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Chapter 7: Problem 58

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 \(\mathrm{kg}\) clings to one end of the stick, and a mouse with mass 0.200 \(\mathrm{kg}\) clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, whit are their speeds as the rod swings through a vertical position?

Short Answer

Expert verified
The speed of both the rat and the mouse as the rod swings through the vertical position is approximately 2.18 m/s.

Step by step solution

01

Identify the System

The wooden rod can be modeled as a rigid body with the two animals at its ends. The axis of rotation passes through the center of the rod.
02

Define Variables

Let mass of the rat, \(m_1 = 0.500\; \mathrm{kg}\) and mass of the mouse, \(m_2 = 0.200\; \mathrm{kg}\). The length of the rod is \(L = 0.800\; \mathrm{m}\) and half-length is \(L/2 = 0.400\; \mathrm{m}\).
03

Conservation of Energy Principle

As the system is released from rest and swings to the vertical position, we can use conservation of mechanical energy. Initial potential energy will be converted into kinetic energy at the lowest point.
04

Calculate Initial Potential Energy

Each animal has gravitational potential energy which is \(U = mgh\). The potential energy for the rat is \(U_1 = m_1 g \frac{L}{2}\) and for the mouse is \(U_2 = m_2 g \frac{L}{2}\). Total potential energy is \(U_{total} = U_1 + U_2\).
05

Total Initial Potential Energy

Substituting the values, \(U_1 = 0.500 \cdot 9.8 \cdot 0.4\) and \(U_2 = 0.200 \cdot 9.8 \cdot 0.4\). Calculate \(U_{total} = U_1 + U_2\).
06

Calculate Final Kinetic Energy

When the rod is vertical, the potential energy is zero and the total energy is converted into kinetic energy. The kinetic energy of the system is \(K = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\). Since \(v_1 = v_2 = v\) at this point, we have \(K = \frac{1}{2}(m_1 + m_2) v^2\).
07

Equate Energy and Solve for Velocity

Set the final kinetic energy equal to the initial potential energy: \(\frac{1}{2}(m_1 + m_2) v^2 = U_{total}\). Solve for \(v\).
08

Perform Calculations

Calculate \(U_{total}\) from Step 5, and use it in the equation from Step 7 to find \(v\). Substituting the masses, solve for \(v^2\) and finally take the square root to find \(v\).\[ v = \sqrt{\frac{U_{total}}{\frac{1}{2}(m_1 + m_2)}} \]
09

Finalize and Verify

Plug in the calculated value of \(U_{total}\) and the given masses, perform the arithmetic carefully to find the final values of \(v\) for both the rat and the mouse.

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Key Concepts

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

Rotational Motion
Rotational motion refers to the movement of an object around a fixed point or axis. In this exercise, the wooden rod is rotating around its central axis. During such motion, different parts of the object travel at different speeds depending on their distance from the axis. Understanding rotational motion involves concepts like angular velocity and angular displacement.

In this scenario, the wooden rod with animals at both ends represents a rotating system where each end travels in a circular path. The speed of the rat and mouse will be determined by how fast the rod rotates as they cling to the ends. The angular velocity of the rod tends to increase as it falls due to gravity. This showcases how rotational motion influences the linear speed of the animals as the system swings from horizontal to vertical.
Rigid Body Dynamics
A rigid body refers to an idealized solid object where the distance between any two particles remains constant during motion. In this exercise, the rod serves as a rigid body. The masses at each end remain fixed relative to the rod's center.

Rigid body dynamics helps in understanding how the forces and torques act on the rod as the overall system moves. When the rod is released, gravitational forces act on both the rat and the mouse. These forces induce a torque, causing the rod to rotate around its center axis. The dynamics of this motion are significantly affected by the position and mass of the two animals.
  • This concept simplifies the calculation of motion by treating the rod and the attached animals as one solid entity, rather than focusing on individual components.
  • Thus, it allows for predicting how the system will behave once released from a given position.
Mechanical Energy
Mechanical energy is a crucial concept that combines both potential and kinetic energy within a physical system. In this scenario, the system's total mechanical energy is conserved when it transitions from a horizontal to a vertical position.

Initially, the system has a specific amount of gravitational potential energy due to the elevated position of the rod. As the rod swings downward, this potential energy is converted into kinetic energy. The total mechanical energy remains constant, provided no external forces like air resistance interfere. This is known as the conservation of mechanical energy.
  • This principle enables us to calculate the speeds of the rat and the mouse when the rod is vertical, by equating the initial potential energy with the kinetic energy at the lowest point.
  • It's a powerful tool for understanding how energy transforms within a system during motion.
Potential and Kinetic Energy
Potential energy and kinetic energy are two fundamental types of energy in mechanics. Potential energy is stored energy based on an object's position relative to a force, such as gravity. In the beginning, the animals on the rod possess gravitational potential energy determined by their height above the ground and their masses.

Kinetic energy, on the other hand, is the energy of motion. As the rod moves towards the vertical position, the system's potential energy becomes kinetic energy. The kinetic energy is greatest at the lowest part of the swing when the potential energy is zero.
  • The initial potential energy of the system is computed using the equation: \( U = mgh \), where \( h \) is the height of the masses from the pivot.
  • As the system swings, every joule of potential energy is converted into joules of kinetic energy, following the conservation of energy.
Understanding these concepts aids in predicting the speed of the animals as the rod completes its motion.

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Most popular questions from this chapter

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\) (a) Show that \(U(x)\) can be written as $$ U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$ Graph \(U(x)\) . Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x),\) the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x)\) . Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0} \text { and } x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

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