Conservation of Angular Momentum of uniform disk

[ash10n]

Homework Statement

A uniform disk turns at 8.6 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed. What is the angular velocity of the combination?
m,r,$$\omega$$_f, are all arbitrary?

Homework Equations

L_i=L_f
L=I*$$\omega$$
I_i=1/2m*r²
I_f=1/2m*r²+1/12m*r²

The Attempt at a Solution

I_f=1/2mr^2+1/12mr^2=7/12mr^2
8.6(1/2mr^2)=$$\omega$$(7/12mr^2)
8.6*1/2=$$\omega$$*7/12
$$\omega$$=4.3/(7/12)=7.37rev/s
Added info
I also tried using a final mass of 2m, with a final answer of 3.69, which was also incorrect
Solution
The length of the rod is equal to 2r, making the final moment of inertia equal to 5/6mr^2, not 7/12mr^2

 

[collinsmark]

I_f=1/2mr^2+1/12mr^2=7/12mr^2

Be careful with the above equation regarding the rod. The moment of inertia for a rod rotating around its center is
I = (1/12)ml²,
where l is the length of the rod. But in this problem, l is not equal to the disk's radius. Rather l is equal to the disk's diameter.

[Edit: Oh, wait. I see you already figured that out. Okay then! ]