Hi, this question was from an archived thread but I have another relevant question that was not answered by any of the posters-- P.S. The answer to c is NOT mg. 1. The problem statement, all variables and given/known data A meter stick (L = 1 m) with mass M = 0.409 kg is supported by two strings, one at the 0 cm mark and the other at the 90 cm mark. c) Now the string at 0 cm is cut. What is the tension in the remaining string immediately thereafter? I got parts a and b figured out, but I can't get c...Can I please get some help... 2. Relevant equations When the left string is cut, the stick pivots about the 90cm mark. Take torques about that point: (1) mgx = I α (Where "x" is the distance from the pivot to the center of mass of the stick; I is the rotational inertia of the stick about the pivot.) Note that the acceleration of the c.m. is related to α by a = αx. Thus you can solve for "a". Now consider the forces acting on the stick: (2) mg - T = ma This will give you T, the tension in the remaining string. 3. The attempt at a solution What formula for I should I use, I = (1/3)ML^2, or I = (1/12)ML^2? Also, should L be equal to "x", the distance from the pivot to the center of mass, or should it be 1meter?