Chain falling on weighing scale-find total reading

[hale2bopp]

Homework Statement

A chain of length l, mass M falls on a weighing scale vertically down. We need to find the reading of the scale when a length 'x' has fallen on the scale.

Homework Equations

F= dP/dt

W=Mg
dm= (M/l)x
dm/dt=m(v^2)/l

The Attempt at a Solution

So initially, I used this approach:
F=dP/dt.
P=Mv, where M is instantaneous mass of rope fallen on the scale.

So, dP/dt=Mdv/dt+vdM/dt.
dv/dt=g, as the rope falls under the influence of the rope.

dM/dt=mv^2/l
Since the centre of mass of the rope has fallen by an amount x/2, the velocity of that portion of rope will be (gx)^(1/2) using energy conservation.

So, F=2Mgx/l.
But this is only impulsive force. To this, we need to add the weight of the body.
So, F=3Mgx/l.

A few doubts I had with this approach:
1) Can we indeed conserve mechanical energy, since normal component of weight is acting upon the rope which hits the ground?
2)If we can, can we conserve the mechanical energy by saying that the change in potential energy of centre of mass of rope (Mgx/2) = (1/2)Mv^2? (This is because, the velocity that you get by this = root (gx). If we just say since the body falls under influence of gravity, it would be root (2gx).
3) And this is the most important bit. When we say that F = M dv/dt + vdm/dt, does this include the weight of the object, since dv/dt=g?

(i.e. is the impulsive force of fall ONLY due to the changing mass or also due to changing velocity?)
Suppose we have an object which has constant mass. We keep it on the table, without any velocity. IT exerts a force=Mg on the table.
Now suppose we take it to some height and leave it. It hits the table with some impulsive force greater than Mg. But, dm/dt is 0 and dv/dt is g, since it is free fall. So where does this impulsive force come from? Is it separate from the weight of the object that it would normally exert on the table??
Another approach: v=root (2gx) for falling body, dv/dt is not considered in the impulsive bit. Therefore, F=mv^2/l, and F=@Mgx/l. To this we add weight. So total reading = 3Mgx/l.

Which of these approaches is correct, and why?

 

[anathema]

I would approach this problem by first making some assumptions and simplifications to better understand the problem. For example, I would assume that the chain is falling in a vacuum, so there is no air resistance or other external forces acting on it. I would also simplify the chain to be a point mass at the center of mass, as this will make the calculations easier.

Using the equation F=ma, we can find the acceleration of the chain as it falls. Since the only force acting on the chain is its weight, we can say that F=mg, where g is the acceleration due to gravity. Therefore, the acceleration of the chain is g.

Next, we can use the equation v^2 = u^2 + 2as to find the velocity of the chain at any given point. In this case, we know the initial velocity is 0, and the acceleration is g. So the equation becomes v^2 = 2gx, where x is the distance the chain has fallen.

Now, we can use the equation F=ma again to find the force acting on the weighing scale. Since we are only interested in the force at a specific point, we can use the instantaneous mass at that point. So the equation becomes F=mg(dm/dt), where dm/dt is the instantaneous mass of the chain at that point, which can be calculated using the equation dm/dt = (M/l)x.

Combining all of these equations, we get F=mg(M/l)x. This is the force acting on the weighing scale at a specific point. To find the reading of the scale, we simply divide this force by the acceleration due to gravity, so the reading would be (M/l)x.

To address your doubts:

1) In this simplified scenario, we can ignore the change in potential energy of the chain as it falls, as it is negligible compared to the kinetic energy. Therefore, conservation of mechanical energy can be applied.

2) We can use the equation v^2 = u^2 + 2as to find the velocity of the center of mass of the chain, which would be (gx)^(1/2). However, this is not the same as the velocity of a specific point on the chain, which would be (2gx)^(1/2). This is because as the chain falls, different points on the chain will have different velocities, depending on their distance from the center of mass.