Question

A particle of mass $50\text{gm}$ is tied to a string of length $30\text{cm}$ is performing circular motion in a vertical plane. What will be the angular speed of particle, if tension in the string becomes zero at the top most position of circular path. (Take $g=10{\text{m/s}}^2$).

• A. $0\text{rad/s}$
• B. $10\sqrt{3}\text{rad/s}$
• C. $\frac{10}{\sqrt{3}}\text{rad/s}$
• D. $5\sqrt{2}\text{rad/s}$

Answer 1

The correct option is C $\frac{10}{\sqrt{3}}\text{rad/s}$

Given: particle of mass $(m=0.05\text{kg})$ and string of length $=30\text{cm}=0.3\text{m}$

$\therefore r=l=30\text{cm}$
Solving it from frame of reference of revolving particle, centrifugal force $(F_{\text{centrifugal}}=m{\omega }^{2}r)$ will act radially-outwards on particle.

$\Rightarrow$ From particle's frame, particle will be at equilibrium, so balancing the forces in radial direction gives:



$\Rightarrow F_{\text{centrifugal}}=T+mg$
$\Rightarrow m{\omega }^{2}r=T+mg$

$\because T=0$ at the topmost position of vertical circular path)

$\therefore mr{\omega }^2=0+mg$
$\Rightarrow mr{\omega }^2=mg$
$\therefore \omega =\sqrt{\frac{g}{r}}=\sqrt{\frac{10}{0.3}}=\frac{10}{\sqrt{3}}\text{rad/s}$