Two identical uniform rods, each of length L and mass m, are rigidly joined at right angles to form an L-shaped body. The body is pivoted at the free end of one rod and is free to oscillate in its own vertical plane. Which expression gives the angular frequency of small oscillations about the stable equilibrium?
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Question

Two identical uniform rods, each of length L and mass m, are rigidly joined at right angles to form an L-shaped body. The body is pivoted at the free end of one rod and is free to oscillate in its own vertical plane. Which expression gives the angular frequency of small oscillations about the stable equilibrium?

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G1 Physics · Accepted Answer

Take the pivot as origin. For the first rod, its centre is at distance L/2 from the pivot along that rod. For the second rod, its centre lies at coordinates corresponding to a distance L along the first rod and L/2 along the second rod. For equal rod masses, the combined centre of mass is at a distance d = L√10/4 from the pivot. Moment of inertia about the pivot: I₁ = mL²/3. For the second rod, I₂ = mL²/12 +m[L²+(L/2)²] = 4mL²/3. Thus, I = 5mL²/3. The total mass is 2m. For a physical pendulum,…