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Question Number 23288 by Tinkutara last updated on 28/Oct/17
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 the chain just after it completely comes off the table.
$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{chain}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{and}\:\mathrm{length} \\ $$$${L}\:\mathrm{is}\:\mathrm{hanging}\:\mathrm{from}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{The}\:\mathrm{chain} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{limiting}\:\mathrm{equilibrium}\:\mathrm{when}\:{l}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{chain}\:\mathrm{over}\:\mathrm{hangs}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{slightly} \\ $$$$\mathrm{disturbed}\:\mathrm{from}\:\mathrm{this}\:\mathrm{position}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chain}\:\mathrm{just}\:\mathrm{after}\:\mathrm{it}\:\mathrm{completely} \\ $$$$\mathrm{comes}\:\mathrm{off}\:\mathrm{the}\:\mathrm{table}. \\ $$
Commented by Tinkutara last updated on 28/Oct/17
Commented by Physics lover last updated on 28/Oct/17
Is it (√(g(L−l))) ?
$${Is}\:{it}\: \\ $$$$\:\:\:\:\:\:\sqrt{{g}\left({L}−{l}\right)}\:\:? \\ $$
Commented by Physics lover last updated on 28/Oct/17
The answer given in the book is (√((gL)/((1+μ)))) ut the answer which i got is same. μ = (l/(L−l )) ⇒ l = ((μL)/((1+μ))) if you substitute l in my answer (√(g(L−l) )) = (√(g{L−((μL)/((1+μ)))})) = (√(g{((L+μL−μL)/((1+μ)))})) = (√((gL)/((1+μ))))
$${The}\:{answer}\:{given}\:{in}\:{the}\:{book}\:{is} \\ $$$$\:\:\:\:\:\:\:\sqrt{\frac{{gL}}{\left(\mathrm{1}+\mu\right)}} \\ $$$${ut}\:{the}\:{answer}\:{which}\:{i}\:{got}\:{is}\:{same}. \\ $$$$\: \\ $$$$\mu\:=\:\frac{{l}}{{L}−{l}\:}\:\:\:\:\:\Rightarrow\:{l}\:=\:\frac{\mu{L}}{\left(\mathrm{1}+\mu\right)} \\ $$$${if}\:\:{you}\:{substitute}\:\:{l}\:\:\:{in}\:\:{my}\:{answer} \\ $$$$\:\:\:\:\sqrt{{g}\left({L}−{l}\right)\:} \\ $$$$=\:\sqrt{{g}\left\{{L}−\frac{\mu{L}}{\left(\mathrm{1}+\mu\right)}\right\}} \\ $$$$=\:\sqrt{{g}\left\{\frac{{L}+\mu{L}−\mu{L}}{\left(\mathrm{1}+\mu\right)}\right\}} \\ $$$$=\:\sqrt{\frac{{gL}}{\left(\mathrm{1}+\mu\right)}} \\ $$$$ \\ $$
Answered by Physics lover last updated on 28/Oct/17
λ = linear mass density = ((M )/L) ⇒ μg(L−l)λ = λgl ⇒ μ = ((l )/((L−l))) let an element dx at a distance x from edge of table. dW_(friction ) = (−μgλdx)∙x as friction ′s direction is opposite to that of dist covered. ⇒ ∫_0 ^( W) dW = − ∫_0 ^((L−l)) μg((M/L))x ∙dx ⇒ W = −((Mgl(L−l))/(2L)) Using Energy conservation ⇒ ∣ΔK.E.∣ = ∣ΔP.E.∣ + W_(friction) ⇒ (1/2) mv^(2 ) = ∣ g(λl)((l/2))−g(λL)((L/2))∣ − ((Mgl(L−l))/(2L)) subtitute λ = (M/L) on solving v = (√(g(L−l)))
$$ \\ $$$$ \\ $$$$\:\:\:\:\lambda\:=\:{linear}\:{mass}\:{density}\:=\:\frac{{M}\:}{{L}} \\ $$$$\:\:\:\Rightarrow\:\mu{g}\left({L}−{l}\right)\lambda\:=\:\lambda{gl} \\ $$$$\:\:\:\Rightarrow\:\mu\:\:=\:\frac{{l}\:}{\left({L}−{l}\right)} \\ $$$$\:\:\: \\ $$$$\:\:\:\:{let}\:{an}\:{element}\:{dx}\:{at}\:{a}\:{distance} \\ $$$$\:\:\:{x}\:{from}\:{edge}\:{of}\:{table}. \\ $$$$\:\:\:{dW}_{{friction}\:} \:\:=\:\left(−\mu{g}\lambda{dx}\right)\centerdot{x} \\ $$$${as}\:{friction}\:'{s}\:{direction}\:{is}\:{opposite} \\ $$$${to}\:{that}\:{of}\:{dist}\:{covered}. \\ $$$$\:\Rightarrow\:\int_{\mathrm{0}} ^{\:\:{W}} {dW}\:=\:\:\:−\:\underset{\mathrm{0}} {\overset{\left({L}−{l}\right)} {\int}}\:\mu{g}\left(\frac{{M}}{{L}}\right){x}\:\centerdot{dx} \\ $$$$\:\:\Rightarrow\:{W}\:=\:−\frac{{Mgl}\left({L}−{l}\right)}{\mathrm{2}{L}} \\ $$$${Using}\:\:{Energy}\:{conservation} \\ $$$$\Rightarrow\:\mid\Delta{K}.{E}.\mid\:=\:\mid\Delta{P}.{E}.\mid\:+\:{W}_{{friction}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\:{mv}^{\mathrm{2}\:} =\:\mid\:{g}\left(\lambda{l}\right)\left(\frac{{l}}{\mathrm{2}}\right)−{g}\left(\lambda{L}\right)\left(\frac{{L}}{\mathrm{2}}\right)\mid\:−\:\frac{{Mgl}\left({L}−{l}\right)}{\mathrm{2}{L}} \\ $$$$\:{subtitute}\:\lambda\:=\:\frac{{M}}{{L}} \\ $$$${on}\:{solving}\: \\ $$$$\:\:{v}\:=\:\sqrt{{g}\left({L}−{l}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Physics lover last updated on 28/Oct/17
Commented by Tinkutara last updated on 28/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by Physics lover last updated on 28/Oct/17
you are welcome. BTW,did you try question no. 14 H section , laws of motion.Its a good one too.
$${you}\:{are}\:{welcome}. \\ $$$${BTW},{did}\:{you}\:{try}\:{question}\:{no}.\:\mathrm{14} \\ $$$$\:{H}\:{section}\:,\:{laws}\:{of}\:{motion}.{Its} \\ $$$${a}\:{good}\:{one}\:{too}. \\ $$
Commented by Physics lover last updated on 28/Oct/17
hmmm.
$${hmmm}. \\ $$$$ \\ $$

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