# Illustrations

Example-1

A stone of mass 0.04 kg is whirled by a string in a horizontal circle of radius 0.5 m at 60 rpm. Find the linear speed of the stone and the tension in the string.

Solution

Frequency
Linear speed
Tension in the string is the centripetal force, which keeps the stone in circular motion. Thus,

*f*= 60 rpm = rps = 1 rps*âˆ´*Angular velocity*Ï‰*= 2*Ï€f*= 2 Ã— 3.14 Ã— 1 = 6.28 rad s^{â€“1}*v*=*rÏ‰*= 0.5 Ã— 6.28 = 3.14 m s^{â€“1}Example-2

An electron revolves in a circular orbit of radius 0.5

*Ã—**10*^{â€“10}*m at a linear speed 2**Ã—**10*^{6}*m s*^{ }^{â€“1}. If the mass of the electron is 9.1*Ã— 10*^{âˆ’31}*kg, calculate the centripetal force.*Solution

Mass of electron,
Radius of the circular orbit,
Linear speed of electron,
= 7.28 Ã— 10

*m*= 9.1 Ã— 10^{â€“31}kg*r*= 0.5 Ã— 10^{â€“10}m*v*= 2 Ã— 106 m s^{â€“1}^{â€“8}N.