Asked by bhdrkn 2 years ago

You are given a uniform flexible chain whose mass is 5 kg and length is 5m, and a small frictionless pulley whose circumference is negligible compares to the length of the chain. Initially the chain is hung over the pulley with nearly equal lengths on both sides, but just unequal enough so that the unstable equilibrium condition will let the chain start to move. After some time, the longer end of the chain is a distance $ l=4\,m $ down from the pulley's axle. Find the acceleration $ a $ of the chain when the chain is at this position. Find the velocity of the $ v $ of the chain when $ l=4\,m $

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bhdrkn

Answered by lancer 2 years ago

Given that $ L=5m,\,l=4m, \, m=5\,kg $

and the linear density of the chain is,

$ \lambda=\frac{m}{L}=\frac{5\,kg}{5\,m}=1\,kg/m $

the mass of the left side of the pulley is,

$ m_{l}=\lambda (L-l) $

and the mass of the right side of the pulley is,

$ m_{r}=\lambda l $

and using Newton's laws on the left side of the pulley,

$ T-mg=ma\,\,\,\to\,\,\,T-\lambda (L-l)g=\lambda (L-l)a $

and on the right side of the pulley,

$ mg-T=ma\,\,\,\to\,\,\,\lambda lg-T=\lambda la $

adding these equations gives,

$ \lambda lg-\lambda(L-l)g=\lambda La $

and solving for a,

$ a=\big[\frac{2l}{L} - 1 \big] g=\big( \frac{2(4\,m)}{5\,m} - 1 \big) (9.8) = 5.88 m/s^2 $

hence the acceleration of the chain at this position is 5.88 m/s2

to find the velocity at this point, we will use the work energy theorem to equate kinetic and potential energies of the chain.

The change in kinetic energy is,

$ \Delta K=\frac 1 2 mv^2=\frac 1 2 \lambda Lv^2 $

the initial position of the chain nearly half hanging on both sides, then the entire chain goes to one side and therefore

$ l_i=\frac L 2 $ and $ l_f=l $, and the chains center of mass is,

$ y_{fcm}=\frac{\lambda l(l/2)+\lambda(L-l)\frac{L-l}{2}}{\lambda L}=\frac{l^2-lL+\frac 1 2 L^2}{L} $

$ y_{icm}=\frac{\lambda \frac L 2 \frac L 4+\lambda \frac L 2 \frac L 4}{\lambda L} $

and the change in potential energy is,

$ \Delta U = \lambda L g(y_{fcm}-y_{icm})=\lambda g [l-\frac 1 2 L]^2 $

finally from the conservation of energy,

$ \Delta K= \Delta U $

$ \frac 1 2 \lambda Lv^2=\lambda g [l-\frac 1 2 L]^2\,\,\,\to\,\,\,v=\sqrt{\frac{2g}{L}}[l-L/2]=\sqrt{\frac{2(9.8)}{5}}[4-5/2]=2.9698\,m/s $

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