IIT JEE , AIEEE Forum - Physics , Mechanics

cvramana

Initially the position of C.M. of the right side hanging mass is at L / 2n from the pulley and that of left side is at L (n - 1) / 2n from the pulley.

Therefore initially the position of C.M. of the entire chain from the pulley is at

Y₁ = [(m L / 2 n²) + (m L (n - 1)² / 2 n²) ] / m

= L (n² - 2n + 2) / 2n².

Finally the position of C.M. of the entire chain from the pulley is

Y₂ = L / 2

There fore the work done by the gravity is

W = m g (Y₂ - Y₁)

= m g L ( n - 1) / n²

This work should be equal to the change in kinetic energy of the chain

m v² / 2 = m g L ( n - 1) / n²

v = Ö (2 g L (n - 1) / n²)

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