Calculate the spring constant k

[zhenyazh]

hi,
i am preparing for the test and have the following question.
as usual i don't see where my mistake is.
an image is attached.

A thin uniform rod has mass M = 0.5 kg and length L= 0.55 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle θ1= 63.0o, and moves through its horizontal position at (B) and up to (C) where it stops with θ2 = 105.0o, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).

this i found. 1.023 m/s

The spring in (A) has a length of 0.11 m and at (B) a length of 0.14 m. Calculate the spring constant k.

ok so i decided to use a and b to calculate.
i decided to use the height of the rod in b as hight zero.
this means that in a the system has two energies. the potential and the spring.
the equation of energy conservation is:
mgh+0.5kx^2=0.5mv^2
thus
-0.5*9.81*0.55/2*cos(63)+0.5*k*0.03^2=0.5*0.5*1.023^2
but it is wrong.

thanks

 

[zhenyazh]

ok.
so i managed so solve it but with a and c and not a and b.
this means of course that i calculate the kinetic energy in b wrong.
can u help and explain how it should be done?
just 0.5mv^2 or just 0.5Iw^2 or their sum?
what do i do when

thanks a lot

 

[zhenyazh]

would it be fair to say
that it has only rotational energy becuase it is fixed at one end?

thanks