Posted by Anon on Tuesday, October 18, 2011 at 8:46pm.
w=2.3/4.3= 1 rad per second.
Now, the moment of Inertia for the rod swinging about an end is... 1/3 ml^2
initial KE= 1/2 I w^2=1/6 ml^2 figure all that out given m, l
now, final PE= mgh were h is to the center of mass. Drawing the figure, then
h+lcosTheta=lor
h= l(1-cosTheta)
finally we have
1/6ml^2=mgl(1-cosTheta)
L/6g=1-cosTheta
costheta= 1- 4.3/6g
theta= arc cos ( above)
check all this.
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