/*! 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 61 Each of the gears A and B has a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Each of the gears A and B has a mass of 675 g and a radius of gyration of 40 mm, while gear C has a mass of 3.6 kg and a radius of gyration of 100 mm. Assume that kinetic friction in the bearings of gears A, B, and C produces couples of constant magnitude 0.15 N?m, 0.15 N?m, and 0.3 N?m, respectively. Knowing that the initial angular velocity of gear C is 2000 rpm, determine the time required for the system to come to rest.

Short Answer

Expert verified
The system will take approximately 13.32 seconds to come to rest.

Step by step solution

01

Convert Units

We begin by converting the mass from grams to kilograms and the radius of gyration from millimeters to meters for gears A, B, and C. Also, convert the initial angular velocity from revolutions per minute (rpm) to radians per second (rad/s) for gear C.Mass of gears A and B: 675 g = 0.675 kgRadius of gyration for gears A and B: 40 mm = 0.04 mMass of gear C: 3.6 kg Radius of gyration for gear C: 100 mm = 0.1 mInitial angular velocity of gear C:2000 rpm = \(2000 \times \frac{2\pi}{60} = \frac{2000 \times \pi}{30} \approx 209.44 \text{ rad/s}\)
02

Calculate Moment of Inertia for Each Gear

Use the formula for the moment of inertia, \( I = m k^2 \), where \( m \) is the mass, and \( k \) is the radius of gyration for each gear.Moment of inertia for gears A and B:\( I_A = 0.675 \times (0.04)^2 = 0.00108 \text{ kg·m}^2 \)Moment of inertia for gear C:\( I_C = 3.6 \times (0.1)^2 = 0.036 \text{ kg·m}^2 \)
03

Sum the Moments and Calculate Angular Deceleration

The total decelerating moment is due to the sum of the frictional forces acting on each gear:Total frictional moment \( \sum M = 0.15 + 0.15 + 0.3 = 0.6 \text{ N·m} \)Calculate the angular deceleration using \( \alpha = \frac{\sum M}{I_{total}} \).Since the gears are coupled, the system as a whole is decelerating:Total moment of inertia of the system \( I_{total} = 2 \cdot 0.00108 + 0.036 = 0.03816 \text{ kg·m}^2 \)Angular deceleration:\( \alpha = \frac{0.6}{0.03816} \approx 15.72 \text{ rad/s}^2 \)
04

Determine the Time to Come to Rest

Use the kinematic equation \( \omega_f = \omega_i + \alpha t \), where \( \omega_f = 0 \) (final angular velocity), \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular deceleration, and \( t \) is time.Solve for \( t \):\( 0 = 209.44 - 15.72 t \)\( t = \frac{209.44}{15.72} \approx 13.32 \text{ seconds} \)

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.

Moment of Inertia
The moment of inertia is a crucial concept in gear mechanics as it measures an object's resistance to rotational motion. Just like mass affects linear motion, moment of inertia affects rotational motion and depends on both mass and the distribution of that mass relative to the axis it rotates around.

It is represented mathematically by the formula \( I = m k^2 \), where \( m \) is the mass and \( k \) is the radius of gyration. The radius of gyration is a hypothetical radius at which the mass of a body can be thought concentrated for rotational purposes. This makes moment of inertia dependent both on the size and the weight of the object.

In this exercise, we calculated the individual moments of inertia for each gear based on their mass and radius of gyration. Knowing these helps in understanding how each gear contributes to the overall rotational dynamics of the system.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around a center or a specific point. In this context, it refers to gears' rotational speed.

It is often expressed in revolutions per minute (rpm) or radians per second (rad/s). In our problem, we converted the initial angular velocity of gear C from 2000 rpm to approximately 209.44 rad/s in order to use it in subsequent calculations efficiently. Conversions from rpm to rad/s involve multiplying by \( \frac{2\pi}{60} \), which adjusts for the gear's rotational nature relative to units in seconds.

Understanding how to handle angular velocity is vital, as it interacts with other factors like moment of inertia to influence the system's overall dynamics.
Kinetic Friction
Kinetic friction is the force that opposes the relative motion between two surfaces in contact while they move over each other. It is crucial in gear mechanics as it can significantly impact the movement and efficiency of a system.

In this exercise, the friction within the gear bearings was accounted for by introducing frictional couples, which are torque values of 0.15 N·m for gears A and B and 0.3 N·m for gear C. These frictional forces worked against the gears' movement, causing the system to decelerate.
  • Important to note, the greater the kinetic friction, the greater the resistance to motion.
  • It not only causes wear and tear over time but also plays a pivotal role in energy losses within a mechanical system.
Accounting for such oppositional forces is key in accurately calculating the system's angular deceleration.
Angular Deceleration
Angular deceleration is the rate at which an object slows its rotation over time. It's essentially the opposite of angular acceleration.

In this context, determining angular deceleration involves the frictional torques acting against the initial angular velocity. Using the formula \( \alpha = \frac{\sum M}{I_{total}} \), where \( \alpha \) is the angular deceleration, \( \sum M \) is the sum of the frictional torques, and \( I_{total} \) is the total moment of inertia, we determined the system's rate of slowing.

The negative value, approximately \( 15.72 \text{ rad/s}^2 \), tells us that the system is losing speed at this rate. This insight is crucial for understanding how long it takes for the system to come to rest. It underscores the integral relation between moment of inertia, frictional forces, and movement resistance.

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

The \(4-\mathrm{kg}\) rod \(A B\) can slide freely inside the 6 -kg tube. The rod was entirely within the tube \((x=0)\) and released with no initial velocity relative to the tube when the angular velocity of the assembly was 5 rad/s. Neglecting the effect of friction, determine the speed of the rod relative to the tube when \(x=400 \mathrm{mm} .\)

A semicircular panel with a radius r is attached with hinges to a circular plate with a radius r and initially is held in the vertical position as shown. The plate and the panel are made of the same material and have the same thickness. Knowing that the entire assembly is rotating freely with an initial angular velocity of \(\omega_{0}\), determine the angular velocity of the assembly after the panel has been released and comes to rest against the plate.

Member \(A B C\) has a mass of \(2.4 \mathrm{kg}\) and is attached to a pin support at \(B\). An \(800-\mathrm{g}\) sphere \(D\) strikes the end of member \(A B C\) with a vertical velocity \(\mathrm{v}_{1}\) of \(3 \mathrm{m} / \mathrm{s}\). Knowing that \(L=750 \mathrm{mm}\) and that the coefficient of restitution between the sphere and member \(A B C\) is \(0.5,\) determine immediately after the impact \((a)\) the angular velocity of member \(A B C,(b)\) the velocity of the sphere.

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 110-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. Knowing that the kinetic friction of the rotor produces a couple with a magnitude of 2.5 lb?ft, determine the number of revolutions that the rotor executes before coming to rest.

A large 3 -lb sphere with a radius \(r=3\) in. is thrown into a light basket at the end of a thin, uniform rod weighing 2 lb and length \(L=10\) in. as shown. Immediately before the impact, the angular velocity of the rod is \(3 \mathrm{rad} / \mathrm{s}\) counterclockwise and the velocity of the sphere is \(2 \mathrm{ft} / \mathrm{s}\) down. Assume the sphere sticks in the basket. Determine after the impact \((a)\) the angular velocity of the bar and sphere, \((b)\) the components of the reactions at \(A .\)

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.