1. The problem statement, all variables and given/known data (Sorry for my bad English) A slippery (frictionless), light horizontal bar rotates about a vertical axis with a constant angular velocity ω. A cylinder with mass m, is initially attatched to a thread with length a and to a spring, which from the beginning has its "natural" length. Suddenly the thread breaks. Now the maximum distance between the cylinder and the axis is b. Determine the spring constant k for the spring. I have uploaded a picture of the problem. Thanks! 2. Relevant equations * T = m*(v^2)*1/2 * -0.5⋅k⋅r^2 * An integration 3. The attempt at a solution I got the correct answer by the following equation, but my teacher says that the "procedure" is wrong. The solution involves an integral. Here is my solution: In the beginning (when the thread is not cut), the cylinder only has the kinetic energy: T1 = 0.5⋅m(ω⋅a)^2. When the thread is cut, the cylinder is rotating with a new radius, b, thus the new kinetic energy is T2= 0.5⋅m(ω⋅b)^2. However, since the thread is cut, the spring makes the energy: -0.5⋅k⋅(b-a)^2. At this stage, the energy is: T2 + the energy from the spring = 0.5⋅m(ω⋅b)^2 - 0.5⋅k⋅(b-a)^2. Then I set the energy from the first stage equal to the energy from the second stage: 0.5⋅m(ω⋅a)^2 = 0.5⋅m(ω⋅b)^2 -0.5⋅k⋅(b-a)^2. I solved for k, and got the correct answer: k = (m⋅(ω^2)⋅(b + a))/(b - a). My teacher said that the procedure is wrong. He said: ⋅The energy is not preserved (as I have assumed). ⋅There should be an integration involved. ⋅The work from the spring is negative. ⋅The Kinetic energy increases. Another teacher said: ⋅We have an external force which rotates the axis. The energy is not preserved. He showed me this integral, which is not complete: ∫m⋅r⋅(ω^2) - k⋅(...) dr = 0 I don't know how to solve this problem, and I really need some help. Thank you!