/*! 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 326 The spring of stiffness \(k\) is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 326

The spring of stiffness \(k\) is compressed and suddenly released, sending the particle of mass \(m\) sliding along the track. Determine the minimum spring compression \(\delta\) for which the particle will not lose contact with the loop-the-loop track. The sliding surface is smooth except for the rough portion of length \(s\) equal to \(R,\) where the coefficient of kinetic friction is \(\mu_{k}\).

Short Answer

Expert verified
The minimum spring compression \( \delta = \sqrt{\frac{5mgR}{k} + \frac{2\mu_k mgR}{k}} \).

Step by step solution

01

Analyzing the Forces on the Particle

At the top of the loop, the forces acting on the particle include the gravitational force \( mg \) and the normal force \( N \). For the particle to just not lose contact, the normal force needs to be zero, i.e., the gravitational force provides the necessary centripetal force. Therefore, the condition at the top of the loop is:\[ mg = \frac{mv^2}{R} \]where \( v \) is the speed of the particle at the top and \( R \) is the radius of the loop.
02

Calculating the Required Speed at the Top

From the equation derived in Step 1:\[ mg = \frac{mv^2}{R} \]we can solve for \( v \):\[ v = \sqrt{gR} \]This is the minimum speed the particle must have at the top of the loop to not lose contact.
03

Energy Conservation Considering Friction

The total mechanical energy at the beginning (spring potential energy) must equal the total mechanical energy at the top of the loop (kinetic and potential energy) plus the work done against friction. Initially, the energy is stored in the spring: \[ \frac{1}{2}k\delta^2 \]At the top of the loop, the kinetic energy is \( \frac{1}{2}mv^2 \), and the potential energy due to height \( 2R \) is \( 2mgR \). Therefore, according to energy conservation and work done by friction, we have:\[ \frac{1}{2}k\delta^2 = \frac{1}{2}m(\sqrt{gR})^2 + 2mgR + \mu_k mgR \]
04

Solving for the Minimum Spring Compression

By substituting \( v \) from Step 2 into the energy equation in Step 3, simplify:\[ \frac{1}{2}k\delta^2 = \frac{1}{2}mgR + 2mgR + \mu_k mgR \]Combine the terms on the right side to:\[ \frac{1}{2}k\delta^2 = \frac{5mgR}{2} + \mu_k mgR \]Now, solve for \( \delta \):\[ \delta = \sqrt{\frac{5mgR}{k} + \frac{2\mu_k mgR}{k}} \]This is the minimum compression needed to ensure the particle remains in contact with the track.

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 91影视!

Key Concepts

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

Spring Compression
Spring compression represents the initial stored energy that launches the particle. When a spring is compressed by a distance \( \delta \), it holds potential energy given by \( \frac{1}{2}k\delta^2 \), where \( k \) is the spring stiffness. This energy is crucial as it determines how much kinetic energy the particle will have once it is released.
  • Understand that a higher compression (\( \delta \)) results in more energy stored in the spring, which in turn increases the particle's speed upon release.
  • The challenge is to find the minimum \( \delta \) such that the particle maintains contact with the loop, not losing contact at its top.
The spring must be compressed sufficiently to not only overcome gravitational forces at the loop's top but also to compensate for energy losses due to friction. Proper spring compression ensures the particle remains on the track throughout its motion.
Energy Conservation
Energy conservation is a crucial concept here, relating the spring's stored energy to the particle's kinetic and potential energies as it moves. Before the particle moves, it holds potential energy in the spring. As the particle is released, this spring energy is transformed into kinetic and potential energy.
  • At the loop's peak, the particle should have sufficient kinetic energy to be sustained in motion, identified via \( \frac{1}{2}m(\sqrt{gR})^2 \).
  • Also, gravitational potential energy (\( 2mgR \)) is present due to the particle's height.
This system's total initial energy (from spring) must equal the sum of kinetic and potential energies at the top of the loop plus any energy lost to friction. This critical balance assures that energy conservation governs how the initial spring compression translates into other energy forms needed for successful loop navigation.
Kinetic Friction
Kinetic friction is the resistive force that acts against the motion of the particle over specific track sections. It is quantified by the force \( \mu_k mg R \), where \( \mu_k \) is the coefficient of kinetic friction.
  • Occurs over a rough portion of the track, which is crucial because energy lost here needs compensation through initial spring energy.
  • Acts to gradually reduce the particle's speed, thereby requiring additional energy input for completing the loop.
When computing the necessary spring compression, friction's work done must be included as part of the energy requirement. Correctly accounting for friction ensures that the overall energy correctly matches the loop's demands.
Loop-the-loop Track
A loop-the-loop track offers a circular path where the particle transitions from horizontal to vertical motion and back. At the circular loop's top, maintaining contact depends on achieving necessary velocity derived from gravitational force needs.
  • The particle's minimum velocity at the top is \( \sqrt{gR} \), balancing gravity and the required centripetal force.
  • This velocity ensures no slipping away or loss of contact, aligning with the centripetal force necessary for stable motion along a loop.
The loop-the-loop poses a dynamic challenge: ensuring velocity constancy sufficient to fulfill all centripetal needs while accounting for energy considerations. Achieving this ensures a successful loop traversal.

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 satellite of Sample Problem \(3 / 31\) has a perigee velocity of \(26140 \mathrm{km} / \mathrm{h}\) at the perigee altitude of \(2000 \mathrm{km} .\) What is the minimum increase \(\Delta v\) in velocity required of its rocket motor at this position to allow the satellite to escape from the earth's gravity field?

A boy of mass \(m\) is standing initially at rest relative to the moving walkway, which has a constant horizontal speed \(u .\) He decides to accelerate his progress and starts to walk from point \(A\) with a steadily increasing speed and reaches point \(B\) with a speed \(\dot{x}=v\) relative to the walkway. During his acceleration he generates an average horizontal force \(F\) between his shoes and the walkway. Write the work-energy equations for his absolute and relative motions and explain the meaning of the term muv.

It is experimentally determined that the drive wheels of a car must exert a tractive force of \(560 \mathrm{N}\) on the road surface in order to maintain a steady vehicle speed of \(90 \mathrm{km} / \mathrm{h}\) on a horizontal road. If it is known that the overall drivetrain efficiency is \(e_{m}=0.70,\) determine the required motor power output \(P\).

The 1.2 -lb sphere is moving in the horizontal \(x-y\) plane with a velocity of \(10 \mathrm{ft} / \mathrm{sec}\) in the direction shown and encounters a steady flow of air in the \(x\) -direction. If the air stream exerts an essentially constant force of 0.2 lb on the sphere in the \(x\) -direction, determine the time \(t\) required for the sphere to cross the \(y\) -axis again.

A car with a mass of \(1500 \mathrm{kg}\) starts from rest at the bottom of a 10 -percent grade and acquires a speed of \(50 \mathrm{km} / \mathrm{h}\) in a distance of \(100 \mathrm{m}\) with constant acceleration up the grade. What is the power \(P\) delivered to the drive wheels by the engine when the car reaches this speed?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Electric Charge Field and Potential

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation

Translational Dynamics

Read Explanation
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.