A non-uniform rod AB of mass M and length L is suspended horizontally by two vertical inextensible strings attached at its ends A and B. The linear mass density of the rod is proportional to the distance from end A. If the string at end A is suddenly cut, the tension in the string at end B immediately after the cut is ( g is acceleration due to gravity)

Step-by-step solution

Let the linear mass density of the rod be = cx , where x is the distance from end A. The total mass of the rod is: M = ₀^ L cx , dx = cL^2 2 c = 2M L^2 The position of the center of mass from A is: x_ cm = 1 M ₀^ L x(cx) , dx = 1 M ( cL^3 3 ) = 2L 3 So, the distance of the center of mass from end B is L - 2L 3 = L 3 . When the string at A is cut, the rod rotates about end B. The moment of inertia of the rod about B is: I_B = ₀^ L (L-x)^2 (cx) , dx = c ₀^ L (L^2x - 2Lx^2 + x^3) , dx I_B = c ( L^4 2 - 2L^4 3 + L^4 4