How Will the Body Move When the Monkey Climbs the Rope?

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The discussion revolves around a physics problem involving a monkey climbing a rope and a body attached to the other end of the rope. Initially, both the monkey and the body are in balance due to equal masses. As the monkey climbs with speed V relative to the rope, the body moves downward, and the analysis shows that both the monkey and the body will have an upward speed of V/2 relative to the ground. The solution involves applying Newton's second law and tension calculations, and the poster seeks confirmation on the correctness of their approach and whether a more efficient solution exists. The conversation highlights the complexities of motion and forces in a pulley system.
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Homework Statement



Here is the drawing which shows the situation:

http://img260.imageshack.us/my.php?image=monkeygn4.jpg

The disc is weightless. There is a rope which is rolled over the disc. A monkey is fastened at the point A. At the point B there is a body bound to rope. The mass of the monkey is equal to the mass of the body, so at the beginning the system is in balance. The monkey starts to move up the rope with speed v in respect of the rope. How will the body (which is bound at the point B) move while the monkey will be climbing up the rope?


The Attempt at a Solution



The monkey and rope are initially at rest, but then the monkey starts moving along the rope at speed V (relative to rope). It doesn't say how long it takes the monkey to reach that speed from rest, but let's call that time "t".

It also doesn't say whether the rope moves or stays stationary while this acceleration is taking place. But let's assume that the rope moves down (on the monkey's side) at speed V_r, while the monkey moves up (relative to the ground) at speed V_m. The monkey's speed V along the rope, is a combination of those two speeds: V = V_m + V_r.

So, the monkey accelerates upward at rate of V_m/t. Also, the only forces acting on the monkey are the rope's tension T, and the monkey's weight mg. So, by Newton's 2nd Law:

Fnet = m*a

T - mg = m*V_m/t

or:

T = m*V_m/t + mg

or:

T = m(V - V_r)/t + mg

Now, the object on the other side of the rope feels this same tension T at the same time. Therefore, its upward acceleration is:

A_o = (T - mg) / m

= (m(V - V_r)/t + mg - mg) / m

= (V - V_r)/t

But we also know that the rope is going up (on the object's side) at speed V_r after t seconds. So another expression for the object's upward acceleration is:

A_o = V_r/t

combining this with the previous equation gives:

V - V_r = V_r

or:

V_r = V/2

Furthermore, since V = V_m + V_r, we have:

V_m = V - V_r = V - V/2 = V/2

So this means the monkey's upward speed (V_m) relative to the ground, is V/2. And the object's upward speed (V_r) is also V/2.

Well, it is what my brain have done to solve this problem. Isn't here any mistakes? And this solution is quite long. Maybe there is a shorter way to reach the answer without using parameters like t (maybe using angular momentum conservation law)?
 
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