/*! 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 37 A uniform cord of length \(25 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniform cord of length \(25 \mathrm{~cm}\) and mass \(15 \mathrm{~g}\) is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

Short Answer

Expert verified
The change in gravitational potential energy is zero.

Step by step solution

01

Define Variables and Concepts

We have a cord of length \(L = 25 \mathrm{~cm} = 0.25 \mathrm{~m}\) and mass \(m = 15 \mathrm{~g} = 0.015 \mathrm{~kg}\). Gravitational potential energy change can be found using the change in height of the cord's center of mass, initially when stuck to the ceiling (center of mass at \(L/2\) or 0.25 m), and finally, when vertically hanging (center of mass at \(L/2\) or 0.125 m).
02

Determine Initial and Final Center of Mass

Initially, the center of mass is halfway along the length of the cord when it is fully horizontal, so initially at \( (L/2 = 0.25/2 = 0.125 \mathrm{~m}) \) above the ceiling. When hanging, the center of mass drops to \( (L/2 = 0.125 \mathrm{~m}) \) from the ceiling (though this is the same measurement numerically, the potential energy changed because of the orientation change).
03

Calculate Change in Gravitational Height

Calculate the change in height for the center of mass in relation to the entire gravitational field. Initially, height of center of mass \( h_i = 0.125 \mathrm{~m} \), and finally also \( h_f = 0.125 \mathrm{~m} \).
04

Calculate Potential Energy Change

The gravitational potential energy \( U = mgh \). Here, changes depend on distribution of mass height. Initially, all parts of the cord are at potential (\(mg \frac{L}{2}\)), when hanging, each differential bit can be integrated from 0 to L vertically. Change is computed via integration of distribution of mass: \(U = \frac{1}{2} mgl\) leading change terms to cancel, seeing it's a dynamic equilibrium shift no net gravitational change discovered.
05

Conclude with No Change in Energy

Ultimately, the exercise's hint urges to inspect distribution/hanging equilibrium differently. Supposed typical hypotheses reflect matched potential each way owing symmetry balance; revealing subtly through cancellation despite an assumed shift.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center of Mass
Understanding the concept of the center of mass can really help make sense of why the gravitational potential energy didn't change for the cord. The center of mass is the point where the mass of an object can be thought of as being concentrated. In simple terms, it's like the balance point of the object.
As the cord moves from the ceiling to hanging vertically, its center of mass changes position relative to the ceiling. Initially, when the cord is laid out horizontally against the ceiling, its center of mass is halfway along its length, so at 0.125 meters from its starting point.
Once it hangs, the center of mass lies at the middle of the cord measuring downwards from the ceiling, also at 0.125 meters. This means, numerically, the center of mass hasn't moved; however, it's how mass is distributed around this center that plays into gravitational energy.
Integral Calculus
Integral calculus comes into play when we need to compute the gravitational potential energy for the entire cord as it changes orientation. Since the cord is evenly distributed, integral calculus provides a way to sum up all the gravitational potential energies of each small part of the cord.
Imagine breaking down the cord into tiny slices. Each of these slices has its own potential energy, and the total energy is the sum of all these little energies.
That's where the integral comes in – it helps us sum up an infinite number of tiny energies for an accurate total. In this situation, while it seems complex, the integration of the uniform distribution along the length results in equations that ultimately showcase how the energies cancel out due to symmetry.
  • Initial laying: total energy was summed from center of mass height down to tiny differentials.
  • Hanging cord needs whole length ensuring integrals uniformly define net positioning through distribution.
Uniform Distribution
Uniform distribution means the mass is spread out consistently over its length. For the cord, this property simplifies how we use our math tools.
When the mass is uniformly distributed, every section of the cord of equal length will have the same mass. This uniformity is why we can use simple calculations to find the center of mass and integrate potential energies.
If the cord's mass weren't uniform, the energy calculations would be more complex. But given our uniform distribution, each slice's mass is simply proportional to its length, helping solve for total gravitational potential more easily.
Using uniform distribution simplifies both center of mass calculation and lets integration techniques rest easier knowing each differential slice contributes equally, overall aiding in contradictions resolving as zero potential energy net difference.

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 \(60.0 \mathrm{~kg}\) 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 \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

A swimmer moves through the water at an average speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force is \(110 \mathrm{~N}\). What average power is required of the swimmer?

The spring in the muzzle of a child's spring gun has a spring constant of \(700 \mathrm{~N} / \mathrm{m}\). To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by \(30^{\circ}\) to the horizontal, a \(57 \mathrm{~g}\) ball is shot to a maximum height of \(1.83 \mathrm{~m}\) above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N}\), the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\), with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\).

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.