Two Rotating Discs: Kinetic Energy Loss Due to Friction

[TG3]

Homework Statement

A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at f initial = 151 rpm. A second disk of mass M2 = 251 g and radius R2 = 8 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed f final.
In the process, how much kinetic energy is lost due to friction?

Homework Equations

K = 1/2 I W^2 + 1/2 mv^2
I =1/2 mr^2
W = V/R
F Final = 103 rpm. (I calculated this in an earlier part and got it right.)

The Attempt at a Solution

Given that I have an equation and data to punch in, this should be really simple. However, to reconcile the fact that there are two radius' in the final solution, I decided to calculate KE for each disc separately. That is correct, right?
Anyways, here goes:
Initial K:
1/2 I W^2 + 1/2 mv^2
1/2 1/2 mr^2 151^2 + 1/2 m (Wr)^2
1/4 35,000 22801 + 1/2 350 2280100
199508750 + 399017500
Initial K = 598526250

Final K: (Done in two parts)

First, the same disc as above:
1/4 35,000 10609 + 1/2 350 1060900
92828750 + 185657500
Final K for first disc= 278486250

Next, for the other disc:
1/4 251 64 103^2 + 1/2 251 (103 8)^2
42605744 + 85211488
Final K for second disc = 127817232

Total Final K= 278486250+127817232
406303482

Change in K = 598526250 - 406303482
Change in K =192222768
However, this is wrong. Where did I slip up?

 

[tiny-tim]

Hi TG3!

(I haven't checked the rest of what you've done, but …)

it would help if you stated the units at all times …

i think you need to change rpm to radians per second