Question

A chain $AB$ of mass $m$ and length $L$ is hanging on a smooth horizontal table as shown in the figure. If it is released from the position shown then the displacement of centre of mass of chain in magnitude, when end $A$ moves a distance $\frac{L}{2}isX\sqrt{2}m$.
Find $X$. ($L=32m$)

• A. $8$
• B. $9$
• C. $10$
• D. $16$

Answer 1

Answer: (A)

$8$

In the first case, we consider two parts one of mass $3m/4$ and length $3L/4$ and other of mass $m/4$ and length $L/4$.
The center of mass of these masses is calculated as

$x_1=\frac{\left(3m/4\right)(-3L/8)+\left(m/4\right)\left(0\right)}{m}=-9L/32=-9$
and
$y_1=\frac{\left(3m/4\right)\left(0\right)+\left(m/4\right)(-L/8)}{m}=-L/32=-1$

Thus the coordinates of the center of mass is $(-9,-1)$

Similarly, when the chain moves by $L/2$ distance we get the new position of the center of mass as:

$x_2=\frac{\left(m/4\right)(-L/8)+\left(3m/4\right)\left(0\right)}{m}=-9L/32=-1$
and
$y_2=\frac{\left(m/4\right)\left(0\right)+\left(3m/4\right)(-3L/8)}{m}=-L/32=-9$

Thus the coordinates of the center of mass in this case is $(-1,-9)$
Thus the distance between the two position of the center of mass is calculated as $8\sqrt{2}$. Thus the value of $X$ is $8$.