Log out Go to App
Go to App

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/s²) 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 small body of mass \(0.50 \mathrm{kg}\) is attached by a \(0.50-\mathrm{m}-\) long cord to a pin set into the surface of a frictionless table top. The body moves in a circle on the horizontal surface with a speed of $2.0 \pi \mathrm{m} / \mathrm{s} .$ (a) What is the magnitude of the radial acceleration of the body? (b) What is the tension in the cord?
Centrifuges are commonly used in biological laboratories for the isolation and maintenance of cell preparations. For cell separation, the centrifugation conditions are typically \(1.0 \times 10^{3}\) rpm using an 8.0 -cm-radius rotor. (a) What is the radial acceleration of material in the centrifuge under these conditions? Express your answer as a multiple of \(g .\) (b) At $1.0 \times 10^{3}$ rpm (and with a 8.0-cm rotor), what is the net force on a red blood cell whose mass is \(9.0 \times 10^{-14} \mathrm{kg} ?\) (c) What is the net force on a virus particle of mass \(5.0 \times 10^{-21} \mathrm{kg}\) under the same conditions? (d) To pellet out virus particles and even to separate large molecules such as proteins, superhigh-speed centrifuges called ultracentrifuges are used in which the rotor spins in a vacuum to reduce heating due to friction. What is the radial acceleration inside an ultracentrifuge at 75000 rpm with an 8.0 -cm rotor? Express your answer as a multiple of \(g\).
A roller coaster car of mass 320 kg (including passengers) travels around a horizontal curve of radius \(35 \mathrm{m}\) Its speed is $16 \mathrm{m} / \mathrm{s} .$ What is the magnitude and direction of the total force exerted on the car by the track?
Grace, playing with her dolls, pretends the turntable of an old phonograph is a merry-go-round. The dolls are \(12.7 \mathrm{cm}\) from the central axis. She changes the setting from 33.3 rpm to 45.0 rpm. (a) For this new setting, what is the linear speed of a point on the turntable at the location of the dolls? (b) If the coefficient of static friction between the dolls and the turntable is \(0.13,\) do the dolls stay on the turntable?
Two gears \(A\) and \(B\) are in contact. The radius of gear \(A\) is twice that of gear \(B\). (a) When \(A\) 's angular velocity is \(6.00 \mathrm{Hz}\) counterclockwise, what is \(B^{\prime}\) s angular velocity? (b) If \(A\) 's radius to the tip of the teeth is \(10.0 \mathrm{cm},\) what is the linear speed of a point on the tip of a gear tooth? What is the linear speed of a point on the tip of \(B\) 's gear tooth?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Circular Motion and Gravitation

Read Explanation

Space Physics

Read Explanation

Engineering Physics

Read Explanation

Energy Physics

Read Explanation

Classical Mechanics

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.

Sign-up for free