A metal rod of length 'L' and mass 'm' is pivoted at one end. A thin disk of mass 'M' and radius 'R' (< L) is attached at its center to the free end of the rod. Consider two ways the disc is attached : (case A). The disc is not free to rotate about its center and (case B) the disc is free to rotate about its center. The rod-disc system performs SHM in vertical plane after being released from the same displacement position. Which of the following statement(s) is (are) true ?
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This problem involves comparing the simple harmonic motion (SHM) of a physical pendulum in two configurations. The restoring torque for small angular displacements θ is τ = -mgh sinθ ≈ -mghθ, where h is the distance from pivot to center of mass. Since the total mass and its position are identical in both cases, the restoring torque is the same: τA = τB.
The angular frequency ω is given by , where I is the moment of inertia about the pivot. In case A, the disk cannot rotate, so its full moment adds to the rod's. In case B, the disk is free to rotate, so only its point mass moment contributes. Therefore, IA > IB, leading to ωA < ωB.
Final Answer: "Restoring torque in case A = Restoring torque in case B" and "Angular frequency for case A < Angular frequency for case B" are true.