Question:
Point masses 2 kg, 3 kg, 4 kg and 5 kg are lying at the points (-1,-1,0), (3,0,0), (0,0,1) and (0,4,0) respectively. The moment of inertia of this system about y-axis will be in kg m\(^{2}\).
Point masses 2 kg, 3 kg, 4 kg and 5 kg are lying at the points (-1,-1,0), (3,0,0), (0,0,1) and (0,4,0) respectively. The moment of inertia of this system about y-axis will be in kg m\(^{2}\).
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For moment of inertia about y-axis, always use distance \(r^2 = x^2 + z^2\).
Updated On: Jun 20, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Formula for moment of inertia about y-axis.
For a particle system: \[ I_y = \sum m(x^2 + z^2) \] Distance from y-axis depends only on x and z coordinates.
Step 2: Compute contribution of each mass.
For 2 kg at (-1,-1,0): \[ I_1 = 2[( -1)^2 + 0^2] = 2(1) = 2 \] For 3 kg at (3,0,0): \[ I_2 = 3(3^2 + 0^2) = 3 \times 9 = 27 \] For 4 kg at (0,0,1): \[ I_3 = 4(0^2 + 1^2) = 4 \] For 5 kg at (0,4,0): \[ I_4 = 5(0^2 + 0^2) = 0 \]
Step 3: Add all contributions.
\[ I_y = 2 + 27 + 4 + 0 = 33 \]
Step 4: Final interpretation.
Moment of inertia depends only on perpendicular distance from axis, not on coordinate along the axis itself.
\[ \boxed{33 \, \text{kg m}^2} \]
For a particle system: \[ I_y = \sum m(x^2 + z^2) \] Distance from y-axis depends only on x and z coordinates.
Step 2: Compute contribution of each mass.
For 2 kg at (-1,-1,0): \[ I_1 = 2[( -1)^2 + 0^2] = 2(1) = 2 \] For 3 kg at (3,0,0): \[ I_2 = 3(3^2 + 0^2) = 3 \times 9 = 27 \] For 4 kg at (0,0,1): \[ I_3 = 4(0^2 + 1^2) = 4 \] For 5 kg at (0,4,0): \[ I_4 = 5(0^2 + 0^2) = 0 \]
Step 3: Add all contributions.
\[ I_y = 2 + 27 + 4 + 0 = 33 \]
Step 4: Final interpretation.
Moment of inertia depends only on perpendicular distance from axis, not on coordinate along the axis itself.
\[ \boxed{33 \, \text{kg m}^2} \]
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