/*! 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 39 A pendulum is \(0.80 \mathrm{m}\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 39

A pendulum is \(0.80 \mathrm{m}\) long and the bob has a mass of 1.0 kg. At the bottom of its swing, the bob's speed is \(1.6 \mathrm{m} / \mathrm{s} .\) (a) What is the tension in the string at the bottom of the swing? (b) Explain why the tension is greater than the weight of the bob.

Short Answer

Expert verified
Answer: The tension in the string at the bottom of the swing is found using the formula \(T = m(a_c+g)\), where \(m\) is the mass of the bob, \(a_c\) is the centripetal acceleration, and \(g\) is the gravitational constant. Using the given values, we can find the tension force \(T\). The tension is greater than the weight of the bob because, at the bottom of the pendulum's swing, it has to counterbalance both the gravitational force and provide the necessary centripetal force for the bob to continue in its circular motion.

Step by step solution

01

Calculate Centripetal Acceleration

We have the speed \(v\) (1.6 m/s) and radius \(r\) (0.80 m) of the pendulum. Using the formula for centripetal acceleration, we get \(a_c = \dfrac{v^2}{r} = \dfrac{(1.6 \,\text{m/s})^2}{0.80 \,\text{m}}\).
02

Apply Newton's Second Law

Newton's second law states that \(F_{net} = ma\), where \(F_{net}\) is the net force acting on the bob, \(m\) is the mass, and \(a\) is the centripetal acceleration. At the bottom of the swing, the net force will be the difference between the tension force (\(T\)) and the gravitational force (\(mg\)), thus \(F_{net} = T - mg\).
03

Solve for Tension Force

We substitute our known values into Newton's second law equation and solve for \(T\): \(ma_c = T-mg \implies T = m(a_c+g)\) Using the given bob's mass \(m\) (1.0 kg), the gravitational constant \(g\) (9.81 m/s虏) and the calculated centripetal acceleration \(a_c\), we can find the tension force \(T\).
04

Explain why tension is greater than the weight of the bob

At the bottom of the pendulum's swing, the tension in the string has to not only counterbalance the gravitational force acting on the bob but also provide the necessary centripetal force for circular motion. This additional force is required to keep the bob moving in a circular path. Consequently, the tension in the string is greater than the weight of the bob at the bottom of the swing.

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!

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

Objects that are at rest relative to Earth's surface are in circular motion due to Earth's rotation. What is the radial acceleration of an African baobab tree located at the equator?
An elevator cable winds on a drum of radius \(90.0 \mathrm{cm}\) that is connected to a motor. (a) If the elevator is moving down at $0.50 \mathrm{m} / \mathrm{s},$ what is the angular speed of the drum? (b) If the elevator moves down \(6.0 \mathrm{m},\) how many revolutions has the drum made?
A velodrome is built for use in the Olympics. The radius of curvature of the surface is \(20.0 \mathrm{m}\). At what angle should the surface be banked for cyclists moving at \(18 \mathrm{m} / \mathrm{s} ?\) (Choose an angle so that no frictional force is needed to keep the cyclists in their circular path. Large banking angles are used in velodromes.)
In a Beams ultracentrifuge, the rotor is suspended magnetically in a vacuum. Since there is no mechanical connection to the rotor, the only friction is the air resistance due to the few air molecules in the vacuum. If the rotor is spinning with an angular speed of $5.0 \times 10^{5} \mathrm{rad} / \mathrm{s}$ and the driving force is turned off, its spinning slows down at an angular rate of \(0.40 \mathrm{rad} / \mathrm{s}^{2} .\) (a) How long does the rotor spin before coming to rest? (b) During this time, through how many revolutions does the rotor spin?
A soccer ball of diameter \(31 \mathrm{cm}\) rolls without slipping at a linear speed of \(2.8 \mathrm{m} / \mathrm{s} .\) Through how many revolutions has the soccer ball turned as it moves a linear distance of \(18 \mathrm{m} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Magnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Space Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Fundamentals of Physics

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.