Question

A body of mass $0.1kg$ when rotated in a circular path of diameter $1.0m$ on a frictionless horizontal plane by means of string, makes $10$ revolutions in $31.4$ seconds. The centripetal force acting on the body will be

• A. $0.2N$
• B. $0.1N$
• C. $2N$
• D. $1N$

Answer 1

Answer: (A)

$0.2N$

Given, $m=0.1kg,r=\frac{1.0}{2}=0.5m$
$T=\frac{31.4}{10}3.14s=\pi sec$
$\therefore F=mr{\omega }^2=mr{\left(\frac{2\pi }{T}\right)}^2=\frac{4{\pi }^{2}mr}{T^2}$
$=\frac{4\times {\pi }^2\times 0.1\times 0.5}{{\pi }^2}$
$=0.2N$.