Physical pendulum wrench oscillations

[squintyeyes]

Question Details:
A 1.44 kg monkey wrench is pivoted at one end and allowed to swing as a physical pendulum. The period of its motion is 0.860 s, and the pivot is 0.290 m from the center of mass of the wrench.

(a) What is the moment of inertia of the wrench?
0.0767 kgm2

If the wrench is initially pulled back to an angle of 0.600 rad from the equilibrium position, (b) what is the angular speed of the wrench as it passes through the equilibrium position?
_______ rad/s

Attempt
x(0)=0.6 = θsin(√(mgd/I)t)+φ
0.6 = θsin(√((mgd/I)(0))+φ)
0.6 = θsin(φ)
0.6/sin(φ) = θ

v(0) = 0 = θ(√(mgd/I))cos(√(mgd/I)t) +φ)
0 = θ(√(mgd/I))cos(√(mgd/I)(0)) +φ)
0 = θ(√(mgd/I))cos(φ)
0 =0.6/sin(φ)(√(mgd/I))cos(φ)

I am stuck here. Did i do something wrong? If i try to solve for φ I can not get a number.

 

[jbsteele]

What is the correct way to solve for angular speed? Thanks in advance. AnswerBy using the conservation of energy, the angular speed at the equilibrium position can be calculated as follows:KE = PE 1/2Iω2 = mgh ω = √(2gh/I) Substituting the given values: ω = √[(2 x 9.81 x 0.290) / 0.0767] ω = 5.07 rad/s