Question

A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \([(\mathrm{MgL}) /(3)]\) (C) \([(\mathrm{MgL}) /(9)]\) (D) \([(\mathrm{MgL}) /(18)]\)

Short answer

The short answer is: (D) \(\frac{(\mathrm{MgL})}{(18)}\)

Step-by-step solution

Identify the length and mass of the hanging part of the chain

Since one-third of the chain is hanging off the edge of the table, the length of the hanging part of the chain is (1/3)L, and the mass of the hanging part of the chain is (1/3)M.

Calculate the potential energy of the hanging part of the chain

The potential energy of the hanging part of the chain can be calculated using the following formula: Potential Energy (PE) = mgh where m is the mass of the object (in this case, the hanging part of the chain), g is the acceleration due to gravity, and h is the height of the object (in this case, the length of the hanging part of the chain). Since the chain is uniform, the center of mass of the hanging part will be at the midpoint of the hanging length i.e., \(\frac{1}{2}\left(\frac{1}{3}L\right) = \frac{L}{6}\) The mass of the hanging part is \(\frac{1}{3}M\), so the potential energy (PE) can be calculated as: PE = \(\frac{1}{3}M \times g \times\frac{L}{6}\) PE = \(\frac{MgL}{18}\)

Determine the work required

Since the potential energy of the hanging part of the chain is equal to the work done in pulling the chain onto the table, the work required is: Work Required = PE = \(\frac{MgL}{18}\) Based on the given options, the correct answer is: (D) \(\frac{(\mathrm{MgL})}{(18)}\)

Key concepts

Potential Energy

Potential energy is a form of energy that represents the potential for an object to do work due to its position or configuration. In the context of gravity, it describes the energy stored in an object because of its height and mass.
For the hanging chain in the problem:

• We consider the mass (\(m\)) of the hanging portion.
• The acceleration due to gravity (\(g\)) acts downwards.
• The height (\(h\)) is the distance of the center of mass from the reference level, which in this case is the table top.

The formula to calculate potential energy is:
\[\text{PE} = mgh\]
For the hanging part of the chain, the center of mass is at a distance \(\frac{L}{6}\) below the table. Thus, the potential energy becomes:
\[\text{PE} = \frac{1}{3}M \times g \times \frac{L}{6} = \frac{MgL}{18}\]
This value tells us the energy needed to lift the chain back onto the table.

Center of Mass

The concept of the center of mass is crucial in understanding how objects behave under force. It's the average position of all the mass in a body. This point is where the force due to gravity can be considered to act.
For a uniform chain, the distribution of mass is even along the length, making calculations straightforward.

• A uniform object’s center of mass is exactly in the middle of its length.
• For the chain hanging over the edge, the center of mass is halfway down the hanging section.

That means the center of mass of the hanging third of the chain is located at \(\frac{L}{6}\) from the top of the table. Understanding this helps us calculate the potential energy accurately, as it simplifies how we sum the gravitational influence on the entire segment.

Uniform Chain

A uniform chain refers to a chain that has equal mass distribution throughout its entire length. This implies that any section of the chain will have a mass proportional to its length.
In this exercise, the uniformity of the chain simplifies calculations:

• The chain has length \(L\) and mass \(M\). Consequently, any part of length \(x\) will have mass \(\frac{Mx}{L}\).
• This proportionality holds true for any segment of the chain, making it easier to relate length and mass.

When one-third of the chain is hanging, it translates into one-third of the total mass. The uniform characteristic means no complex distributions are involved, making potential energy and work calculations straightforward. This uniformity ensures that the center of mass and subsequent calculations can be approached with simple linear assumptions.