What is the Angular Speed of the Spool After the Bucket Falls?

[Touchme]

Use conservation of energy to determine the angular speed of the spool shown in Figure P8.36 after the 3.00 kg bucket has fallen 4.65 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

I used conservation of energy.
(KEr + KEt + mgh)i = (KEr + KEt + mgh)f
0 + 0 + (3 x 9.8 x 4.65) = [.5(.5Mr^2)w^2] + .5(Mv^2) + 0
136.71 = (.25Mr^2)w^2 + .5Mv^2
136.71 = (.25 x 5 x v^2) + (.5 x 5 x v^2)
v = 7.39
w = 12.324

The ans is not correct, what am i doing wrong?

Thank you for lookin but I manage to see my mistake. I used the wrong mass for the KEt

 

[Nate810]

and KEr. The mass for the KEt and KEr should be the mass of the bucket only not the spool and the bucket. 0 + 0 + (3 x 9.8 x 4.65) = [.5(.5Mb^2)w^2] + .5(Mbv^2) + 0136.71 = (.25Mb^2)w^2 + .5Mbv^2136.71 = (.25 x 3 x v^2) + (.5 x 3 x v^2)v = 6.753w = 14.06

 

[kyle8921]

term. It should have been the mass of the bucket (3 kg) rather than the total mass (5 kg). The correct calculation would be:

(KEr + KEt + mgh)i = (KEr + KEt + mgh)f
0 + 0 + (3 x 9.8 x 4.65) = [.5(.5Mr^2)w^2] + .5(Mv^2) + 0
136.71 = (.25Mr^2)w^2 + .5(3v^2) + 0
136.71 = (.25 x 5 x v^2) + 1.5v^2
v = 7.39 m/s
w = 12.324 rad/s

This is the correct answer for the angular speed of the spool after the bucket falls. It is important to carefully consider all the terms and their corresponding masses when applying conservation of energy equations.