/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A Texas cockroach of mass \(0.17... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Texas cockroach of mass \(0.17 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(15 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

Short Answer

Expert verified
(a) 7.4 rad/s. (b) No, mechanical energy is not conserved.

Step by step solution

01

Define Variables Based on the Problem Statement

Firstly, we need to identify the given variables in the problem. We are given:- Mass of the cockroach, \( m = 0.17 \text{ kg} \).- Radius of the lazy Susan, \( R = 0.15 \text{ m} \).- Rotational inertia of the lazy Susan, \( I = 5.0 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \).- Speed of the cockroach, \( v = 2.0 \text{ m/s} \).- Initial angular speed of the lazy Susan, \( \omega_0 = 2.8 \text{ rad/s} \).
02

Calculate Angular Momentum of the Cockroach

The cockroach's initial angular momentum is calculated about the axis of the lazy Susan.The formula for angular momentum, \( L \), is \( L = m \cdot v \cdot R \).Substituting in the given values:\[ L = 0.17 \cdot 2.0 \cdot 0.15 = 0.051 \text{ kg} \cdot \text{m}^2/s \].Since the cockroach runs counterclockwise and the lazy Susan turns clockwise, we'll assign the cockroach's angular momentum a positive value by convention.
03

Calculate Total Initial Angular Momentum

The initial angular momentum of the system (cockroach + lazy Susan) is the sum of their individual angular momenta.Lazy Susan's initial angular momentum is:\[ L_{susan} = I \cdot (-\omega_0) = 5.0 \times 10^{-3} \cdot (-2.8) = -0.014 \text{ kg} \cdot \text{m}^2/s \].Total initial angular momentum:\[ L_{total} = L + L_{susan} = 0.051 - 0.014 = 0.037 \text{ kg} \cdot \text{m}^2/s \].
04

Apply Conservation of Angular Momentum

Since no external torques are acting on the system, the angular momentum is conserved.Thus, the final angular momentum of the lazy Susan (after the cockroach stops) must equal the initial total angular momentum of the system.Let \( \omega' \) be the final angular speed of the lazy Susan. Therefore:\[ I \cdot \omega' = L_{total} \].
05

Solve for the Final Angular Speed

Rearranging for \( \omega' \):\[ \omega' = \frac{L_{total}}{I} = \frac{0.037}{5.0 \times 10^{-3}} = 7.4 \text{ rad/s} \].Thus, the angular speed of the lazy Susan after the cockroach stops is \( 7.4 \text{ rad/s} \).
06

Determine Mechanical Energy Conservation

Check if mechanical energy is conserved by comparing initial and final mechanical energy.Initial kinetic energy:\[ KE_{initial} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega_0^2 = \frac{1}{2} \times 0.17 \times 2^2 + \frac{1}{2} \times 5.0 \times 10^{-3} \times 2.8^2 = 0.34 + 0.0196 = 0.3596 \text{ J} \].Final kinetic energy (only the lazy Susan rotates):\[ KE_{final} = \frac{1}{2} I \omega'^2 = \frac{1}{2} \times 5.0 \times 10^{-3} \times (7.4)^2 = 0.1369 \text{ J} \]. Since initial kinetic energy (0.3596 J) is greater than final kinetic energy (0.1369 J), mechanical energy is not conserved due to transformation to other forms of energy (e.g., thermal due to the cockroach stopping).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rotational Inertia
Rotational inertia, sometimes called the moment of inertia, is a concept that reflects how much an object resists being rotated. It's like the rotational equivalent of mass. The larger the rotational inertia, the harder it is to change the object's rotational speed. For instance, in the problem, the lazy Susan has a rotational inertia of \(5.0 \times 10^{-3} \text{ kg} \cdot \text{m}^2\).
This inertia depends on both the mass of the lazy Susan and how this mass is distributed about the axis of rotation. The further the mass is from the axis, the greater the rotational inertia.
  • It's crucial for understanding how easily a body can rotate.
  • Different shapes have different rotational inertias, even if they have the same mass and size.
  • Think of a door: it’s harder to open if you push near the hinge because the mass (your hand force) is closer to the rotational axis.
Understanding rotational inertia helps explain why unbalanced washing machines shake vigorously – the clothes form an uneven distribution of mass! Increasing the distance of mass from the axis increases the rotational inertia, contributing to the motion's resistance.
Conservation of Angular Momentum
The principle of conservation of angular momentum states that if no external torques act on a system, the total angular momentum remains constant. Angular momentum is defined as \( L = I \cdot \omega \), where \( I \) is the rotational inertia and \( \omega \) is the angular velocity.
In our exercise, the system comprises the lazy Susan and the cockroach. Initially, both have angular momentum, with the cockroach's contribution being \( m \cdot v \cdot R \). The lazy Susan's initial angular momentum is calculated using \( I \cdot (-\omega_0)\).
  • No external forces mean the total angular momentum before the cockroach stops must equal the total after.
  • This conservation principle helps us deduce the lazy Susan’s final speed once the cockroach halts.
This principle is what keeps figure skaters spinning fast when they pull in their arms; they're simply reducing their rotational inertia to increase their angular velocity to conserve angular momentum!
Mechanical Energy Conservation
Mechanical energy conservation focuses on the sum of kinetic and potential energies in a system remaining constant, assuming no losses occur due to non-conservative forces such as friction or air resistance. Kinetic energy is given by \( \frac{1}{2} I \omega^2 \) for rotational motions.
In the original exercise, mechanical energy isn’t conserved. Initially, both the cockroach and the lazy Susan have kinetic energy. When the cockroach stops, its energy doesn't all transfer to the lazy Susan; instead, some dissipates as heat or sound due to non-conservative processes.
  • Initial kinetic energy was higher than the final energy, showing energy left the system.
  • This energy loss often happens in real-world scenarios.
  • It's crucial to understand energy transformation to accurately predict system behavior.
While we might usually hope for energy to remain in mechanical form, transformations to other forms are common and impact many everyday activities, from driving cars to spinning carnival rides.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks \(A\) and \(C\). Another belt runs around a central hub on disk \(A\) and the rim of disk \(B .\) The belts move smoothly without slippage on the rims and hub. Disk \(A\) has radius \(R ;\) its hub has radius \(0.5000 R ;\) disk \(B\) has radius \(0.2500 R ;\) and disk \(C\) has radius \(2.000 R .\) Disks \(B\) and \(C\) have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk \(C\) to that of disk \(B\) ?

A solid sphere of weight \(36.0 \mathrm{~N}\) rolls up an incline at an angle of \(30.0^{\circ}\). At the bottom of the incline the center of mass of the sphere has a translational speed of \(4.90 \mathrm{~m} / \mathrm{s}\). (a) What is the kinetic energy of the sphere at the bottom of the incline? (b) How far does the sphere travel up along the incline? (c) Does the answer to (b) depend on the sphere's mass?

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm}\). Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the \((\mathrm{g})\) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, (k) top, and (1) bottom of each tire?

Is an overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3.00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p}\) ?

At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has the position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity is given by \(\vec{v}=\left(-6.0 t^{2} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}\) for \(t \geq 0\) in seconds. About the origin, what are (a) the particle's angular momentum \(\vec{L}\) and \((\mathrm{b})\) the torque \(\vec{\tau}\) acting on the particle, both in unit-vector notation and for \(t>0 ?\) About the point \((-2.0 \mathrm{~m},-3.0 \mathrm{~m}, 0)\), what are (c) \(\vec{L}\) and (d) \(\vec{\tau}\) for \(t>0 ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.