Question:
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is \(\frac{x}{12} ML^2\) kg m\(^2\). The value of x is ______ .
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is \(\frac{x}{12} ML^2\) kg m\(^2\). The value of x is ______ .
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Parallel Axis Theorem: \(I = I_{cm} + Md^2\). Ensure the axis through CM is parallel to the required axis. For a T-shape, the top rod is effectively a point mass \(M\) at distance \(L\) plus its own spin inertia \(\frac{ML^2}{12}\).
Updated On: Feb 24, 2026
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Correct Answer: 17
Solution and Explanation
Step 1: System Configuration:
Rod 1 (Vertical): Mass \(M\), Length \(L\), axis at end P.
Rod 2 (Horizontal): Mass \(M\), Length \(L\), attached at its midpoint to the top of Rod 1. Distance from P to Rod 2's center is \(L\). Step 2: Calculation:
Moment of Inertia of Rod 1 about end P: \[ I_1 = \frac{ML^2}{3} = \frac{4ML^2}{12} \] Moment of Inertia of Rod 2 about axis P (using Parallel Axis Theorem): \(I_{CM} = \frac{ML^2}{12}\) (about its own center parallel to axis P - actually perpendicular to length). Distance \(d = L\). \[ I_2 = I_{CM} + Md^2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12} \] Total Moment of Inertia: \[ I_{total} = I_1 + I_2 = \frac{4ML^2}{12} + \frac{13ML^2}{12} = \frac{17ML^2}{12} \] Comparing with \(\frac{x}{12} ML^2\): \[ x = 17 \] Step 4: Final Answer:
17
Rod 1 (Vertical): Mass \(M\), Length \(L\), axis at end P.
Rod 2 (Horizontal): Mass \(M\), Length \(L\), attached at its midpoint to the top of Rod 1. Distance from P to Rod 2's center is \(L\). Step 2: Calculation:
Moment of Inertia of Rod 1 about end P: \[ I_1 = \frac{ML^2}{3} = \frac{4ML^2}{12} \] Moment of Inertia of Rod 2 about axis P (using Parallel Axis Theorem): \(I_{CM} = \frac{ML^2}{12}\) (about its own center parallel to axis P - actually perpendicular to length). Distance \(d = L\). \[ I_2 = I_{CM} + Md^2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12} \] Total Moment of Inertia: \[ I_{total} = I_1 + I_2 = \frac{4ML^2}{12} + \frac{13ML^2}{12} = \frac{17ML^2}{12} \] Comparing with \(\frac{x}{12} ML^2\): \[ x = 17 \] Step 4: Final Answer:
17
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