Question

One end of the chain falls through a hole in its support and pulls the remaining links after it in a steady flow it the links which are initially at rest acquire the velocity of the chain suddenly and without frictional resistance or interference from the support or from adjacent links Choose the incorrect statement (when x = o, then v = 0) (length of the chain is L and p is the mass per unit length of the chain)

• A. the velocity v of the chain as a function of x is $\sqrt{\frac{2gx}{3}}$
• B. the acceferation of a of the falling chain as a function of x is $\frac{g}{3}$
• C. the energy Q lost from the system as the last link leaves the plafform is $\frac{pg{L}^2}{6}$
• D. tension at the middle point of falling chain is $\frac{pgx}{3}$

Answer 1

The correct option is D tension at the middle point of falling chain is $\frac{pgx}{3}$

Since chain is loose at the surface so no tension will generate at the starting point from where chain comes out from heaps so Using the FBD of falling part of chain

$P_1=pxv$ $\Rightarrow P_1=p(x+dx)(v+dv)$

$dp=pv+pgxdv+pgvdx+pgdxdv-pgxv=pxdv+pvdx$

$F_x=\frac{dp}{dt}=px\frac{dv}{dt}+pv\frac{dx}{dt}\Rightarrow pgxdt=p(xdv+vdx)=pd\left(vx\right)$

$gxdt=d\left(vx\right)$

multiplying both side by vx we have $\Rightarrow g{x}^{2}vdt=xvd\left(xv\right)\Rightarrow g{\int }_0^x{x}^{2}dx={\int }_0^{xv}xvd\left(xv\right)\Rightarrow \frac{g{x}^3}{3}=\frac{x^2{v}^2}{2}\Rightarrow v=\sqrt{\frac{2gx}{3}}$

$a=\frac{dv}{dt}=\sqrt{\frac{2g}{3}}\times \frac{1}{2\sqrt{x}}\frac{dx}{dt}=\sqrt{\frac{2g}{3}}\times \frac{1}{2\sqrt{x}}\times \sqrt{\frac{2gx}{3}}=\frac{g}{3}$

$\frac{pxg}{2}-T=px\times \frac{g}{3}\Rightarrow T=pxg[\frac{1}{2}-\frac{1}{3}]=\frac{pgx}{6}$

Heat develop $=-\Delta U-K=-[-\frac{pg{L}^2}{2}-0]-\frac{1}{2}pL\frac{2gL}{3}$

$=\frac{pg{L}^2}{2}-\frac{pg{L}^2}{3}=\frac{pg{L}^2}{6}$