Question

A uniform chain of length $l$ and mass $m$ lies on a smooth horizontal table with its length perpendicular to the edge of the table and small part overhanging. The chain starts sliding down from rest due to the weight of the hanging part. The acceleration and velocity of the chain when length of the hanging portion is $x$
A) $\dfrac{{gx}}{l},\sqrt {\dfrac{{g{x^2}}}{l}} $
B) $\dfrac{{gx}}{l},\sqrt {gx} $
C) $\dfrac{{gx}}{l},\sqrt {gl} $
D) $\dfrac{{gx}}{l},\sqrt {g(l - x)} $

Answer 1

Hint:First calculate the force of the hanging part by figuring out the mass of that part using a unitary method. Then, when force is calculated, use the formula of force to calculate acceleration and then further calculate velocity by differentiating it.

Formula Used:
Force, $F = ma$
Where, $m$ is the mass of the object, $a$ is the acceleration of the object.

Complete step by step solution:
First, we need to calculate the mass of the hanging part of the chain.
We are given that the length of the chain is $l$ and its mass is $m$ therefore, this mass of the chain is divided over the entire $l$ length of the chain. Hence the mass of a unit length of the chain will be $\dfrac{m}{l}$
We are also given that the length of chain hanging is $x$. Therefore, the mass of the hanging chain will be equal to $\dfrac{{mx}}{l}$ .
Now, the force experienced by this part of the chain can be calculated using $F = ma$
Where, $m$ is the mass of the object, $a$ is the acceleration of the object.
In this case, acceleration will be acceleration due to gravity, or $g$. Hence, $F = \dfrac{{mx}}{l}g$
Let acceleration of the chain at any instant be $a$. Therefore, $a = \dfrac{F}{m}$
Or, $a = \dfrac{{\dfrac{{mx}}{l}g}}{m} = \dfrac{{gx}}{l}$
This will be the acceleration of the chain at any instant of time.
Now, we know that acceleration is the derivative of velocity.
Therefore, $\dfrac{{gx}}{l} = \dfrac{{dv}}{{dt}} = \dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}}$
Where, $v$ is the velocity of the chain at any instant of time.
$ \Rightarrow \dfrac{{gx}}{l} = \dfrac{{dv}}{{dx}} \times v$ (since differential of distance is velocity)
We can write this as $\dfrac{{gx}}{l}dx = vdv$
Integrating both sides, we get
$\dfrac{{g{x^2}}}{{2l}} = \dfrac{{{v^2}}}{2}$ or, ${v^2} = \dfrac{{g{x^2}}}{l}$
$ \Rightarrow v = \sqrt {\dfrac{{g{x^2}}}{l}} $
This will be the velocity of the chain at any instant of time.

Therefore, Option A: $\dfrac{{gx}}{l},\sqrt {\dfrac{{g{x^2}}}{l}} $ is the correct option.

Note:Using a unitary method for calculating the mass of a specific length of chain is a compulsory step otherwise we will get the answer corresponding to the entire length of the chain. Pay extra care to derivatives and integrals as you solve the question.