1. The problem statement, all variables and given/known data The 8 lb uniform rod AB shown in the figure rotates in the vertical plane about a pin at C. The spring attached to the rod at D has a spring constant of 1.25 lb/in., and its unstretched length is 24in. If the rod is released from rest in the position shown, determine its angular velocity when it is in the vertical position. 2. Relevant equations U = mgh U = .5*k(x^2) U = .5*Iω^2 I = I_cm + mr^2 3. The attempt at a solution Initial length = sqrt(36^2+20^2) = 41.183 in Initial stretch = 41.183 - 24 = 17.183 in Since the movement will tend towards lower energy (i.e. the spring will compress), the spring will become less stretched as the rod goes towards its vertical position (as opposed to stretching further into the opposite vertical position). Final length = sqrt(26^2+10^2) = 27.857 in Final stretch = 27.857 - 24 = 3.857 in As the spring becomes less stretched, elastic potential energy is converted into other forms of energy (rotational or gravitational potential). Initial elastic potential energy = (.5)(k)(x)^2 = .5(1.25 lb/in)(17.183^2) = 184.53 lb*in Final elastic potential energy = .5(1.25 lb/in)(3.857^2) = 9.298 lb*in Gravitational potential energy is gained as the rod goes into its vertical position. Final U = mgh = (8 lb)(386.4 in/s^2)(10 in) = 30912 lb*in Rotational energy is gained = .5*Iω^2 Final I = ml^2/12 + mr^2 = (8 lb)(40 in)^2/12 + (8 lb)(10 in)^2 = 13600 lb*in^2 Initial energy = final energy 184.53 lb*in = 9.298 lb*in + 30912 lb*in + .5(13600)ω^2 Solve for ω ω = imaginary...? So, obviously, I'm screwing up somewhere. Probably with the units, but I don't know which is wrong. Is there anything I'm not accounting for? What value should I be using for gravity? Thank you so much.