Question

A block is fastened at one end of a wire and is rotated in a vertical circle of radius $R$. Determine the ratio of change in length of the wire at the lowest point to that at the highest point of the circle. Assume that speed of the block at highest and lowest points is the same ($v$).

• A. $\frac{v+gR}{v-gR}$
• B. $\frac{v-gR}{v+gR}$
• C. $\frac{v^2+gR}{v^2-gR}$
• D. $\frac{v^2-gR}{v^2+gR}$

Answer 1

The correct option is C $\frac{v^2+gR}{v^2-gR}$

Case 1: When the block is at the lowest point of the circle.


$\Rightarrow T_l-mg=\frac{m{v}^2}{R}$
$\Rightarrow T_l=\frac{m{v}^2}{R}+mg$

Case 2: When the block is at the highest point of the circle.


$\Rightarrow T_h+mg=\frac{m{v}^2}{R}$
$\Rightarrow T_h=\frac{m{v}^2}{R}-mg$
Now, $\Delta L=\frac{FL}{AY}\text{where}Y$ is the Young's modulus

Change in length at the lowest point
$\Delta L_l=\frac{(\frac{m{v}^2}{R}+mg)\times R}{AY}$
$\Delta L_l=\frac{(m{v}^2+mgR)}{AY}......\left(1\right)$
Similarly,
Change in length at highest point $\Delta L_h$
$\Delta L_h=\frac{(\frac{m{v}^2}{R}-mg)R}{AY}$
$\Delta L_h=\frac{(m{v}^2-mgR)}{AY}......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$\frac{\Delta L_l}{\Delta L_h}=\frac{\frac{m(v^2+gR)}{AY}}{\frac{m(v^2-gR)}{AY}}$
$\Rightarrow \frac{\Delta L_l}{\Delta L_h}=\frac{v^2+gR}{v^2-gR}$
Hence, option (c) is the correct answer.