Question

A uniform rod of mass m-6 kg has length L = 60 cm . Instead of pivoting it at its end, we pivoted it at L/3 from the end as shown in the figure, we now set the rod into small angle oscillations. What is the period of the oscillation given that the moment of inertia about the pivot point is ml2/9? L/3 Pivot L С.М. 2L/3

Answer 1

Take moment about pivot tau = I lambda mg sin Theta L/6 = ML^2/g alpha Rightarrow alpha = 3 Mg sin Theta/2 L M w^2 Theta = 3 g Theta/2 L w = squareroot 3 g/2 L and we know that T = 2 pi/w = 2 pi squareroot 2 L/3g d = L/2 - L/3 = L/6 = L/6 I = ML^2/12 + ML^2/36 = 4ML^2/36 = ML^2/9