A uniform flexible chain of length $l = 2.5 ext{ m}$ and mass $m = 1.5 ext{ kg}$ is placed on a sm

Question

A uniform flexible chain of length $l = 2.5	ext{ m}$ and mass $m = 1.5	ext{ kg}$ is placed on a smooth horizontal table such that a fraction $\eta = 0.6$ of its length hangs over the table edge. The chain is released from rest. Taking $g = 10	ext{ m/s}^2$, find the velocity of the chain (in $	ext{m/s}$) at the exact instant the trailing end of the chain leaves the table.

Professor PC Thomas Classes · Answer

We apply conservation of mechanical energy, taking the table surface as the zero level of gravitational potential energy ($U = 0$). Initially, only the hanging part of length $\eta l$ and mass $\eta m$ contributes to potential energy. Its center of mass is at depth $\frac{\eta l}{2}$ below the table edge, so $U_i = -(\eta m)g \left(\frac{\eta l}{2}ight) = -\frac{1}{2} m g l \eta^2$. When the entire chain leaves the table, its length is $l$, mass is $m$, and its center of mass is at depth $\frac{l}{2}$, giving $U_f = -mg \left(\frac{l}{2}ight) = -\frac{1}{2} m g l$. Since $K_i = 0$, energy conservation gives $K_f + U_f = U_i \implies \frac{1}{2} m v^2 - \frac{1}{2} m g l = -\frac{1}{2} m g l \eta^2$. Solving for velocity gives $v = \sqrt{gl(1 - \eta^2)}$. Substituting $g = 10	ext{ m/s}^2$, $l = 2.5	ext{ m}$, and $\eta = 0.6$, we get $v = \sqrt{10 	imes 2.5 	imes (1 - 0.36)} = \sqrt{25 	imes 0.64} = \sqrt{16} = 4	ext{ m/s}$.