Question

A particle of mass $1\text{kg}$ is moving in a circle of radius $0.36\text{m}$ such that its projection on diameter executes SHM. In $0.1\text{sec}$ interval, particle undergoes angular displacement of ${30}^{\circ }$. Find the force acting on particle at position $B$ if it starts from $A$.

• A. $0.2{\pi }^2\text{N}$
• B. ${\pi }^2\text{N}$
• C. $0.3{\pi }^2\text{N}$
• D. $0.1{\pi }^2\text{N}$

Answer 1

The correct option is B ${\pi }^2\text{N}$

Given in uniform circular motion,
Radius of the circle, $R=0.36\text{m}$
Angular displacement in $0.1\text{sec}=\frac{\pi }{6}\text{rad}$
So, angular speed, $\omega =\frac{d\theta }{dt}$
$\Rightarrow \omega =\frac{\frac{\pi }{6}}{0.1}=\frac{5\pi }{3}\text{rad/s}$
From projection of particle on diameter, $O$ is the mean position and $A$ and $B$ are extreme positions. Thus, force on particle at extreme position (i.e. at $B$),
$F=m{\omega }^{2}R[a={\omega }^{2}x]$
$\Rightarrow F=1\times {\left(\frac{5\pi }{3}\right)}^2\times 0.36={\pi }^2\text{N}$

$\begin{array}{c}\text{Why this question?}\\ \text{Concept - Projection of particle going in uniform}\\ \text{circular motion on its diameter executes SHM with}\\ \text{angular frequency equal to angular speed of uniform}\\ \text{circular motion.}\end{array}$