A uniform chain of length 3 m and mass 3 kg overhangs a smooth table with 2 m laying on the table. I

Q: Why do we only consider the center of mass of the hanging part?

For a uniform object, the gravitational potential energy can be calculated by assuming its entire mass is concentrated at its center of mass. The part on the table has its CM on the datum, so its PE is zero.

Q: Why is the potential energy negative?

We chose the table surface as our reference level ($U=0$). Since the hanging part is below this level, its height is negative relative to the datum, resulting in a negative potential energy.

Q: What happens to the potential energy of the part of the chain on the table?

It remains zero as long as it is on the table. As the chain slips, mass transfers from the $U=0$ level to a lower level, which is why the total potential energy of the system decreases.

Q: Can we solve this using Newton's laws and integration?

Yes! The force pulling the chain down is the weight of the hanging part, which increases as it slips. You would set up a differential equation $F = m(x)g = M \frac{dv}{dt}$ and integrate. However, the energy method is much faster and less prone to calculus errors.

Q: What if the table had friction?

If the table was rough, mechanical energy would not be conserved. You would need to calculate the work done by friction, which is the integral of $-\mu \lambda x g \, dx$, and use the Work-Energy Theorem: $W_{\text{friction}} = \Delta K + \Delta U$.

Question

A uniform chain of length 3 m and mass 3 kg overhangs a smooth table with 2 m laying on the table. If $k$ is the kinetic energy of the chain in joule as it completely slips off the table, then the value of $k$ is ......... .
(Take, $g = 10	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

Have you ever watched a chain slowly slip off a table and wondered exactly how fast it's moving when the last link finally falls? It's a mesmerizing piece of physics that perfectly demonstrates the elegant power of energy conservation.

In this problem, we are dealing with a uniform chain of length $3 	ext{ m}$ and mass $3 	ext{ kg}$. It starts with $2 	ext{ m}$ resting peacefully on a smooth table, while the remaining $1 	ext{ m}$ dangles over the edge.

Our mission is to find the kinetic energy of the chain the exact moment it completely slips off the table. Let's dive into the mechanics of this beautiful system.

### The Master Principle: Conservation of Energy

The very first thing we must notice is the word **smooth**. A smooth table means there is absolutely zero friction.

When there are no non-conservative forces like friction or air resistance doing work on our system, the total mechanical energy remains perfectly constant. This is the **Law of Conservation of Mechanical Energy**.

Mathematically, we can state this as:

$K_i + U_i = K_f + U_f$

Here, $K$ represents kinetic energy and $U$ represents gravitational potential energy. The subscripts $i$ and $f$ stand for the initial and final states, respectively.

### Tracking the Center of Mass

To calculate the potential energy, we need a reference point. Let's choose the surface of the table as our **datum level**, meaning $U = 0$ exactly on the table.

Initially, the chain is at rest, so our initial kinetic energy is simply:

$K_i = 0$

Now, what about the initial potential energy $U_i$? The $2 	ext{ m}$ of chain lying on the table is exactly at the datum level, so it contributes nothing to the potential energy. We only need to worry about the $1 	ext{ m}$ hanging part.

Since the chain is uniform, its mass per unit length is $\lambda = 3 	ext{ kg} / 3 	ext{ m} = 1 	ext{ kg/m}$. Therefore, the $1 	ext{ m}$ hanging part has a mass of $1 	ext{ kg}$.

For a uniform object, we can assume all its mass is concentrated at its **center of mass**. The center of mass of the $1 	ext{ m}$ hanging part is exactly in its middle, which is $0.5 	ext{ m}$ below the table.

Because it is below our datum, the height is negative. Let's calculate the initial potential energy:

$U_i = m_{	ext{hang}} \cdot g \cdot h_{	ext{cm},i}$

$U_i = 1 \cdot 10 \cdot (-0.5) = -5 	ext{ J}$

### The Final State

Now, fast forward to the moment the chain completely slips off. The entire $3 	ext{ m}$ length is now hanging vertically in the air.

The total mass is $3 	ext{ kg}$. Where is its center of mass now? It's right in the middle of the $3 	ext{ m}$ chain, which is $1.5 	ext{ m}$ below the table.

Let's calculate the final potential energy:

$U_f = M \cdot g \cdot h_{	ext{cm},f}$

$U_f = 3 \cdot 10 \cdot (-1.5) = -45 	ext{ J}$

The final kinetic energy is exactly what we are trying to find. Let's call it $k$.

### The Final Calculation

We have all our puzzle pieces. Let's plug them back into our master energy conservation equation:

$K_i + U_i = K_f + U_f$

$0 - 5 = k - 45$

Now, it's just a matter of simple algebra. We move the $-45$ to the other side of the equation:

$k = 45 - 5$

$k = 40 	ext{ J}$

And there we have it! The kinetic energy of the chain as it completely slips off the table is exactly **$40 	ext{ Joules}$**.

### The Way Forward: What if there was friction?

This problem was beautifully straightforward because the table was smooth. But what if the table was rough?

If friction were present, mechanical energy would no longer be conserved. As the chain slips, the length of the chain remaining on the table constantly decreases. This means the normal force, and consequently the kinetic friction force, would be variable!

To solve that, you would need to use the **Work-Energy Theorem** ($\Delta K + \Delta U = W_{	ext{friction}}$) and set up an integral to calculate the total work done by the variable friction force. It's a fantastic thought experiment to test your calculus skills!

Enter Numerical Value:

Visualized Solution

$Analyzing the Initial Setup$

$The Master Principle$

$Initial Energy State$

$Final Energy State$

$Applying Conservation$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Conservation of Mechanical Energy

The Master Principle

Tracking the Center of Mass

The Final State

The Final Calculation

The Way Forward

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