a uniform chain of length l is placed on a smooth table with half of its length hanging over the edge , when released from rest it starts falling down . find the velocity of the chain when the last link leaves the table

To solve the problem of a uniform chain falling off a table, we can apply the principles of energy conservation. The scenario involves a chain of length \( l \) with half of it, \( \frac{l}{2} \), hanging over the edge of the table. When the chain is released, it begins to fall, and we want to determine its velocity at the moment the last link leaves the table.

Understanding the Setup

Imagine the chain is uniform, meaning its mass is evenly distributed along its length. Let's denote the mass per unit length of the chain as \( \lambda \). Therefore, the total mass \( m \) of the chain can be expressed as:

• Mass \( m = \lambda l \)

Initially, when the chain is at rest, the center of mass of the hanging part is at a height of \( \frac{l}{4} \) from the edge of the table. As the chain falls, the potential energy of the hanging part converts into kinetic energy.

Calculating Potential Energy

When the chain is released, the potential energy \( PE \) of the hanging part can be calculated as follows:

• Potential Energy \( PE = mgh \)

Here, \( h \) is the height of the center of mass of the hanging part, which is \( \frac{l}{4} \). Thus, the potential energy of the hanging part is:

• Potential Energy \( PE = \lambda l \cdot g \cdot \frac{l}{4} = \frac{\lambda g l^2}{4} \)

Kinetic Energy at the Moment of Leaving the Table

When the last link of the chain leaves the table, all the potential energy has been converted into kinetic energy \( KE \). The kinetic energy of the entire chain can be expressed as:

• Kinetic Energy \( KE = \frac{1}{2} mv^2 \)

Substituting the mass \( m = \lambda l \) into the kinetic energy equation gives:

• Kinetic Energy \( KE = \frac{1}{2} \lambda l v^2 \)

Applying Conservation of Energy

According to the conservation of energy principle, the potential energy lost by the chain equals the kinetic energy gained. Therefore, we can set the potential energy equal to the kinetic energy:

• \(\frac{\lambda g l^2}{4} = \frac{1}{2} \lambda l v^2\)

Notice that \( \lambda \) cancels out from both sides, simplifying our equation:

• \(\frac{g l^2}{4} = \frac{1}{2} l v^2\)

Solving for Velocity

Now, we can solve for \( v \) by rearranging the equation:

• Multiply both sides by 2: \( \frac{g l^2}{2} = l v^2 \)
• Divide both sides by \( l \): \( \frac{g l}{2} = v^2 \)
• Taking the square root gives us: \( v = \sqrt{\frac{g l}{2}} \)

Final Result

Thus, the velocity of the chain when the last link leaves the table is:

• Velocity \( v = \sqrt{\frac{g l}{2}} \)

This result shows how gravitational potential energy transforms into kinetic energy as the chain falls, illustrating the principles of mechanics in a clear and engaging manner.