A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal z axis, being free to swing in the xy plane (x horizontal, y vertically down). Its mass is m, its CM (center of mass) is a distance a from the pivot, and its moment of inertia (about the z axis) is I. Write down the equation of motion d/dt(Lz) = Torque_z and assuming the motion is confined to small angles, find the period of this compound pendulum.
T = 2pi/w
Lz = mvr
Torque = r cross F = I*(angular accel)
The Attempt at a Solution
Hmm, I think I am close to getting the right answer, but just can't quite get it.
I know Torque = I*alpha = I*a*d^2/dt^2(phi)
I also know that Lz = m*(a*d/dt(phi))*a = m*a^2*d/dt(phi)
Taking the deriv. of Lz seems to yield an equation where the phi's cancel out.
I also know that I = 1/3*MR^2 for a rod, but in this problem, I don't think that I need to use it.
Hmm, I need to somehow work gravity into the equation... but I can't quite figure out how.
Does anyone have any hints?