4. A uniform chain of length 𝑚 is placed on horizontal frictionless table so that a [10] length 𝑝 of the chain dangles over the side. How long will it take for the chain
to slide off the table?

To determine how long it takes for a uniform chain of length \( m \) with a portion \( p \) hanging off the edge of a frictionless table to slide off completely, we can analyze the forces and motion involved in this scenario. The key concept here is that the chain will accelerate due to the gravitational force acting on the part that is hanging off the table.

Understanding the Problem

Imagine a chain lying on a table, with a segment of it hanging off the edge. The part of the chain that is hanging will experience a gravitational force, causing it to accelerate downward. As this happens, more of the chain will slide off the table, increasing the length of the hanging portion.

Setting Up the Equations

Let’s denote the total length of the chain as \( m \) and the length hanging off the table as \( p \). The mass per unit length of the chain can be expressed as \( \lambda = \frac{M}{m} \), where \( M \) is the total mass of the chain. The gravitational force acting on the hanging part of the chain is given by:

• Force \( F = \lambda g p \)

Here, \( g \) is the acceleration due to gravity. The acceleration \( a \) of the chain can be derived from Newton's second law:

• Acceleration \( a = \frac{F}{M} = \frac{\lambda g p}{M} = \frac{g p}{m} \cdot p = \frac{g p^2}{m} \

Finding the Time to Slide Off

As the chain slides off, the length of the hanging part increases. We can express the length of the chain hanging off the table at any time \( t \) as \( p(t) \). The relationship between the length hanging and the time can be modeled using kinematics. The distance \( d \) that the chain slides off can be described by the equation:

• Distance \( d = \frac{1}{2} a t^2 \)

Since the chain will slide off completely when the entire length \( m \) has fallen, we can set \( d = m \) and substitute for \( a \):

• Setting \( m = \frac{1}{2} \cdot \frac{g p^2}{m} \cdot t^2 \)

Rearranging this gives us:

• Time \( t^2 = \frac{2m^2}{g p^2} \)
• Thus, \( t = \sqrt{\frac{2m^2}{g p^2}} \)

Final Expression

So, the time it takes for the chain to slide off the table completely can be expressed as:

Time \( t = \sqrt{\frac{2m^2}{g p^2}} \)

This formula shows that the time depends on the total length of the chain and the length that is initially hanging off the table, as well as the acceleration due to gravity. The more chain that is hanging, the quicker it will slide off due to the increased gravitational force acting on it.