A-uniform-chain-of-mass-M-and-length-L-is-hanging-from-the-table-The-chain-is-in-limiting-equilibrium-when-l-length-of-chain-over-hangs-It-is-slightly-disturbed-from-this-position-Find-the-speed-of

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Question Number 23288 by Tinkutara last updated on 28/Oct/17

$$\mathrm{A}\mathrm{uniform}\mathrm{chain}\mathrm{of}\mathrm{mass}M\mathrm{and}\mathrm{length}$$$$L\mathrm{is}\mathrm{hanging}\mathrm{from}\mathrm{the}\mathrm{table}.\mathrm{The}\mathrm{chain}$$$$\mathrm{is}\mathrm{in}\mathrm{limiting}\mathrm{equilibrium}\mathrm{when}l\mathrm{length}$$$$\mathrm{of}\mathrm{chain}\mathrm{over}\mathrm{hangs}.\mathrm{It}\mathrm{is}\mathrm{slightly}$$$$\mathrm{disturbed}\mathrm{from}\mathrm{this}\mathrm{position}.\mathrm{Find}\mathrm{the}$$$$\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{chain}\mathrm{just}\mathrm{after}\mathrm{it}\mathrm{completely}$$$$\mathrm{comes}\mathrm{off}\mathrm{the}\mathrm{table}.$$

Commented by Tinkutara last updated on 28/Oct/17

Commented by Physics lover last updated on 28/Oct/17

$$Isit$$$$\sqrt{g(L-l)}?$$

Commented by Physics lover last updated on 28/Oct/17

$$Theanswergiveninthebookis$$$$\sqrt{\frac{gL}{(1+\mu )}}$$$$uttheanswerwhichigotissame.$$$$$$$$\mu =\frac{l}{L-l}\Rightarrow l=\frac{\mu L}{(1+\mu )}$$$$ifyousubstitutelinmyanswer$$$$\sqrt{g(L-l)}$$$$=\sqrt{g\{L-\frac{\mu L}{(1+\mu )}\}}$$$$=\sqrt{g\left\{\frac{L+\mu L-\mu L}{(1+\mu )}\right\}}$$$$=\sqrt{\frac{gL}{(1+\mu )}}$$$$$$

Answered by Physics lover last updated on 28/Oct/17

$$$$$$$$$$\lambda =linearmassdensity=\frac{M}{L}$$$$\Rightarrow \mu g(L-l)\lambda =\lambda gl$$$$\Rightarrow \mu =\frac{l}{(L-l)}$$$$$$$$letanelementdxatadistance$$$$xfromedgeoftable.$$$${dW}_{friction}=(-\mu g\lambda dx)\cdot x$$$$asfriction{}^{\prime }sdirectionisopposite$$$$tothatofdistcovered.$$$$\Rightarrow {\int }_0^{W}dW=-\underset{0}{\stackrel{(L-l)}{\int }}\mu g\left(\frac{M}{L}\right)x\cdot dx$$$$\Rightarrow W=-\frac{Mgl(L-l)}{2L}$$$$UsingEnergyconservation$$$$\Rightarrow \mid \mathrm{\Delta }K.E.\mid =\mid \mathrm{\Delta }P.E.\mid +W_{friction}$$$$\Rightarrow \frac{1}{2}{mv}^2=\mid g\left(\lambda l\right)\left(\frac{l}{2}\right)-g\left(\lambda L\right)\left(\frac{L}{2}\right)\mid -\frac{Mgl(L-l)}{2L}$$$$subtitute\lambda =\frac{M}{L}$$$$onsolving$$$$v=\sqrt{g(L-l)}$$$$$$$$$$$$$$$$$$

Commented by Physics lover last updated on 28/Oct/17

Commented by Tinkutara last updated on 28/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by Physics lover last updated on 28/Oct/17

$$youarewelcome.$$$$BTW,didyoutryquestionno.14$$$$Hsection,lawsofmotion.Its$$$$agoodonetoo.$$

Commented by Physics lover last updated on 28/Oct/17

$$hmmm.$$$$$$

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