- #1
NTesla
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- Homework Statement
- A chain AB of length l is located in a smooth horizontal tube so that its fraction of length h hangs freely and touches the surface of a table with its end B . At certain moment the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube ?
- Relevant Equations
- I've worked out the equation, but could somebody please let me know whether I've made any mistake or the equation is ok ? Also please let me know why the equation is wrong if it is wrong at any place.
My solution: For the horizontal portion of the chain: let at any instant the length of chain inside the tube is x, and at that instant the chain in the tube is having a velocity v. Then, at any instant:
##F = \frac{\mathrm{d} p}{\mathrm{d} t}##
##p##= mass of the chain in the tube at the instant t, multiplied by its velocity at that instant, i.e ##v##.
mass of the total length ##l## of the chain = ##m##.
mass of chain inside the tube = ##\frac{m}{l}\times x##
velocity of the chain inside the tube = ##-\frac{\mathrm{d} x}{\mathrm{d} t}##
Now ##F= \frac{\mathrm{d} (mv)}{\mathrm{d} t} (General\;equation)##
##\therefore F= \frac{m\mathrm{d} v}{\mathrm{d} t} + \frac{v\mathrm{d} m}{\mathrm{d} t} (General\;equation)##
##\therefore## ##F = \frac{mx}{l}\frac{\mathrm{d} v}{\mathrm{d} t} + \frac{vm}{l}\frac{\mathrm{d} x}{\mathrm{d} t} (In\;the\;present\;question)##For the hanging part of the chain(which is of length h):
##\frac{mhg}{l} - F=\frac{mh}{l}\frac{\mathrm{d} v}{\mathrm{d} t}##
##\therefore \frac{mhg}{l}-\frac{mh}{l}\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{m}{l}\left ( x\frac{\mathrm{d} v}{\mathrm{d} t}-v^2 \right )##
##\Rightarrow hg=x\frac{\mathrm{d} v}{\mathrm{d} t}+h\frac{\mathrm{d} v}{\mathrm{d} t}-v^2 ##
##\Rightarrow hg=-xv\frac{\mathrm{d} v}{\mathrm{d} x}-hv\frac{\mathrm{d} v}{\mathrm{d} x}-v^2##
Please let me know, if there is any mistake till here. If there is, then also let me know what is that mistake, and your reasoning.
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