# Chain problem - velocity as the chain becomes completely vertical

## 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?

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

haruspex