Question

A small bob is connected to one end of two identical massless strings each of length $12\text{cm}$. The other ends of the strings are fixed to a vertical rod as shown in the figure. If the tension in the first string $T_1$ is 2 times that of the tension in the second string $T_2$, then what will be the angular velocity of the bob?

• A. $17\text{rad/s}$
• B. $23\text{rad/s}$
• C. $29.4\text{rad/s}$
• D. $32.2\text{rad/s}$

Answer 1

The correct option is A $17\text{rad/s}$

FBD of the bob:


From FBD, using $\sum F_x=m{a}_x;\sum F_y=m{a}_y$
We get,
$T_1\cos {30}^{\circ }=T_2\cos {30}^{\circ }+mg$ ... (1)
($\because a_y=0$)
$T_1\sin {30}^{\circ }+T_2\sin {30}^{\circ }=m{a}_x$ ... (2)
$\therefore$ Components of tension in radial direction provides necessary centripetal force.
i.e $a_x=\frac{v^2}{r}$
$\therefore T_1\sin {30}^{\circ }+T_2\sin {30}^{\circ }=\frac{m{v}^2}{r}$... (3)

Dividing (3) $÷$ (1), we get
$\frac{T_1\sin {30}^{\circ }+T_2\sin {30}^{\circ }}{T_1\cos {30}^{\circ }-T_2\cos {30}^{\circ }}=\frac{v^2}{rg}$
Given $T_1=2T_2$
$\therefore \frac{3T_2\sin {30}^{\circ }}{T_2\cos {30}^{\circ }}=\frac{v^2}{rg}$
or $3\tan {30}^{\circ }=\frac{v^2}{rg}$

From question figure, $r=12\sin {30}^{\circ }=6\text{cm}$
$\therefore 3\tan {30}^{\circ }=\frac{r{\omega }^2}{g}$
$\Rightarrow {\omega }^2=\frac{3g}{r}\tan {30}^{\circ }$
$=\frac{3\times 10}{6\times {10}^{-2}}\times \frac{1}{\sqrt{3}}$
$\Rightarrow {\omega }^2=289$
$\Rightarrow \omega =17\text{rad/s}$