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Chapter 8: Problem 24

A man of mass \(m\) clings to a rope ladder suspended below a balloon of mass \(M\); see Fig. \(8-25\). The balloon is stationary with respect to the ground. (a) If the man begins to climb the ladder at speed \(v\) (with respect to the ladder), in what direction and with what speed (with respect to the ground) will the balloon move? (b) What is the state of the motion after the man stops climbing?

Short Answer

Expert verified
(a) Balloon moves at \(\frac{m v}{M}\) in the opposite direction. (b) Both will be stationary once the man stops climbing.

Step by step solution

01

- Understand the Scenario

To solve this problem, start by understanding that both the balloon and the man are initially stationary. When the man begins to climb at speed \(v\) relative to the ladder, there will be an equal and opposite reaction on the balloon due to the conservation of momentum.
02

- Apply Conservation of Momentum

Since the system is initially at rest, the total initial momentum is zero. When the man climbs, he and the balloon must move in opposite directions such that the total momentum remains zero. Let the velocity of the balloon with respect to the ground be \(V\).
03

- Momentum Equation Setup

Set up the equation for the conservation of momentum: \[ m v + M(-V) = 0 \]
04

- Solve for Balloon's Velocity

Solve the equation for \(V\): \[ V = \frac{m v}{M} \]The balloon will move in the opposite direction to the man鈥檚 climb.
05

- State of Motion After Man Stops Climbing

After the man stops climbing, both the velocities of the man and the balloon will again become zero because momentum is conserved, and the system must return to a state of no motion relative to the ground.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balloon Dynamics
Understanding balloon dynamics is essential to solve this exercise. When dealing with a balloon, it鈥檚 important to consider how forces and motions affect it due to its ability to float in the air.
  • A balloon鈥檚 motion is largely influenced by external factors like wind and internal changes like weight shifts.
  • In this problem, the balloon is initially stationary, which means there鈥檚 a balance of forces acting on it.
  • As the man climbs the ladder, he exerts a force due to his movement, influencing the entire balloon system.
The important takeaway is that the climbing creates a reaction force which causes the balloon to move in the opposite direction relative to his motion on the ladder.
Hence, any movement by the man has an immediate and equal effect on the balloon, making it move in the opposite direction to conserve the total momentum of the system.
Relative Motion
Relative motion is the concept of movement in relation to another object. In this problem, the man鈥檚 climb speed is relative to the ladder, but we need to consider how it translates to motion relative to the ground.
  • The man climbs up with speed \( v \) relative to the ladder, but this speed affects the balloon鈥檚 movement as well.
  • By conservation of momentum, the speed of the balloon with respect to the ground must be calculated to maintain the overall zero initial momentum.
When we analyze relative motion, we consider both the velocity of the man and the balloon.
It鈥檚 crucial to understand that due to conservation of momentum, the balloon will move in the opposite direction to the man's climb at a speed determined by the ratio of their masses given by \[ V = \frac{m v}{M} \].
Stationary Systems
A stationary system means there鈥檚 no net movement relative to a reference point鈥攊n this case, the ground. At the start of the exercise, the balloon and the man are stationary.
  • This implies the total momentum of the system is zero.
  • Regardless of the forces within the system (man climbing), the momentum must remain conserved for the system to naturally return to its initial stationary state after external action ceases.
When the man climbs, although the balloon moves to keep momentum conserved, the system reaches a new temporary state.
Once the man stops climbing, the system (balloon plus man) will again come to rest, thereby maintaining an overall zero momentum state consistent with its being stationary relative to the ground.
It helps to remember that in a real-world scenario, the balloon鈥檚 movement would reflect all external and internal forces to stay dynamically balanced in the air.

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Most popular questions from this chapter

At a certain instant, four particles have the \(x y\) coordinates and velocities given in the following table. At that instant, what are (a) the coordinates of their center of mass and (b) the velocity of their center of mass? $$\begin{array}{cccc} \text { Particle } & \text { Mass (kg) } & \text { Position (m) } & \text { Velocity (m/s) } \\ \hline 1 & 2.0 & 0,3.0 & -9.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}} \\ 2 & 4.0 & 3.0,0 & 6.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{i}} \\ 3 & 3.0 & 0,-2.0 & 6.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}} \\ 4 & 12 & -1.0,0 & -2.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{i}} \\ \hline \end{array}$$

A \(2.00 \mathrm{~kg}\) block is released from rest over the side of a very tall building at time \(t_{1}=0 .\) At time \(t_{2}=\) \(1.00 \mathrm{~s}\), a \(3.00 \mathrm{~kg}\) block is released from rest at the same point. The first block hits the ground at \(t_{3}=5.00 \mathrm{~s}\). Plot, for the time interval \(t_{1}=0\) to \(t_{4}=6.00 \mathrm{~s}\), (a) the position and (b) the speed of the center of mass of the two-block system. Take \(y=0\) at the release point.

At \(t_{1}=0\), a \(1.0 \mathrm{~kg}\) jelly jar is projected vertically upward from the base of a 50 -m-tall building with an initial velocity of \(40 \mathrm{~m} / \mathrm{s}\). At the same instant and directly overhead, a \(2.0 \mathrm{~kg}\) peanut butter jar is dropped from rest from the top of the building. How far above ground level is the center of mass of the two-jar system at \(t_{2}=3.0 \mathrm{~s}\) ?

A stone is dropped at \(t_{1}=0 .\) A second stone, with twice the mass of the first, is dropped from the same point at \(t_{2}=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t_{3}=300 \mathrm{~ms}\) ? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

You are on an iceboat on frictionless, flat ice; you and the boat have a combined mass \(M\). Along with you are two stones with masses \(m_{A}\) and \(m_{B}\) such that \(M=6.00 m_{A}=12.0 m_{B}\). To get the boat moving, you throw the stones rearward, either in succession on together, but in each case with a certain speed \(v^{\text {rel }}\) relative to the boat after the stone is thrown. What is the resulting speed of the boat if you throw the stones (a) simultaneously, (b) in the order \(m_{A}\) and then \(m_{B}\), and (c) in the order \(m_{B}\) and then \(m_{A}\) ?
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