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A chain AB of length lis located on a smooth horizontal table so that its fraction of length h hange freely with end B on the table. At a certain moment, the end A of the chain is set free. With what velocity with this end of the chain slip off the table? |
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Answer» Solution :Let m be the mass per unit length of the CHAIN. Let at any instant x be the length of hanging PART of chain and T the tension in chain. Then x mg-T= x ma ....... (1) and T=(l-x)ma...... (2) ADDING these equations, we get x mg = [x m + (l -x) m ]a = l ma ( or) ` a=(x)/(l) g , " As a " (dv)/(dt) implies (dv)/(dt)= (x)/(l) g ` Multiplying both sides by `(dx)/(dt) , ` we get ` (dx)/(dt) (dv)/(dt) = (x)/(l) g(dx)/(dt) or vdv = (x)/(l) g dx ` INTEGRATING both sides , we get ` int_0^(v) v dv = (g)/(l) int_h^(l) x dx ` `[(v^(2))/(2)]_0^(v)=(g)/(l)[(x^(2))/(2)]_(H)^(l)=(g)/(l)[(l^(2)-h^(2))/(2)] :. v=sqrt( {(g(l^(2)-h^(2)))/(l)})` |
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