Question

A ball of mass $2\text{kg}$ is tied to a string $3\text{m}$ long and is rotated at a constant speed of $30\text{m/s}$ in a vertical circle. The ratio of the tensions at the top and bottom is (Take $g=10{\text{m/s}}^2$)

• A. $\frac{39}{41}$
• B. $\frac{11}{18}$
• C. $\frac{39}{31}$
• D. $\frac{29}{31}$

Answer 1

The correct option is D $\frac{29}{31}$

Tension in the string at top of circle
$T_{\text{top}}=\frac{m{v}^2}{r}-mg\dots \dots \left(1\right)$


Tension in the string at bottom of circle
$T_{\text{bottom}}=\frac{m{v}^2}{r}+mg\dots \dots \left(2\right)$


Divide $\left(1\right)$ by $\left(2\right)$
$\Rightarrow \frac{T_{\text{top}}}{T_{\text{bottom}}}=\frac{\frac{m{v}^2}{r}-mg}{\frac{m{v}^2}{r}+mg}=\frac{\frac{v^2}{r}-g}{\frac{v^2}{r}+g}$
$=\frac{\frac{30\times 30}{3}-10}{\frac{30\times 30}{3}+10}=\frac{\frac{900}{3}-10}{\frac{900}{3}+10}=\frac{300-10}{300+10}$
$\therefore \frac{T_{\text{top}}}{T_{\text{bottom}}}=\frac{290}{310}=\frac{29}{31}$