A uniform chain of length l and mass m lies on a smooth table. A very small part of this chain hangs from the table. It begins to fall under the weight of hanging end. Find the velocity of chain when the length of hanging part becomes y.

Asked by devanshu_jain | 19th Oct, 2010, 07:21: PM

Expert Answer:

Dear Student,

Let d = mass per length = m/l

When length of hanging part is y, the velocity = v    . . (I)

Table surface is frictionless.

Kinetic energy at (I) =  1/2 mv2 = Loss of potential energy.

Loss of potential energy =  d*y*g*y/2           .  .  . (dygy - dygy/2)

(centre of mass of hanging chain is at y/2)

Hence

1/2 mv2  =  m/l * y2 * g/2

v=  g * y2/ l

v   =  y √(g/l)

We hope that clarifies your query.
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Answered by  | 22nd Oct, 2010, 01:32: AM

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