A simple pendulum of length l and a uniform rod of length L , pivoted at its upper end, are both displaced by the same small angle ₀ from the vertical and released from rest. If the magnitudes of their initial angular accelerations are found to be exactly equal, which of the following relations between L and l is correct?

Step-by-step solution

For the simple pendulum of length l , the magnitude of the initial angular acceleration is ₁ = g l ₀ . For the uniform rod of length L pivoted at its upper end, the center of mass is at a distance of L 2 from the pivot. The restoring torque for a small angular displacement ₀ is = mg ( L 2 ) ₀ . The moment of inertia of the rod about the pivot is I = mL^2 3 . The magnitude of its initial angular acceleration is ₂ = I = mg ( L 2 ) ₀ mL^2 3 = 3g 2L ₀ . Equating the two angular accelerations: g l ₀ = 3g 2L ₀ . Solving