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Chapter 8: Problem 64

A uniform log of length \(2.50 \mathrm{~m}\) has a mass of \(91.0 \mathrm{~kg}\) and is floating in water. Standing on this log is a \(72.0-\mathrm{kg}\) man, located \(22.0 \mathrm{~cm}\) from one end. On the other end is his daughter \((m=20.0 \mathrm{~kg})\), standing \(1.00 \mathrm{~m}\) from the end. a) Find the center of mass of this system. b) If the father jumps off the log backward away from his daughter \((v=3.14 \mathrm{~m} / \mathrm{s}),\) what is the initial speed of \(\log\) and child?

Short Answer

Expert verified
Answer: The center of mass of the system is approximately 0.872 m from the left end of the log. The initial speed of the log and the child is approximately -2.04 m/s, moving in the opposite direction of the father's jump.

Step by step solution

01

Determine the center of mass of the system for part (a)

To find the center of mass of the system, we can consider the log, the man, and the child as individual masses at specific locations. Let the left end of the log be the origin (0). The position of the man is 0.22 m from the origin, the position of the daughter is at 2.50 m - 1.00 m = 1.50 m from the origin, and the center of mass of the log lies in its middle, i.e., at 1.25 m. So, the equation for the center of mass is given by: \(x_{cm} = \frac{m_{man}x_{man} + m_{daughter}x_{daughter} + m_{log}x_{log}}{m_{man} + m_{daughter} + m_{log}}\)
02

Calculate the center of mass

Now, we can plug in the given values into the formula: \(x_{cm} = \frac{(72)(0.22) + (20)(1.5) + (91)(1.25)}{72 + 20 + 91}\) \(x_{cm} = \frac{15.84 + 30 + 113.75}{183}\) \(x_{cm} = \frac{159.59}{183}\) \(x_{cm} \approx 0.872~m\) The center of mass is approximately at 0.872 m from the origin, which is the left end of the log.
03

Set up the equation for conservation of momentum for part (b)

When the father jumps off the log, the total momentum of the system is conserved. Before the jump, the total momentum is 0. After the jump, the father, child, and log will have their own existing momenta. So, we can write the equation for conservation of momentum as: \(m_{man}v_{man} + m_{daughter}v_{daughter} + m_{log}v_{log} = 0\) where \(v_{man}\), \(v_{daughter}\), and \(v_{log}\) are the velocities of man, daughter, and log, respectively, after the jump. Since both the daughter and the log would have the same initial velocity, we can rewrite the equation as: \(-m_{man}v_{man} = (m_{daughter} + m_{log})v\) where v represents the initial speed of the log and the child.
04

Calculate the initial speed of the log and the child

Now we can plug in the given values into the momentum equation: \(-(72)(3.14) = (20+91)v\) Solving for v, we get: \(v = \frac{-226.08}{111}\) \(v\approx -2.04~m/s\) The initial speed of the log and child is approximately -2.04 m/s. The negative sign implies that they will move in the opposite direction of the father's jump.

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is foundational in physics. It suggests that in a closed system, the total momentum before any event is equal to the total momentum after the event, provided no external forces interfere. This principle is elegantly applied in the scenario where the father jumps off the log.

In the exercise, right before the father leaps, the system鈥攃omposed of the man, the daughter, and the log鈥攈as no net external momentum (assuming negligible resistance from the water). Consequently, when the father jumps backward, his momentum must equal in magnitude and oppose the momentum gained by the daughter and the log. This is described by the equation:
  • \(m_{man}v_{man} + m_{daughter}v_{daughter} + m_{log}v_{log} = 0\)
  • Rearranging gives: \(-m_{man}v_{man} = (m_{daughter} + m_{log})v\)
This equation enables us to solve for the velocity of the log and daughter, showing how interconnected forces are balanced within the system.
Physics Problem Solving
When tackling physics problems, especially involving rigid bodies and other dynamic systems, a methodical approach is crucial. Start by identifying known quantities and relationships, then work systematically to apply the right principles.

Take, for example, the calculation of the center of mass. The location of each component (the man, the daughter, and the log) is crucial for determining the center of mass position. By choosing the origin鈥攕uch as one end of the log鈥攜ou can use the formula:
  • \(x_{cm} = \frac{m_{man}x_{man} + m_{daughter}x_{daughter} + m_{log}x_{log}}{m_{man} + m_{daughter} + m_{log}}\)
This formula allows you to weight each position by the respective mass, giving the overall center of mass position. It's a pastiche of physics principles applied in a systematic way, showcasing both calculation and conceptual understanding.
Dynamics of Rigid Bodies
Understanding the dynamics of rigid bodies involves analyzing how these bodies move and interact under the influence of forces and torques. In the exercise scenario, the log can be considered a rigid body, not deforming under the man's or daughter's weight.

Crucial to problems involving rigid bodies is the identification of pivot points, force vectors, and how these affect motion. The log, being a uniform material, has its center of mass right at 1.25 m, halfway along its length, which simplifies calculations. This center isn't just a mere mathematical necessity but also the point where external forces, like gravity, effectively act, making it crucial for balancing calculations.
  • Consider how the father's jump influences the motion of the log and his daughter鈥攏ot only through linear momentum but potential rotational dynamics as well.
  • The absence of external torques (assuming calm waters) in the initial setup ensures that our focus remains solely on linear dynamics.
This approach highlights a more tangible understanding of how rigid bodies behave in a practical context.

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Most popular questions from this chapter

The USS Montana is a massive battleship with a weight of \(136,634,000 \mathrm{lb}\). It has twelve 16 -inch guns, which are capable of firing 2700 -lb projectiles at a speed of \(2300 \mathrm{ft} / \mathrm{s}\). If the battleship fires three of these guns (in the same direction), what is the recoil velocity of the ship?

Can the center of mass of an object be located at a point outside the object, that is, at a point in space where no part of the object is located? Explain.

The Saturn \(V\) rocket, which was used to launch the Apollo spacecraft on their way to the Moon, has an initial mass \(M_{0}=2.80 \cdot 10^{6} \mathrm{~kg}\) and a final mass \(M_{1}=8.00 \cdot 10^{5} \mathrm{~kg}\) and burns fuel at a constant rate for \(160 .\) s. The speed of the exhaust relative to the rocket is about \(v=2700 . \mathrm{m} / \mathrm{s}\). a) Find the upward acceleration of the rocket, as it lifts off the launch pad (while its mass is the initial mass). b) Find the upward acceleration of the rocket, just as it finishes burning its fuel (when its mass is the final mass). c) If the same rocket were fired in deep space, where there is negligible gravitational force, what would be the net change in the speed of the rocket during the time it was burning fuel?

A baseball player uses a bat with mass \(m_{\text {bat }}\) to hit a ball with mass \(m_{\text {ball }}\). Right before he hits the ball, the bat's initial velocity is \(35.0 \mathrm{~m} / \mathrm{s}\), and the ball's initial velocity is \(-30.0 \mathrm{~m} / \mathrm{s}\) (the positive direction is along the positive \(x\) -axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that \(m_{\text {bat }}\) is much greater than \(m_{\text {ball }}\), so the center of mass of the two objects is essentially at the bat.

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