InterviewSolution
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InterviewSolution
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Consider a uniform chain of mass `m` and length `L`, the chain is released when hanging length is `L_0`. Find the speed with which the chain leaves the table. |
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Answer» `K_1=0`,`U_1=-(1)/(2)((m)/(L)L_0)gL_0` `K_2=(1)/(2)mv^2`,`U_2=-(mgL)/(2)` By mechanical energy conservation `K_1+U_1=K_2+U_2` `0-(mgL_0^2)/(2L)=(1)/(2)mv^2-(mgL)/(2)` `v^2=gL-(gL_0^2)/(L)=(g(L^2-L_0^2))/(L)` `v=sqrt((g(L^2-L_0^2))/(L))` |
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