Have a problem with this Rotational Motion question

[Shuo Xue]

Homework Statement

[/B]
A wheel, of radius 200mm, rolls over the top of a hill with a speed of 20m/s and negligible friction losses. (I = 1/2mr^2)

Homework Equations

[/B]
Find the speed of the wheel when it is 10m below the top.

The Attempt at a Solution

[/B]
mgh = 1/2mv^2 + 1/2IW^2

W= v/r

mgh = 1/2mv^2 + 1/2(1/2mr^2)(v/r)^2
mgh = 1/2mv^2 + 1/4mv^2
gh = 3/4mv^2
v^2 = 4gh/3
v^2 = 4(9.81)(10)/3
v = 11.4m/s

I got v = 11.4 m/s
but my answer is incorrect as it is different from the answer given which is 23m/s.
I want to know the correct solution.

I've also tried searching for the height first.

h = (3/4v^2)/g
h = 30.58m

and then 10m below from the top

h = 20.58m

v^2 = 4gh/3
v^2 = 4(9.81)(20.58)/3
v = 16.41m/s

Which is still far from the answer also.

 

[Shuo Xue]

h = 30.58m

I tried adding 10m to h instead of subtracting 10m

so, h = 40.58

v^2 = 4gh/3
v^2 = 4(9.81)(40.58)/3
v = 23m/s

Doing it this way, I got the correct answer.

 

[haruspex]

I tried adding 10m to h instead of subtracting 10m

Do you now understand why that is correct?

 

[Shuo Xue]

The wheel is already rolling on the top of a hill
my mistake is I assume the initial kinetic energy is 0 but it is actually not.
So initial energy is mgh + 1/2mv^2 + 1/2IW^2mgh + 1/2mv^2 + 1/4mv^2 = 1/2mv^2 + 1/4mv^2

gh + 1/2v^2 + 1/4v^2 = 1/2v^2 + 1/4v^2

gh + 3/4v^2 = 3/4v^2

(9.81)(10) + (3/4)(20)^2 = 3/4v^2

398.1 = 3/4v^2

v^2 = 4(398.1)/3
v^2 = 530.8
v = 23m/s

Here is another way to answer the question.
This solution is more understandable for me than the earlier solution.

 

[Shuo Xue]

Do you now understand why that is correct?

Although I got the correct answer for that, I still don't understand why do I have to add 10m to the height that I got instead of subtracting 10m to it.

 

[Shuo Xue]

I also cannot assume that the final potential energy is 0 so

I think this is the most understandable solution.

mgh + 1/2mv^2 + 1/2IW^2 = mgh + 1/2mv^2 + 1/2IW^2

mgh + 1/2mv^2 + 1/2mv^2 = mgh + 1/2mv^2 + 1/2mv^2

gh + 3/4v^2 = gh + 3/4v^2

(9.81)(30.58) + (3/4)(20^2) = (9.81)(20.58) + 3/4v^2

(9.81)(30.58) - (9.81)(20.58) + 300 = 3/4v^2

3/4v^2 = (9.81)(30.58-20.58) + 300

3/4v^2 = (9.81)(10) + 300

3/4v^2 = 398.1
v^2 = 530.8
v = 23m/s