Question

A string of length $0.5\text{m}$ is fixed at one end and connected with a bob of mass $m$ at the other end. The string makes $\frac{4}{\pi }{\text{revolution s}}^{-1}$ around a vertical axis through its fixed end in a horizontal plane. The angle of inclination of the string with vertical axis is
(Take $g=10{\text{m/s}}^2$)

• A. ${\cos }^{-1}\left(\frac{5}{10}\right)$
• B. ${\cos }^{-1}\left(\frac{5}{12}\right)$
• C. ${\cos }^{-1}\left(\frac{5}{14}\right)$
• D. ${\cos }^{-1}\left(\frac{5}{16}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{5}{16}\right)$

Given:
Length of string, $l=0.5\text{m}$
Angular speed of motion, $\omega =\frac{4}{\pi }{\text{revolution s}}^{-1}=\frac{4}{\pi }\times 2\pi =8\text{rad/s}$
FBD of the bob-


Here, we have applied a pseudo force (centrifugal force) of magnitude $mr{\omega }^2$.
Now, from equilibrium of forces,
$T\cos \theta =mg$ ............$\left(1\right)$
$T\sin \theta =mr{\omega }^2=ml\sin \theta {\omega }^2$
$\Rightarrow T=ml{\omega }^2$ ............$\left(2\right)$
On dividing $\left(1\right)$ by $\left(2\right)$,
$\cos \theta =\frac{g}{l{\omega }^2}$
$\Rightarrow \cos \theta =\frac{10}{0.5\times 8^2}=\frac{5}{16}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{16}\right)$

Tips: Newton's laws are valid only in inertial frames. In non-inertial frames, a pseudo force (like centrifugal force) has to be applied.