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A-uniform-flexible-chain-of-length-3-2-m-rests-on-a-fixed-smooth-sphere-of-radius-R-2-pi-m-such-that-one-end-A-of-chain-is-on-the-top-of-the-sphere-while-the-other-end-B-is-hanging-freely-Chain-




Question Number 22370 by Tinkutara last updated on 16/Oct/17
A uniform flexible chain of length (3/2) m  rests on a fixed smooth sphere of  radius R = (2/π) m such that one end A of  chain is on the top of the sphere while  the other end B is hanging freely. Chain  is held stationary by a horizontal  thread PA. Calculate the acceleration  of chain when the horizontal string PA  is burnt. (g = 10 m/s^2 )
Auniformflexiblechainoflength32mrestsonafixedsmoothsphereofradiusR=2πmsuchthatoneendAofchainisonthetopofthespherewhiletheotherendBishangingfreely.ChainisheldstationarybyahorizontalthreadPA.CalculatetheaccelerationofchainwhenthehorizontalstringPAisburnt.(g=10m/s2)
Commented by mrW1 last updated on 17/Oct/17
AC^(⌢) =((πr)/2)=(π/2)×(2/π)= 1 m  BC=(3/2)−1=(1/2) m  ρ=mass each meter  a=acceleration  T_0 =tension at point C, at θ=0  (1/2)ρg−T_0 =(1/2)ρa  ⇒T_0 =(1/2)ρ(g−a)    −dT+ρgdscos θ=aρds  −dT=ρds(a−gcos θ)  −dT=ρr(a−gcos θ)dθ  −∫_T_0  ^( 0) dT=ρr∫_0 ^(π/2) (a−gcos θ)dθ  T_0 =ρr[aθ−gsin θ]_0 ^(π/2)   T_0 =ρr[((aπ)/2)−g]  (1/2)ρ(g−a)=ρ(2/π)(((aπ)/2)−g)  (g−a)=(4/π)(((aπ)/2)−g)  g−a=2a−(4/π)g  ⇒a=((4+π)/(3π))g ≈ 7.58 m/s^2
AC=πr2=π2×2π=1mBC=321=12mρ=masseachmetera=accelerationT0=tensionatpointC,atθ=012ρgT0=12ρaT0=12ρ(ga)dT+ρgdscosθ=aρdsdT=ρds(agcosθ)dT=ρr(agcosθ)dθT00dT=ρr0π2(agcosθ)dθT0=ρr[aθgsinθ]0π2T0=ρr[aπ2g]12ρ(ga)=ρ2π(aπ2g)(ga)=4π(aπ2g)ga=2a4πga=4+π3πg7.58m/s2
Commented by Tinkutara last updated on 16/Oct/17
Commented by Tinkutara last updated on 17/Oct/17
Thank you very much Sir!
ThankyouverymuchSir!
Answered by revenge last updated on 18/Oct/17
Commented by revenge last updated on 18/Oct/17
Consider a small element of mass dm on the sphere which  has length Rdθ. Mass of this small element is g sin θ dm.  dm=λdx or dm=λRdθ; where λ=((2M)/3) kg/m.  Total mass of such elements=∫_0 ^(π/2) (g sin θ)(λRdθ)  =λRg∫_0 ^(π/2) sin θ dθ=λRg=((4Mg)/(3π))  and mass of hanging part which has length (1/2) m=((Mg)/3).  Acceleration of chain=((Total weight(force))/(Total mass))=((((4Mg)/(3π))+((Mg)/3))/M)  =((4g)/(3π))+(g/3)=(((π+4)g)/(3π))
ConsiderasmallelementofmassdmonthespherewhichhaslengthRdθ.Massofthissmallelementisgsinθdm.dm=λdxordm=λRdθ;whereλ=2M3kg/m.Totalmassofsuchelements=π20(gsinθ)(λRdθ)=λRgπ20sinθdθ=λRg=4Mg3πandmassofhangingpartwhichhaslength12m=Mg3.Accelerationofchain=Totalweight(force)Totalmass=4Mg3π+Mg3M=4g3π+g3=(π+4)g3π
Commented by Tinkutara last updated on 18/Oct/17
Thank you very much Sir!
ThankyouverymuchSir!

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