A flexible chain of length L slides off the edge of a frictionless table. Initia

Question

A flexible chain of length L slides off the edge of a frictionless table. Initially a length y(initial) hangs over the edge. (a) Find the acceleration of the chain as a function of y. (b) show that the velocity as the chain becomes completely vertical is
v = (g(L- (y(initial))^2/L))

Note: This problem requieres integration

// the diagram is better than this one. The chain of course is consistant.

Explanation / Answer

let the length of the chain hanging = y length of the chain on the table = L-y let the total mass of the chain = M as mass is uniformly distributed over the length of the chain mass fraction of the chain hanging = M*y/L mass fraction of the chain on table = M*(L-y)/L force acting on the total chain is only due to the gravitational force acting on the hanging part of the chain force acting = mass * acceleration = force due to gravity on y length of chain => M*a = M*y/L *g where a is acceleration g is acceleration due to gravity cancelling M => a = y*g/L a) acceleration of chain is y*g/L acceleration is function of y as g and L are constants b)velocity ? as know external forces are acting and the gravitational force is conservative force energy of chain remains constant let the table be as reference of zero potential energy => intially a chain of length y(intial) hangs intial energy = kinetic energy + potential energy = 0 + (-M*(y(intial))/L * y(intial)/2 * g = -M* (y(intial))^2 *g/2L as length of chain completely moves down energy = kinetic energy + potential energy = Mv^2 /2 + (-M* L^2 *g/2L) = Mv^2/2 - M*L*g/2 equating energies intially and finally Mv^2/2 - M*L*g/2 = -M* (y(intial))^2 *g/2L Mv^2/2 = M*L*g/2 -M* (y(intial))^2 *g/2L cancelling M and 2 in denominator on both sides v^2 = L*g - y(intial)^2 *g/L = g(L - y(intial)^2 /L) => v = f(g(L - y(intial)^2 /L))

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