A rod of length L is pivoted at one of its ends and rotated in a horizontal plane with a uniform angular velocity. Let T₁ and T₂ represent the tensions at distances L/4 and 3L/4 , respectively, from the pivoted end.

Step-by-step solution

Let the mass per unit length of the rod be and its angular velocity be . Consider an element of length dx at a distance x from the pivoted end. The centripetal force required for this element is dF = ( dx) ^2 x . The tension T(x) at a distance x from the pivot provides the necessary centripetal force for the remaining portion of the rod from x to L . T(x) = _ x ^ L ^2 x' dx' = ^2 2 (L^2 - x^2) Since T(x) is a decreasing function of x , the tension is greater at points closer to the pivot. Given x₁ = L 4 and x₂ = 3L