91Ó°ÊÓ

Given, the travelling path of the object is a circle of radius $5cm$. and in $20$seconds a central angle of role="math" localid="1647239440098" $\frac{1}{3}$radian is swept out.

Therefore,role="math" localid="1647239448898" $\mathrm{θ}=\frac{1}{3}radians$and$t=20seconds$and$r=5cm$

• The angular speed $\mathrm{Ó\neg }$of this object is the angle $\mathrm{θ}$swept out, divided by the elapsed time $t$; that is, localid="1647237995882" $\mathrm{Ó\neg }=\frac{\mathrm{θ}}{t}$
  
• If an object moves around a circle of radius $r$at a constant speed. If $s$is the distance traveled in time $t$around this circle of radius $r$and the angle made is localid="1647241946750" $\mathrm{θ}$, then the linear speed $v$of the object is defined as $v=\frac{s}{t}$and the arc length $s=r\mathrm{θ}$, that is $v=\frac{r\mathrm{θ}}{t}=r·\mathrm{Ó\neg }$

The arc length $s=5·\frac{1}{3}=\frac{5}{3}cm$

Substituting the values in the mentioned formulae, we get,

Angular speed $\mathrm{Ó\neg }=\frac{{\displaystyle \frac{1}{3}}}{20}=\frac{1}{60}radians/second$

and the linear speed$v=5·\frac{1}{60}=\frac{1}{12}cm/second$