Oscillation and Moment of inertia

[tempneff]

Homework Statement

A pendulum is constructed of a solid metal sphere of
mass M = 4.00 kg, attached to a thin metal rod of mass
m = 1.00 kg and length L = 40.0 cm. The pivot point for
the pendulum is at the upper end of the thin rod. The
pendulum oscillates through a small angle with a period
of T = 1.40 s. Find the radius of the sphere.

Homework Equations

$$\omega^2=\frac{I}{mgd}\hspace{15pt}\omega=\frac{2\∏}{T}\\I=\frac{1}{3}M_{rod}L^2+\frac{2}{5}M_{sphere}R^2(L+R)^2$$Center of Mass$$=\frac{m_1d_1+m_2d_2...m_nd_n}{m_{total}}$$
For d I used the distance from the center of mass to the pivot point$$\frac{L}{2}+R$$

The Attempt at a Solution

I have filled in all known values leaving only R, then solved for R using the quadratic formula.
My professor gave use 10.8 cm as an answer. I am getting ~30cm. This should be straight forward right? Chug and Plug?

 

[ehild]

Check your equations. Neither the one for the moment of inertia nor that for d (distance of CM from the pivot) are correct.

ehild

 

[tempneff]

Okay so this is my final equation, but I'm still not getting it right. What am I doing wrong.
$\omega^2=\frac{\frac{1}{3}L^2+\frac{2}{5}M_{sph}R^2+M_{sph}(L+R)^2}{M_{total}g(\frac{\frac{1}{2}LM_{rod}+RM_{sph}}{M_{total}})}$

 

[ehild]

The CM is still wrong. How far is the centre of the sphere from the pivot?
The mass of the rod is missing in the numerator. ehild