<< Mentor's note: The full solution should not have been posted, since no correct solution was available. The advocated approach here would have been to try to guide the OP to the answer. If ever considering providing a complete solution for any reason, please contact a mentor for approval before posting. This post has not been deleted as the OP has already read it and the continued discussion in the thread refers to it. >>
sarthak sharma said:
please can you explain how did you get this answer...
Normally I wouldn't give a detailed answer in this sub-forum on PF, but since the topic this thread is about solution given to you in the book, and the book's solutions have some obvious mistakes anyway [Edit: not to mention its approach to the solution doesn't even match the problem statement], I do not think that it is inappropriate to introduce an alternate solution in this case.
I may skip a few details in the steps though for you to work through yourself.
Anyhoo, my solution didn't use conservation laws. It's all just kinematics.
Define the mass, m as being only the mass of the part of the rope that has fallen though the hole. Since the rest of the rope is sitting there quietly, piled up in its sleepy lump, minding its own business, we can ignore it for most of this derivation.
The only force acting on this part of the rope is the force due to gravity, mg. This gravitational force is equal to that part of the rope's rate of change in momentum.
\vec F = \frac{d \left( m \vec v \right)}{dt}
If the mass of this part of the rope was constant, the result would be equal to the familiar \vec F = m \vec a. But that's not the case here since that part of the rope's mass changes as a function of time. So we need to use the chain rule.
\vec F = \left( m \right) \left( \frac{d \vec v}{dt} \right) + \left( \frac{d m}{dt} \right) \left( \vec v \right)
Now let's make our mg substitution. And while we're at it, let's switch over to differential equation notation with the dots indicating time-based derivatives.
mg = m \ddot x + \dot m \dot x
The next step is to make appropriate substitutions to represent the differential equation with the variables given in the problem statement. Or at least change them to make them jive with the defintions given by your book's solution, of the rope has length
L, mass
M, and the length of the rope hanging through the hole is
x.
m = M\frac{x}{L}
\dot m = M\frac{ \dot x}{L}
which gives, after a little simplification,
xg = x \ddot x + \left( \dot x \right)^2
The solution to which is
x = \frac{1}{2} \left( \frac{g}{3} \right) t^2
I'll skip the differential equation solution process, But you can check it yourself to make sure it fits. Take the derivative to find
v and another derivative to find
a.
\dot x = v = \frac{g}{3}t
\ddot x = a = \frac{g}{3}
Plug those functions into the differential equation and you can see that the solution is valid. [Edit: and you can check that it satisfies our initial conditions of
x = 0 and
v = 0 when
t = 0.]
The next step is to solve for the time
T it takes for the rope to fall through a distance
L (substitute in
L for
x and
T for
t and solve for
T)
.
T = \sqrt{\frac{6L}{g}}
Plop that into the velocity equation and we finish with
v = \sqrt{\frac{2gL}{3}}
And there ya' have it.
This derivation did not use any conservation laws. I'd be interested to hear the conservation of momentum idea. Perhaps model the energy loss coming from an infinite number of infinitesimal inelastic collisions (causing each small section of rope to jump from 0 to a finite speed almost instantaneously, like what happens with inelastic collisions).
(Again, normally such a detailed solution [by someone other than the OP] as given above is in violation of the forum rules. But in this case the OP's question was about a solution given in a textbook, and that given solution was incorrect on several counts. I don't think it's inappropriate in this case to provide an alternative.)
