer: \[ \boxed{\frac{1}{4}} \]