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A uniform rope of total length \(2 a\) hangs in equilibrium over a smooth nail. A very small impulse causes the rope to slowly roll off the nail. Find the velocity of the rope as it just clears the nail. Assume the rope is prevented from lifting off the nail and is in free fall.

Short Answer

Expert verified
The velocity of the rope as it just clears the nail is given by \( v = \sqrt{2ga} \), where \(g\) is the gravitational acceleration and \(a\) is half the length of the rope.

Step by step solution

01

Determine the Potential Energy

First, we need to determine the potential energy of the rope when it is hanging over the smooth nail. We can consider the midpoint of the rope as its center of mass, and hence, the rope is suspended at a height of \( a \) over the nail. The mass of the rope is given by \( M \). Thus, the potential energy of the rope, denoted as \( P.E. = Mgh \), where \( g \) is the gravitational acceleration and \(h\) represents height. For the rope hanging over the nail, the potential energy is given by: \( P.E. = Mga \)
02

Determine the Kinetic Energy of the rope

As the rope just clears the nail, it possesses kinetic energy. We need to determine how much kinetic energy the rope has at this moment. The kinetic energy, denoted as \( K.E. = \frac{1}{2} Mv^2 \), where \( v \) is the velocity we want to find.
03

Apply Conservation of Mechanical Energy

We can now use conservation of mechanical energy to equate the potential energy when the rope is hanging over the nail to the kinetic energy when it just clears the nail. This gives us the following equation: \( Mga = \frac{1}{2} Mv^2 \)
04

Solve for the velocity of the rope

Now, solve for the velocity, \( v \), of the rope when it just clears the nail: Divide both sides of the equation by \( M \): \( ga = \frac{1}{2} v^2 \) Now, multiply both sides of the equation by 2: \( 2ga = v^2 \) Finally, take the square root of both sides to find the velocity: \( v = \sqrt{2ga} \) Thus, the velocity of the rope as it just clears the nail is given by: \( v = \sqrt{2ga} \)

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Key Concepts

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

Potential Energy
When we talk about potential energy in the context of a physical object, we are referring to the energy stored in that object due to its position in a gravitational field. In the exercise about the sliding rope, this concept is central to understanding how the rope's position above the ground contains energy that can be converted into motion, or kinetic energy.

Consider a rope with a mass of \(M\) hanging over a nail. The potential energy of the rope is found using the expression \(P.E. = Mgh\), where \(g\) is the acceleration due to gravity and \(h\) is the height of the rope’s center of mass above the ground. For a rope of length \(2a\), this height is simply \(a\), leading to a potential energy of \(Mga\).

This stored energy is like a reserve, waiting to be released. It's not yet doing anything active, but it has the potential to do work, such as moving the rope, once it's in motion.
Kinetic Energy
Kinetic energy is the energy of motion. If potential energy is the stored energy due to an object's position, kinetic energy is the energy it has due to its movement. The formula for kinetic energy (\(K.E.\)) is \(\frac{1}{2}Mv^2\), where \(M\) is the mass of the object and \(v\) is its velocity.

In our exercise, as the rope falls and clears the nail, all of the potential energy we discussed previously is transformed into kinetic energy. This is because the rope moves with a certain velocity, and its mass is still \(M\). The principle of conservation of mechanical energy allows us to relate this kinetic energy back to the potential energy the rope had when it was hanging still over the nail. By equating the two, we can solve for the rope's velocity as it just clears the nail.
Uniform Rope Dynamics
The dynamics of a uniform rope involve how it moves and changes shape under various forces. In our problem, we assume the rope is uniform, which simplifies the calculation as the mass is evenly distributed along its length. This uniformity allows us to calculate easily the center of mass and assume that, initially, the gravitational force acts through this point.

When the rope receives an impulse and starts to slip off the nail, the uniformity ensures that the acceleration is the same throughout the length of the rope. As the rope falls, every part of it accelerates downward at the rate of \(g\), Earth's gravitational acceleration. Understanding uniform rope dynamics is central to predicting how the rope will act under the influence of gravity and other forces, and it allows us to apply the conservation of mechanical energy reliably to find the rope's velocity as it clears the nail.

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Most popular questions from this chapter

A cannon in a fort overlooking the ocean fires a shell of mass \(M\) at an elevation angle \(\theta\) and muzzle velocity \(v_{0} .\) At the highest point, the shell explodes into two fragments (masses \(m_{1}+m_{2}=M\) ), with an additional energy \(E\), traveling in the original horizontal direction. Find the distance separating the two fragments when they land in the ocean. For simplicity, assume the cannon is at sea level.

Consider a multistage rocket of \(n\) stages, each with exhaust speed \(u\). Each stage of the rocket has the same mass ratio at burnout \(\left(k=m_{i} / m_{f}\right) .\) Show that the final speed of the \(n\) th stage is \(n u \ln k\)

A new single-stage rocket is developed in the year \(2023,\) having a gas exhaust velocity of \(4000 \mathrm{m} / \mathrm{s}\). The total mass of the rocket is \(10^{5} \mathrm{kg}\), with \(90 \%\) of its mass being fuel. The fuel burns quickly in 100 s at a constant rate. For testing purposes, the rocket is launched vertically at rest from Earth's surface. Answer parts (a) through (d) of the previous problem.

The center of gravity of a system of particles is the point about which external gravitational forces exert no net torque. For a uniform gravitational force, show that the center of gravity is identical to the center of mass for the system of particles.

The force of attraction between two particles is given by $$\mathbf{f}_{12}=k\left[\left(\mathbf{r}_{2}-\mathbf{r}_{1}\right)-\frac{r}{v_{0}}\left(\mathbf{\dot { r }}_{2}-\dot{\mathbf{r}}_{1}\right)\right]$$ where \(k\) is a constant, \(v_{0}\) is a constant velocity, and \(r \equiv\left|\mathbf{r}_{2}-\mathbf{r}_{1}\right| .\) Calculate the internal torque for the system; why does this quantity not vanish? Is the system conservative?

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