Chain problem - velocity as the chain becomes completely vertical

[hi im nimdA]

Homework Statement

A flexible chain of length L slides off the edge of a frictionless table. Initially a length y0 hangs over the edge. Using energy methods, show that the velocity as the chain becomes completely vertical is v=sqrt(g(L-((y0)^2)/L)).

Relevant Equations

(1)Ki + Ui = Kf + Uf (2) K = 1/2 mv^2 (3) U = mgy

To start this problem, I used equation (1) K_i + U_i = K_f + U_f Then, using (2) and (3) and knowing that the initial velocity is 0, I have m_igy_i = \frac{1}{2} m_fv_f^2 + m_fgy_f The mass of the hanging part of the rope is $\frac{y_0}{L} m$. Additionally, I set the face of the table as y = 0. Therefore, I end up with this equation \frac{y_0}{L} mg(-y_0) = \frac{1}{2}mv^2 + mg(-L) Using algebra to solve for $v$, \frac{-gy_0^2}{L} = \frac{1}{2}v^2 -gL gL - \frac{gy_0^2}{L} = \frac{1}{2}v^2 g(L-\frac{y_0^2}{L}) = \frac{1}{2}v^2 From here it doesn't look like I'll be getting the right answer. Where did I go wrong?

 

[hurreechunder]

The centre of mass is at L/2 and (yo/2) respectively

 

[haruspex]

@hurreechunder has identified your error, but I really dislike this question.
The chain will not become vertical. As the horizontal section leaves the table it will have horizontal momentum. There is nothing to translate the KE of that into vertical motion.
To get the given answer, there would need to be a smooth curved duct deflecting the chain downwards.