Question

A $10\text{kg}$ mass and a $0.5\text{kg}$ mass hang at two ends of a string that passes over a smooth tube as shown in the figure. The $0.5\text{kg}$ mass moves on a circular path in a horizontal plane about vertical axis. The length of string connecting the $0.5\text{kg}$ mass to the top of the tube is $4\text{m}$ and it makes ${30}^{\circ }$ with the vertical. Then what should be the frequency of revolution (in $\text{cycles/sec}$) of $0.5\text{kg}$ mass so that $10\text{kg}$ mass remains stationary ?

• A. $\frac{\sqrt{5g}}{4\pi }$
• B. $\frac{\sqrt{5g}}{2\pi }$
• C. $\frac{\sqrt{109g}}{2\pi }$
• D. $\frac{\sqrt{109g}}{4\pi }$

Answer 1

The correct option is B $\frac{\sqrt{5g}}{2\pi }$

The FBD of $10\text{kg}$ and $0.5\text{kg}$ have been shown below:



For $10kg$ mass:-
$T=10g-\left(1\right)$

For $0.5\text{kg}$ mass:-
$T\cos {30}^{\circ }=0.5g$ and
$T\sin {30}^{\circ }=m{\omega }^{2}r=0.5{\omega }^{2}r$
(provides required centripetal force)

$r=4\sin {30}^{\circ }$ (from geometry of figure in question)
$\therefore T\sin {30}^{\circ }=0.5\times 4\sin {30}^{\circ }\times {\omega }^2$
or $T=2{\omega }^2$
$\Rightarrow 10g=2{\omega }^2$ (from $\left(1\right)$)
$\Rightarrow {\omega }^2=5g$

Now, the formula which relates angular speed and frequency is given by:
$f=\frac{1}{T}=\frac{\omega }{2\pi }$
$\therefore f=\frac{\sqrt{5g}}{2\pi }$