/*! 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

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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 approximately \(0.001875 \cdot g\) Joules.

Step by step solution

01

Understand the Initial and Final States

Initially, the cord is horizontal, implying all parts are at the same height relative to the floor. Finally, when hanging vertically, one end of the cord remains attached to the ceiling, and the other end hangs down, indicating a change in height for differential segments of the cord.
02

Define the Variables

Let the height from the ceiling be denoted by \( y \). The cord has a linear mass density \( \lambda = \frac{m}{L} \), where mass \( m = 15 \text{ g} = 0.015 \text{ kg} \) and length \( L = 0.25 \text{ m} \). So, \( \lambda = \frac{0.015}{0.25} = 0.06 \text{ kg/m} \).
03

Differential Mass Segment

Consider an infinitesimally small segment of the cord \( dy \) at a distance \( y \) from the ceiling. The mass of this segment is \( dm = \lambda \, dy = 0.06 \, dy \).
04

Gravitational Potential Energy of a Differential Segment

The gravitational potential energy \( dU \) of this segment at height \( y \) is given by \( dU = dm \cdot g \cdot y = 0.06 \cdot y \cdot g \cdot dy \).
05

Calculate Total Change in Gravitational Potential Energy

Integrate the potential energy from the ceiling (\( y = 0 \)) to the bottom of the cord (\( y = 0.25 \)) to find the total potential energy in the vertical state.\[U = \int_0^{0.25} 0.06 \cdot g \cdot y \, dy = 0.06 \cdot g \cdot \left[ \frac{y^2}{2} \right]_0^{0.25}\]\[U = 0.06 \cdot g \cdot \left( \frac{0.25^2}{2} \right) = 0.06 \cdot g \cdot 0.03125 \approx 0.001875 \cdot g \text{ J}\]
06

Initial Gravitational Potential Energy

When stuck horizontally, the centre of mass of the cord is at height \( y = 0 \), so initial potential energy \(U_i = 0\).
07

Conclusion: Total Change in Potential Energy

Since initially, \( U_i = 0 \) and in the final state, \( U_f \approx 0.001875 \cdot g \), the change in potential energy is \( \Delta U = U_f - U_i = 0.001875 \cdot g \).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. When the cord is initially horizontal on the ceiling, its gravitational potential energy is zero because it is at height zero. When it hangs vertically, the cord's segments are at varying heights, accumulating gravitational potential energy as a result.

To calculate the change in gravitational potential energy, we determine how much energy each small segment gains when the cord is hanging. By summing the energy of all these segments, we find the total change.
  • The differential segment at height \(y\) from the ceiling contributes \(dU = dm \cdot g \cdot y\) to the gravitational potential energy, where \(g\) is the acceleration due to gravity.
  • This energy integrates over the length of the cord from its full height \(25\ \text{cm}\) to zero.
This integration provides the total change in gravitational potential energy when the orientation of the cord changes.
Integral Calculus
Integral calculus is vital in understanding and calculating changes in gravitational potential energy, especially when dealing with varying heights or distributions, like in this cord problem.

The potential energy of a uniform cord hanging from the ceiling can be calculated by integrating over its length. By dividing the cord into infinitesimally small pieces, we individually calculate the potential energy for each piece and then use integration to sum them.
  • The formula \(U = \int_0^{0.25} 0.06 \cdot g \cdot y \, dy\) integrates the function of potential energy with respect to height \(y\), covering the entire length from ceiling to floor.
  • Integration effectively adds up all the potential energies of these segments, providing the total energy change when the cord shifts from horizontal to vertical.
This approach highlights how integral calculus is a powerful tool in physics, allowing us to handle complex variable scenarios with precision.
Linear Mass Density
Linear mass density is a critical concept in solving this problem. It represents the mass per unit length of an object, such as the cord. Understanding this helps us distribute mass correctly along the cord's length, which is crucial for calculating gravitational potential energy.

In the problem, we use linear mass density \(\lambda\), defined as \(\lambda = \frac{m}{L}\), where \(m\) is the total mass, and \(L\) is the length of the cord. This allows us to find the mass of each tiny segment of the cord.
  • For the cord, \(\lambda = 0.06 \text{ kg/m}\), meaning each meter of the cord weighs 0.06 kg.
  • We calculate segment mass by \(dm = \lambda \, dy\), meaning even small slices of the cord at distance \(dy\) from the starting point have a precise mass, allowing accurate calculations of energy.
Linear mass density is vital for correctly applying calculus in mechanical problems like this, ensuring each part's mass is appropriately accounted for in the calculations.

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Most popular questions from this chapter

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