Physics problem?

Question

Hello,

I need to solve one physics problem which is described below.

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?

I think that angular momentum conservation law should be used to solve this problem.

Thanks in advance.

Edward, you incorrectly understood my question. The rope can slip.

Answer 1

I don't think angular momentum plays a part--because the only thing that's rotating is the pulley; but it is weightless, hence has no angular momentum.

The monky 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 (V_m may be positive, zero, or negative; we don't know yet). 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.

This means the monkey and the object will rise at the same rate, and always stay level with each other. Note that doesn't matter how much V is; so whether the monkey climbs fast or slow, the object will always rise (or fall) so as to stay level with the monkey.

Answer 2

Yes you are on the right track
I'll assume that the rope was thrown over the pulley and does not slip as monkey is climbing that rope. The mass of the monkey is counteracted my the inertia of the pulley. If so

Here are the basic eequations
W=mg
T=I A
where
I=0.5 m R^2
A= (V/R)/t

F=W-T/R=ma
m-mass of the monkey
a -acceleration of the monkey