Laws of Motion related doubt

Question

A uniform chain of length 'l' is placed on a smooth horizontal table, such that half of its length hangs over one edge. It is released from rest, the velocity with which it leaves the tables is ______?

Correct answer is √3gl/4

 

Please explain

.

Answer 1

Invoke Conservation of Energy ..... Change in PE + Change in KE = 0  .... dPE = dKE     d=delta or change

1. The half of the chain off the table falls a distance L/2. It's change in potential energy dPE is rho (L/2) g (L/2)
    Note;  [PE = mgh; rho L = mass, m]
2. The half on the table.... the part right at the edge falls a distance L/2, but a part that is at the end doesn't fall at all. The part that start's on the table and falls, will fall an average distance of L/4. Thus dPE, the potential energy change for this part is rho (L/2) g (L/4) = rho g (L)^2 / 8

dPE = dKE  = 1/2mv^2  

1/2 mv^2 = (rho L^2) g /4  + (rho L^2) g /8 = (3/8) rho L^2 g = (3/8) mgL   [ rho L = m ]

 

1/2 mv^2 = (3/8) mgL

 

v = sqrt(3gL/4)

 

Answer 2

This is essentially the same solution that Paul gave.  I just wanted to put the explanation in symbols, expressing potential energy in terms of the elevation of each half of the chain. 

 

K = kinetic energy

U = gravitational potential energy

 

From energy conservation,

 

K₁ + U₁ = K₂ + U₂

 

K₁ = 0

K₂ = mv²/2

 

In K₂, m = TOTAL mass of the chain, since the whole chain moves at the same speed.

 

With the reference elevation at the top of the table, the center of mass of the hanging HALF of the chain is at -L/4.  The potential energy of this half of the chain is

 

U₁ = -(m/2)g(L/4) = -mgL/8

 

U₂ has 2 parts.  One part is from the already hanging half of the chain that falls L/2 below its initial elevation of -L/4 down to -(3/4)L, and one part is from the half that was lying on the table, but whose center of mass has now translated down by L/4 to -L/4. 

 

U₂ = - (m/2)g(3/4)L - (m/2)g(L/4) = -3mgL/8 - mgL/8 = -mgL/2

 

Put it all together.

 

K₁ + U₁ = K₂ + U₂

 

0 - mgL/8 = mv²/2 - mgL/2

 

v²/2 = (3/8)gL

 

v = √(3gL/4)