Question

A piece of stone of mass $0.2kg$ is tied to the end of a string $1m$ long and whirled in a horizontal circle about the other end of the string. If it performs $90$ revolutions per minute, calculate

(a) angular velocity

(b) centripetal acceleration

(c) centripetal force and

(d) the tension in the string

Answer 1

Mass, $m=0.2$ $kg$

Radius, $r=1$ $m$

Angular speed, $\omega =90$ $rpm$

$⟹\omega =\frac{90}{60}$

$⟹\omega =1.5$ $rps$

$⟹\omega =3\pi$ $rad/s$

$\left(a\right)$ Angular velocity, $\omega =3\pi$ $rad/s$

$\left(b\right)$ Centripetal acceleration. $a=r{w}^2$

$⟹a=1\times 3\pi \times 3\pi$

$⟹a=9{\pi }^2$ $m/s^2$

$\left(c\right)$ Centripetal Force, $F=ma$

$⟹F=0.2\times 9{\pi }^2$

$⟹F=1.8{\pi }^2$ $N$

$\left(d\right)$ Tension in string, $=$ Centripetal Force

$⟹T=F=1.8{\pi }^2$ $N$