A uniform chain of mass M and lengthL is hanging from the table. The chain is in limiting equilibrium when l length of chain overhangs. It is slightly disturbed from this position. Find the speed of the chain just after it completely comes off the table

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Solution

Now consider the part that is originally on the table. A piece that's right at the edge falls a distance L/2, but a piece that's at the end doesn't fall at all. On average the part that start's on the table falls a distance L/4. Therefore the potential energy change for this part is (L/2)(L/4)

Set this change in potential energy to 1/2mv², and remember that pL = m:

1/2mv² = (pgL²)/4 + (pg L²)/8 = (3/8) pg L² = (3/8) mgL

v = (3gL/4)

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