Question

A $4\text{kg}$ ball tied to the end of a cord of length $1\text{m}$ is swinging in a vertical circle. The maximum speed at which it can swing, if the cord can sustain a maximum tension of $163.6\text{N}$ will be: [Take $g=9.8{\text{m/s}}^2$]

• A. $v=\sqrt{11.5}\text{m/s}$
• B. $v=\sqrt{31.1}\text{m/s}$
• C. $v=\sqrt{18}\text{m/s}$
• D. $4\text{m/s}$

Answer 1

The correct option is B $v=\sqrt{31.1}\text{m/s}$


If we solve from the frame of the swinging ball, then a centrifugal force will act in radially outward direction.
$F_{\text{centrifugal}}=\frac{m{v}^2}{r}=\frac{m{v}^2}{l}...\left(i\right)$

Maximum tension will occur at the bottommost position, where weight of the ball $mg$ and $F_{\text{centrifugal}}$ will add up, as both are acting vertically downwards.

Applying equilbrium condition $a=0$ from ball's frame,
$T=mg+\frac{m{v}^2}{l}$
$\therefore T_{\text{max}}=mg+\frac{m{v}^2}{l}...\left(ii\right)$
Given, maximum tension in string,
$T_{\text{max}}=163.6\text{N}$
$\Rightarrow \frac{m{v}^2}{l}=T_{\text{max}}-mg$
or, $\frac{m{v}^2}{l}=163.6-4\times 9.8$
$\frac{4\times v^2}{1}=163.6-(4\times 9.8)$
$\therefore v=\sqrt{31.1}\text{m/s}$
This corresponds to the maximum speed of the bob without breaking the cord.