Question

An elastic string carrying a body of mass $m$ at one end extends by $1.5cm$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

• A. $3.0cm$
• B. $4.5cm$
• C. $1.5cm$
• D. $9.0cm$

Answer 1

Answer: (D)

When the body is at rest, the extension is $1.5cm$ while force acting on it is $mg$
Force when it reaches bottom is $F_b=\frac{m{v}^2}{r}+mg$
Now using energy balance between topmost point and bottom most point ,
$\frac{1}{2}m{v}^2=\frac{1}{2}m(\sqrt{rg}{)}^2+mgh$ (h is the height difference between the top and bottom point ; $\sqrt{gr}$ is critical velocity at topmost point )
$v^2=rg+2gh$
or, $v^2=rg+4gr=5gr$
So, $F_b=6mg$
Using this in: $\frac{e_2}{e_1}=\frac{F_2}{F_1}$, we get:
$e_2=1.5\times 6=9.0cm$