Question

A uniform chain of total mass m is laid out straight on a frictionless table and held stationary so that one-quarter of its length, L = 1.31 m, is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-quarter of its length remains on the table.​

Answer 1

let us consider that potential energy on the table is zero. therefore the part hanging will have the negative potential energy.
Consider a part of length dx of the chain at the depth x from the table.
mass will be dm = m/L dx
therfore its potential energy = -m/L gxdx
Now integrate it from 0 to L/4 as initial quarter length is hanging
on integration we get Initial potential energy = -mgL/32
Now when the 3/4 of the length is hanging then the potential energy will be
by intergrating under the limits of 0 to 3/4L
final potential energy = -9mgL/32
change in potential energy = kinetic energy
Initial - final = (1/2)mV²
8mgL/32 = (1/2)mV²
on solving we get
V = 2.535 m/s