A balloon of mass M with a light rope and a m | vedclass

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(A) Since there is no external force acting on the system (balloon + monkey) in the vertical direction,the position of the center of mass $(COM)$ of t

Vedclass · Accepted Answer

$\frac{mL}{m + M}$

Vedclass · Answer

(A) Since there is no external force acting on the system (balloon + monkey) in the vertical direction,the position of the center of mass $(COM)$ of the system remains unchanged.Let the balloon descend by a distance $y$. As the monkey climbs the rope of length $L$,the monkey moves upward by a distance $(L - y)$ relative to the ground.Taking the initial position of the $COM$ as the reference point,the condition for the $COM$ to remain stationary is:$M \cdot y = m \cdot (L - y)$$My = mL - my$$My + my = mL$$y(M + m) = mL$$y = \frac{mL}{M + m}$Therefore,the distance by which the balloon descends is $\frac{mL}{M + m}$.

$A$ balloon of mass $M$ with a light rope and a monkey of mass $m$ are at rest in mid-air. If the monkey climbs up the rope and reaches the top of the rope,the distance by which the balloon descends will be (Total length of the rope is $L$):

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