A thin, uniform rod of mass M is supported by two vertical strings. Find the tension
in the remaining string immediately after one of the strings is severed
I = 1/3MR^2 (thin rod rotating at end)
The Attempt at a Solution
So I was thinking:
Torque = Moment of Inertia * angular accel
R*Mg = (1/3*MR^2)(acm/R)
3g = acm
then F = ma
macm = Mg - T
Unfortunately this gives me the incorrect answer (it is supposed to be 1/4Mg).
Does anyone see where I am going wrong? It seems like this should be a rather straightforward problem.