Energy conservation/rolling object

[Epiphone]

Homework Statement

A ring of mass 2.4kg, inner radius 6.0cm, and outer radius 8.0cm is rolling (without slipping) up an inclined plane that makes an angle of theta=36.9 with the horizontal. At the moment the ring is x=2.0 m up the plane its speed is 2.8 m/s. the ring continues up the plane for some additional distance and then rolls back down. Assuming that the plane is long enough so that the ring does not roll off the top end, how far up the plane does it go?

Homework Equations

K = .5Iw^2 + .5mv^2
I = MR^2

The Attempt at a Solution

I tried to solve this using energy conservation.
K = .5Iw^2 + .5mv^2
after you plug in the moment of inertia in variables, you can cancel out the radius i think to get:

K = .5mv^2 + .5mv^2
K = mv^2

it struck me odd that translational KE would equal rotational KE, and why would they give me 2 values for R if R cancels out?

plugging in values for m and v, you get:
K = 18.816J

then i used energy conservation:
U+K = Ufinal

using trig to find the U value at x = 2

35.32+18.816 = mgh
h = 2.302

using trig again to find the length up incline
my final answer is 3.834m up the incline. I really don't think its right. any help checking would be greatly appreciated.

 

[Epiphone]

anyone?
sorry if bumping is taboo, but i have to go soon!
this is my last hope!

 

[KvnBushi]

The Attempt at a Solution

I tried to solve this using energy conservation.
K = .5Iw^2 + .5mv^2
after you plug in the moment of inertia in variables, you can cancel out the radius i think to get:

K = .5mv^2 + .5mv^2
K = mv^2

Your first equation for Kinetic Energy is incorrect.

The Kinetic Energy of a rolling object is $$\frac{1}{2} M w^2 + frac{1}{2} Mv^2$$
where 'w' is rotational speed and 'v' is translational speed.

I would take another look at the book until it makes more sense to you.