/*! 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 19 A ring A of mass \(m\) is thread... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 19

A ring A of mass \(m\) is threaded on to a smooth fixed horizontal straight wire. The ring is attached to one end of a light elastic string whose other end is fixed to a point \(\mathrm{B}\) at a height \(h\) above the wire. Initially the ring is vertically below \(\mathrm{B}\). In this position it is given a velocity \(v\) along the wire. The string has a natural length \(h\) and modulus of elasticity \(m g\). Show that the angle \(\theta\) between \(\mathrm{AB}\) and the wire when the ring first comes to instantaneous rest, is given by $$ \sin \theta\left(\frac{v}{\sqrt{g h}}+1\right)=1 $$

Short Answer

Expert verified
\( \sin \theta \left( \frac{v}{\sqrt{gh}} + 1 \right) = 1 \)

Step by step solution

01

Define the problem variables

Introduce the given parameters: mass of the ring is denoted as m, velocity as v, height as h, and modulus of elasticity as mg. The angle between AB and the wire when the ring comes to rest is denoted as \(\theta\).
02

Use the energy conservation principle

At the initial position directly below point B, the ring has kinetic energy due to its velocity v. When the ring moves to the point where it comes to instantaneous rest, all this kinetic energy is converted into potential energy in the elastic string. The initial kinetic energy is \( \frac{1}{2} m v^2 \).
03

Calculate the elastic potential energy

The potential energy stored in the elastic string when it is stretched by a distance l is given by \( \frac{1}{2} k (l - h)^2 \), where k is the modulus of elasticity. Here, k = mg, and the length when the ring comes to rest can be represented using trigonometry as \( l = \frac{h}{\sin \theta} \).
04

Express stretch in terms of angle

The total stretch of the elastic string when the ring is at an angle \( \theta \) to the wire is \( l - h = \frac{h}{\sin \theta} - h \).
05

Set up the energy conservation equation

According to the conservation of energy principle, the initial kinetic energy equals the potential energy stored in the stretched elastic string:
06

Substitute and simplify

Substitute the expressions for kinetic and potential energy into the equation: \[ \frac{1}{2} m v^2 = \frac{1}{2} mg \left( \frac{h}{\sin \theta} - h \right)^2 \] Simplify the equation to express it in terms of \( \theta \) and solve:
07

Solve for \( \sin \theta \)

Further simplify the equation: \[ v^2 = g h \left( \frac{1}{\sin \theta} - 1 \right)^2 \] Take the square root of both sides and rearrange to find the final expression: \[ \frac{v}{\sqrt{gh}} + 1 = \frac{1}{\sin \theta} \] Therefore, \( \sin \theta \left( \frac{v}{\sqrt{gh}} + 1 \right) = 1 \).

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.

Energy Conservation in Physics
In physics, the principle of energy conservation is fundamental. It states that energy cannot be created or destroyed, only transformed from one form to another. When solving mechanics problems, understanding this law is crucial. For instance, in the given exercise, kinetic energy is converted into elastic potential energy. Initially, the ring has kinetic energy because it moves with velocity v. As it moves, this energy converts into potential energy stored in the elastic string. By comprehending how energy transforms, you can grasp why the ring stops moving at a certain point.
Elastic Potential Energy
Elastic potential energy refers to the energy stored in elastic materials when they are stretched or compressed. For the ring in our exercise, this energy translates into stretching the elastic string. The formula for elastic potential energy is:
\( \frac{1}{2} k (l - h)^2 \)
.
Here, k is the string's modulus of elasticity, h is the natural length, and l is the stretched length. Essentially, the more you stretch the string, the more potential energy it stores. This stored energy is what ultimately balances the kinetic energy of the moving ring, bringing it to rest.
Kinetic Energy
Kinetic energy is the energy that an object possesses because of its motion. It鈥檚 given by the formula:
\( \frac{1}{2} m v^2 \)
, where m is the mass and v is the velocity. In the provided problem, the ring starts with a kinetic energy that will be fully transformed into elastic potential energy when the ring stops.
An understanding of kinetic energy is crucial in mechanics, as it helps in predicting how objects will move and interact. When you analyze the ring's movement, recognizing its initial kinetic energy helps in setting up the energy conservation equation.
Trigonometric Relationships
Trigonometry helps in dealing with angles and lengths in physics problems. In our problem, it鈥檚 used to express the stretched length of the string in terms of angle \(\theta\). The length of the string when the ring is at rest can be represented as:
\( l = \frac{h}{\sin \theta} \)
.
This relationship arises from understanding the components of the triangle formed. Knowing these relationships allows you to link the physical stretch of the elastic string to the angle \(\theta\), providing a way to solve the energy conservation equation.
Modulus of Elasticity
The modulus of elasticity (or elastic modulus) is a measure of a material's ability to resist deformation. It is denoted by k in physics. In the exercise, the string's modulus of elasticity is given as mg. This value is crucial when calculating the stored potential energy in the string.
Essentially, the higher the modulus, the stiffer the string, and the more force it takes to stretch it. By understanding the modulus of elasticity, you can better grasp how the elastic string in the problem behaves under force, which is key to solving for the angle \(\theta\).

