What is the rotational velocity of the crank?

[Lauren Wright]

Homework Statement

You accidentally knock a full bucket of water off the side of the well. The bucket plunges 13 m to the bottom of the well. Attached to the bucket is a light rope that is wrapped around the crank cylinder. The cylinder has a radius of 0.085 m and inertia of 4.0 kg. The inertia of the bucket plus water is 12 kg.
How fast is the handle turning when the bucket hits the bottom?
the answer should be in terms of Wv (unit s^-1)

Homework Equations

V(omega)=s/r (where s is distance traveled around circle and r is radius)
Wv(rotational velocity)=(change in V)/(change it time)
Wv=(Vt (tangential velocity))/r
Ac (centripetal acceleration)=(V^2)/r
Ug(gravitational potential energy)=mgh (where h is change in height)
K=1/2(I)(Wv)^2
I=(mR^2)/2 (where R is radius and I is the rotational inertia of a cylinder)
*there could be more but I think I covered everything

The Attempt at a Solution

https://goo.gl/ILj9hk <----- url for a picture of my work, ignore bottom page, it is for a different problem
I know the answer is not 8.14 s^-1, which is what I originally got[/B]

 

[JBA]

Lauren, you say that you know your answer is wrong; so, do you know the correct answer?

 

[J Hann]

You should be able to solve this using conservation of energy.
Initially, you have potential energy only.
Finally, you have the kinetic energy of the linear motion of the bucket and water plus
the rotational kinetic energy of the cylinder.
Apparently you are not asked to consider the rotational KE of the crank.

 

[haruspex]

url for a picture of my work,

Too hard to read. Please do not post images of your algebra. Only use images for extracts from textbooks and diagrams. Take the trouble to type your algebra into the post. That makes it much easier to read and easier to comment on specific items.
Also, avoid plugging in numbers until the final step. Keeping everything symbolic (creating variable names as necessary for given data) has many advantages.