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Chapter 6: Problem 42

A uniform chain of total mass \(m\) is laid out straight on a frictionless table and held stationary so that one-third of its length, \(L=1.00 \mathrm{~m},\) is hanging vertically over the edge of the table. The chain is then released. Determine the speed of the chain at the instant when only one-third of its length remains on the table.

Short Answer

Expert verified
Answer: The speed of the chain at the instant when only one-third of its length remains on the table is approximately \(1.14 \mathrm{m/s}\).

Step by step solution

01

Identify the initial and final conditions

Initially, one-third of the chain's length is hanging over the edge of the table, and the chain is stationary. At the final condition, there is still one-third of the chain's length on the table and the entire chain now has some speed \(v\).
02

Find the expression for the change in potential energy

Let's first find the mass per unit length of the chain, \(\mu\). This can be calculated as: \(\mu = \frac{m}{L}\). When the chain is released, the mass of the chain hanging over the edge decreases from one-third to zero. The change in potential energy, \(\Delta U\), can be calculated as: \(\Delta U = \int_{0}^{\frac{L}{3}} mgh \mathrm{d}x = \int_{0}^{\frac{L}{3}} \mu g x \mathrm{d}x\).
03

Calculate the change in potential energy

Integrate the expression for \(\Delta U\) with respect to \(x\): \(\Delta U = \mu g \int_{0}^{\frac{L}{3}} x \mathrm{d}x = \mu g \left[\frac{1}{2}x^2 \right]_{0}^{\frac{L}{3}} = \mu g \left[\frac{1}{2}\left(\frac{L}{3}\right)^2\right]\).
04

Apply the conservation of mechanical energy

The total mechanical energy of the system is conserved, so the loss in potential energy is equal to the gain in kinetic energy: \(\Delta U = \Delta K\). The kinetic energy of the chain is given by: \(K = \frac{1}{2}mv^2\). Hence, we have the equation: \(\mu g \left[\frac{1}{2}\left(\frac{L}{3}\right)^2\right] = \frac{1}{2}mv^2\).
05

Solve for the final speed of the chain

Now, we will solve for \(v\) by rearranging the equation and substituting in the given values for \(m\) and \(L\): \(v^2 = \frac{2\mu g \left[\frac{1}{2}\left(\frac{L}{3}\right)^2\right]}{m} = \frac{2g\left(\frac{L}{3}\right)^2}{L}\). Since we know \(L = 1.00 \mathrm{m}\) and \(g = 9.81 \mathrm{m/s^2}\), we can compute the final speed \(v\): \(v = \sqrt{\frac{2(9.81 \mathrm{m/s^2})\left(\frac{1.00 \mathrm{m}}{3}\right)^2}{1.00\mathrm{m}}} \approx 1.14 \mathrm{m/s}\). So, the speed of the chain at the instant when only one-third of its length remains on the table is approximately \(1.14 \mathrm{m/s}\).

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Key Concepts

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

Conservation of Mechanical Energy
When we talk about the conservation of mechanical energy, we're referring to a fundamental principle that states the total mechanical energy of an isolated system remains constant if the only forces doing work are conservative. In other words, energy cannot be created or destroyed, it can only be transformed from one form to another.

For the chain speed physics problem, this principle is key. Initially, the chain has potential energy due to gravity. As it falls, that potential energy is converted into kinetic energy, the energy of motion. Since no external work is done on the chain and there is no friction to dissipate the energy, the initial potential energy of the chain is equal to the final kinetic energy, allowing us to set up an equation expressing the conservation of mechanical energy.
Potential Energy Calculation
The concept of potential energy is related to the position of an object in a force field, such as the gravitational field of the Earth. The potential energy calculation for an object at height h is given by the formula \( U = mgh \), where m is the mass, g is the acceleration due to gravity, and h is the height above a reference point.

In the chain problem, we calculate the change in potential energy as the chain falls. This involves integrating the potential energy over the portion of the chain moving from the table to a lower height. Understanding this change is crucial, as it gives us the amount of energy available to be converted into kinetic energy.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is calculated by the equation \( K = \frac{1}{2}mv^2 \), where m is the mass and v is the velocity of the object.

In our chain problem, we're interested in finding the speed of the chain, which means we need to calculate the kinetic energy of the chain as it moves. Once we've determined the loss of potential energy as the chain falls, we can equate it to the gained kinetic energy to solve for the chain's velocity. This interplay between potential and kinetic energy is beautifully illustrated in this problem.
Uniform Chain Dynamics
The term uniform chain dynamics refers to the motion and behavior of a chain or similar object that has a consistent mass per unit length and is subjected to uniform gravitational force. In problems like this, we make the assumption that the chain does not stretch and that its mass is distributed evenly along its length.

Understanding uniform chain dynamics is essential for solving our problem, as it allows us to treat different segments of the chain independently while knowing that the physical properties like mass per unit length stay constant. This uniformity simplifies the calculation of both potential and kinetic energies as the chain moves, and brings us closer to uncovering the speed at which the chain will fall.

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Most popular questions from this chapter

A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \mathrm{~cm}\) from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length of \(4.00 \mathrm{~m}\) and is inclined at \(30.0^{\circ} .\) Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of \(0.350 .\) When the spring is compressed, the mass is \(1.50 \mathrm{~m}\) from the bottom of the plane. a) What is the speed of the mass as it reaches the bottom of the plane? b) What is the speed of the mass as it reaches the top of the plane? c) What is the total work done by friction from the beginning to the end of the mass's motion?

The molecular bonding in a diatomic molecule such as the nitrogen \(\left(\mathrm{N}_{2}\right)\) molecule can be modeled by the Lennard Jones potential, which has the form $$ U(x)=4 U_{0}\left(\left(\frac{x_{0}}{x}\right)^{12}-\left(\frac{x_{0}}{x}\right)^{6}\right) $$ where \(x\) is the separation distance between the two nuclei and \(x_{0}\), and \(U_{0}\) are constants. Determine, in terms of these constants, the following: a) the corresponding force function; b) the equilibrium separation \(x_{0}\), which is the value of \(x\) for which the two atoms experience zero force from each other; and c) the nature of the interaction (repulsive or attractive) for separations larger and smaller than \(x_{0}\).

a) If you jump off a table onto the floor, is your mechanical energy conserved? If not, where does it go? b) A car moving down the road smashes into a tree. Is the mechanical energy of the car conserved? If not, where does it go?

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

In 1896 in Waco, Texas, William George Crush, owner of the K-T (or "Katy") Railroad, parked two locomotives at opposite ends of a 6.4 -km-long track, fired them up, tied their throttles open, and then allowed them to crash head- on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming that each locomotive weighed \(1.2 \cdot 10^{6} \mathrm{~N}\) and its acceleration along the track was a constant \(0.26 \mathrm{~m} / \mathrm{s}^{2},\) what was the total kinetic energy of the two locomotives just before the collision?
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