/*! 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 70 The string is \(L=120 \mathrm{~c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 70

The string is \(L=120 \mathrm{~cm}\) long, has a ball attached to one end, and is fixed at its other end. A fixed peg is at point \(P\). Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distance \(d \mathrm{ex}\) ceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)

Short Answer

Expert verified
The distance \(d\) must exceed 80 cm for the ball to swing completely around the peg.

Step by step solution

01

Understanding the Problem

The problem involves a pendulum with a peg at a point below the pivot. The string length is 120 cm, and we need to determine the minimum distance from the pivot to the peg, denoted as \(d\), such that the ball can swing completely around the peg.
02

Analyzing Motion and Energy Conservation

When released, the ball converts potential energy at the top into kinetic energy at the bottom. After catching on the peg, it must have enough kinetic energy to swing entirely around it. At the top of the circle, the ball's speed should be such that it still has tension to maintain circular motion. This happens when its speed is at least the minimum required for circular motion.
03

Applying Energy Conservation to Determine Speed

When the string catches on the peg, the length from the peg to the ball changes to \(L - d\). The potential energy at the top is \(mgh\), where \(h = L\). At the lowest point, all energy is kinetic: \(\frac{1}{2}mv^2 = mg(L)\). Solving for \(v\), we get \(v = \sqrt{2gL}\).
04

Minimum Speed for Complete Revolution

For the ball to complete a circle around the peg, at the top of the swing (with length \(L-d\)), it must have a speed \(v_{min} = \sqrt{g(L-d)}\). By equating kinetic energy at the lowest point to the sum of potential and kinetic energy at the highest point of the circle, we get \(\frac{1}{2}m(\sqrt{2gL})^2 \geq mg(2(L-d)) + \frac{1}{2}mv_{min}^2\).
05

Solving for Distance \(d\)

Simplifying the expression from Step 4 yields \(2gL \geq 2g(L-d) + g(L-d)\) because \(v_{min}^2 = g(L-d)\). Rearranging gives us \(2gL \geq 3g(L-d)\), leading to \(L \geq 3(L-d)\). Solving for \(d\), we find that \(d > \frac{2L}{3}\). Substituting \(L = 120 \text{ cm}\), \(d > 80 \text{ cm}\).

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.

Pendulum Motion
Pendulum motion refers to the back and forth movement of an object suspended from a fixed point. Imagine a playground swing or a clock's pendulum. This motion is periodic, meaning it repeats in a regular cycle.

Key features include:
  • Period: Time it takes to complete one full swing.
  • Amplitude: Maximum distance from the rest position.
  • Frequency: Number of swings per unit time.
The exercise's pendulum swings, initially descending under gravity, from its maximum amplitude. After catching on the peg, it follows a circular path around the peg.

This movement incorporates simple harmonic motion principles initial release and the complex dynamics of circular motion after the peg is engaged.
Energy Conservation
Energy conservation in physics states that energy cannot be created or destroyed, only transformed from one form to another. This principle is central to understanding pendulum motion.

In our pendulum problem:
  • At the start, the ball has maximum potential energy because of its height above the lowest point.
  • As it swings downwards, this potential energy converts to kinetic energy.
  • When the ball reaches the bottom, all potential energy has changed to kinetic energy.
Once the ball catches on the peg, this energy transformation allows it to swing around the peg. To do this, it must retain enough kinetic energy to overcome gravitational forces at the top of its circular path.
Circular Motion
Circular motion involves the movement of an object along the circumference of a circle or circular path. It requires a force directing towards the center of the circle, known as centripetal force. In our scenario:

  • The ball, upon engaging with the peg, initiates a circular path.
  • Centripetal force is crucial, to maintain this motion, which is provided by the tension in the string.
  • At the circle鈥檚 top, the ball requires sufficient speed to keep moving.
The ball's minimum speed at the top is partly determined by its weight and the tension required to ensure it doesn't fall off its circular path. This condition is described by the equation: \[ v_{min} = \sqrt{g(L-d)} \] This equation ties into the need for continued motion around the peg.
Kinetic Energy
Kinetic energy is the energy of motion, calculated by the formula \( KE = \frac{1}{2}mv^2 \). When the pendulum is at its lowest point, all its energy is kinetic. This energy is critical in allowing the pendulum to swing up and around the peg.

  • Initially, kinetic energy is gained as potential energy decreases.
  • This energy allows the pendulum to move in a circular path when needed.
  • The minimum kinetic energy at the top of the circular path equation ensures that gravitational pull won't halt the motion.
The transformation of energy types illustrates how kinetic energy in a pendulum sustains motion and enables it to swing around obstacles like pegs. Maintaining enough kinetic energy at the top ensures the pendulum continues its path smoothly and completely.

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

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm} ?\)

A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}\) (Fig. 8-57). A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of \(16 \mathrm{~J}\). (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m}\) ? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m}\) ?

A \(60.0 \mathrm{~kg}\) circus performer slides \(4.00 \mathrm{~m}\) down a pole to the circus floor, starting from rest. What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of \(500 \mathrm{~N}\) ?

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-42). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm}\), is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Fluids

Read Explanation

Nuclear Physics

Read Explanation

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

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.