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

A particle of weight \(W\) is attached to a point \(C\) of an unstretched elastic string \(\mathrm{AB}\), where \(\mathrm{AC}=4 a / 3, \mathrm{CB}=4 a / 7\). The ends \(\mathrm{A}\) and \(\mathrm{B}\) are then attached to the extremities of a horizontal diameter of a fixed hemispherical bowl of radius \(a\). and the particle rests on the smooth inner surface, the angle BAC being \(30^{\circ}\). Show that the modulus of elasticity of the string is \(W\) and determine the reaction of the bowl on the particle. (U of L)

A light elastic string, of unstretched length \(a\) and modulus of elasticity \(W\), is fixed at one end to a point on the ceiling of a room. To the other end of the string. is attached a particle of weight \(W\). A horizontal force \(P\) is applied to the particle and in equilibrium it is found that the string is stretched to three times its natural length. Calculate: (a) the angle the string makes with the horizontal, (b) the value of \(P\) in terms of \(W\). If, instead, \(P\) is not applied horizontally find the least value of \(P\) which in equilibrium will make the string have the same inclination to the horizontal as before. Deduce that the stretch length of the string is \(\frac{3}{2} a\) in this case and find the inclination of \(P\) to the vertical. (U of L)

Two fixed points \(\mathrm{A}\) and \(\mathrm{B}\) on the same horizontal level are \(20 \mathrm{~cm}\) apart. \(\mathrm{A}\) light elastic string, which obeys Hooke's Law, is just taut when its ends are fixed at A and B. A block of mass \(5 \mathrm{~kg}\) is attached to the string at a point \(\mathrm{P}\) where \(\mathrm{AP}=15 \mathrm{~cm}\). The system is then allowed to take up its position of equilibrium with P below AB and it is found that in this position the angle APB is a right angle. If \(\angle \mathrm{BAP}=\theta\), show that the ratio of the extensions of \(\mathrm{AP}\) and \(\mathrm{BP}\) is $$ \frac{4 \cos \theta-3}{4 \sin \theta-1} $$ Hence show that \(\theta\) satisfies the equation $$ \cos \theta(4 \cos \theta-3)=3 \sin \theta(4 \sin \theta-1) $$

A particle of mass \(m\) slides a distance \(d\) down a plane inclined at \(\theta\) to the horizontal. The work done by the normal reaction \(R\) is: (a) \(R d\) (b) \(m g d \cos \theta\) (c) 0 (d) \(m g d \sin \theta\).

A ring of mass \(m\) can slide freely on a smooth wire in the shape of a circle of diameter \(2 a\), which is fixed in a vertical plane. The ring is fastened to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(m g\). The other end of the string is attached to the lowest point of the wire. The ring is held at the highest point of the wire and is slightly disturbed from rest. Find the velocity of the ring: (a) when it is level with the centre of the circular wire, (b) when the string first becomes slack, (c) when the string makes an acute angle \(\theta\) with the upward vertical.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

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.