/*! 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 40 A 35.0 -kg child swings on a rop... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 35.0 -kg child swings on a rope with a length of \(6.50 \mathrm{m}\) that is hanging from a tree. At the bottom of the swing, the child is moving at a speed of \(4.20 \mathrm{m} / \mathrm{s} .\) What is the tension in the rope?

Short Answer

Expert verified
Answer: To find the tension in the rope, we must first calculate the change in height (\(h\)) between the highest and lowest points of the swing using the formula \(h = L - \sqrt{L^2 - \frac{1}{4}m^2v^4 / m^2g^2}\). Then, we can find the radius of the circular path (\(R\)) at the lowest point of the swing, using the equation \(R = \frac{1}{2}mv^2 / mg\). Finally, we can find the tension in the rope using the formula \(T = mg + \frac{mv^2}{R}\), where \(m\) is the mass of the child, \(g\) is the acceleration due to gravity, \(v\) is the speed of the child, and \(R\) is the radius of the circular path. After finding the values for \(h\) and \(R\), we can substitute them into the tension formula and solve for the tension (\(T\)) in the rope.

Step by step solution

01

Identify known values and variables

In this exercise, we are given the mass of the child (\(m = 35.0\,\text{kg}\)), the length of the rope (\(L = 6.50\,\text{m}\)), and the speed of the child at the bottom of the swing (\(v = 4.20\,\text{m/s}\)). We are asked to find the tension in the rope, which we will denote as \(T\).
02

Calculate gravitational potential energy

At the highest point in the swing, the child has no kinetic energy and maximum potential energy. At the lowest point, they have maximum kinetic energy and the least potential energy. The change in the gravitational potential energy, \(\Delta U\), at the lowest point in the swing can be found using: \(\Delta U = mgh\) where \(m\) is the mass of the child, \(g\) is the acceleration due to gravity (approximately \(9.81\,\text{m/s}^2\)), and \(h\) is the change in height between the highest and lowest points. The change in height can be found using the Pythagorean theorem: \(h = L - \sqrt{L^2 - R^2}\), where \(R\) is the radius of the circular path at the lowest point. At the lowest point, \(R = \frac{1}{2}mv^2 / mg\), so we have: \(h = L - \sqrt{L^2 - \frac{1}{4}m^2v^4 / m^2g^2}\)
03

Find the net force on the child at the lowest point

At the lowest point in the swing, the net force on the child will be the centripetal force acting towards the center of the circular path, given by: \(F_\text{net} = F_\text{c} = \frac{mv^2}{R}\) The net force on the child is also the difference between the tension in the rope and the gravitational force acting on the child: \(F_\text{net} = T - mg\)
04

Calculate the tension in the rope

Now we can calculate the tension in the rope by setting the net force on the child equal to the centripetal force: \(T - mg = \frac{mv^2}{R}\) Solving for the tension gives: \(T = mg + \frac{mv^2}{R}\) Using the known values for the mass of the child, the length of the rope, and the speed of the child, we can calculate the tension in the rope: \(T = (35.0\,\text{kg})(9.81\,\text{m/s}^2) + \frac{(35.0\,\text{kg})(4.20\,\text{m/s})^2}{R}\) After finding the change in height \(h\) and the radius of the circular path \(R\) using the formulas in Steps 2 and 3, we can plug in those values and calculate the tension, \(T\), in the rope.

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 91Ó°ÊÓ!

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

Find the orbital radius of a geosynchronous satellite. Do not assume the speed found in Example \(5.9 .\) Start by writing an equation that relates the period, radius, and speed of the orbiting satellite. Then apply Newton's second law to the satellite. You will have two equations with two unknowns (the speed and radius). Eliminate the speed algebraically and solve for the radius.
In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle's velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle's speed, \(F=k v .\) Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period. (c) Show that the radius of the particle's circular path is proportional to the speed.
Grace is playing with her dolls and decides to give them a ride on a merry-go- round. She places one of them on an old record player turntable and sets the angular speed at 33.3 rpm. (a) What is their angular speed in rad/s? (b) If the doll is \(13 \mathrm{cm}\) from the center of the spinning turntable platform, how fast (in \(\mathrm{m} / \mathrm{s}\) ) is the doll moving?
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 child swings a rock of mass \(m\) in a horizontal circle using a rope of length \(L\). The rock moves at constant speed \(v\). (a) Ignoring gravity, find the tension in the rope. (b) Now include gravity (the weight of the rock is no longer negligible, although the weight of the rope still is negligible). What is the tension in the rope? Express the tension in terms of \(m, g, v, L,\) and the angle \(\theta\) that the rope makes with the horizontal.
See all solutions

Recommended explanations on Physics Textbooks

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.