Angular momentum of turntable problem

[Vagabond7]

Homework Statement

A 1.9kg , 20 cm-diameter turntable rotates at 150rpm on frictionless bearings. Two 480g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick.

what is the objects angular velocity in rpm right after this event?

Homework Equations

L=Iω where L is angular momentum, I is the moment of inertia and ω is the angular velocity
moment of inertia for a disk rotating about it's center 1/2*m*r^2

The Attempt at a Solution

Ok, so it seems like a simple conservation problem, so I tried to solve it like one.

I final*ω final = I initial * ω initial
ergo
ω final = (I initial * ω initial)/ I final

convert 150 rpm in rad/s, (150/60)*2\pi =15.71

since the formula for the moment of inertia "I" for a disk rotating about the center is 1/2*m*r^2
I just plug in the values

(.5*1.9kg*.1^2m*15.71 rad/s)/ (.5*(1.9+.96kg)*.1^2m) = 10.44 rad/s

converting back to rpm (10.44/ 2\pi )*60 = 99.66 rpm rounded to 100 rpm.

But this solution is incorrect. What have I done wrong?

 

[voko]

You seem to believe that the final moment of inertia is that of a more massive disk, but that is not write. You should treat the system as a disk with two additional masses at its rim.

 

[Vagabond7]

I see what you're saying, but I have no idea how to account for that in the math. Another hint please?

 

[mfb]

Do you know the moment of inertia for a point-mass at a radius r? Just use this (for both masses).

 

[voko]

You can consider the disk and the masses separately, and sum their ang. mom.

 

[Vagabond7]

Thank you much guys, those tips got me to where I needed to go!

Edit: Oh yeah, I made "I final" equal the sum of the moments of inertia of the disk, and the two point masses. Thanks a bunch.