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 a 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 ? (Take $g=10{\text{m/s}}^2)$

• A. $\frac{5}{2\pi \sqrt{2}}$
• B. $\frac{10}{\pi \sqrt{2}}$
• C. $\frac{3}{2\pi \sqrt{2}}$
• D. $\frac{5}{\pi \sqrt{2}}$

Answer 1

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

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



For $10\text{k}g$ 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)

$\therefore T\sin {30}^{\circ }=0.5\times 4\sin {30}^{\circ }\times {\omega }^2$

$\Rightarrow T=2{\omega }^2$

$\Rightarrow 10g=2{\omega }^2$ (from eqaution $\left(1\right)$)

$\Rightarrow {\omega }^2=5g$

Now, the formula which relates angular speed and frequency is given by:

$f=\frac{\omega }{2\pi }$

$\therefore f=\frac{\sqrt{5g}}{2\pi }=\frac{\sqrt{5\times 10}}{2\pi }=\frac{5}{\pi \sqrt{2}}\text{cycles/sec}$

Hence, option $\left(d\right)$ is the correct answer.