Question

The balloon, the light rope and the monkey shown in the figure (9-E7) are at rest in the air. If the monkey reaches the top of the rope, by what distance does the balloon descend? Mass of the balloon = M,mass of the monkey = m and the length of the rope ascended by the monkey = L.

Answer 1

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ANSWER::

Monkey and Balloon are at rest initially.

Centre of mass is at 'P'.

Monkey descends through a distance 'L'

Shifted centre of mass (t₀) = (m x L + M x 0) / M + m = mL / (M+m)

Therefore , the balloon descends through a distance mL / (M+m) from P

Hope it helps!

Answer 2

Explanation:

In the beginning, monkey and balloon are at rest.

The balloon's mass = M,  

A monkey's mass = m  

and  

The length of the rope the monkey climbed = L.

The mass center is at ' P. '

Monkey walks down a distance ' L '

Mass center shift (t0) = $\frac{(m \times L+M \times 0)}{M+m}$

                                   = $\frac{m L}{M+m}$

Consequently, the balloon goes down a distance $\left(\frac{m L}{M+m}\right)$ from P .