Answer:
[tex]\omega=177.09\ rad.s^{-1}[/tex]
Explanation:
Given :
Linear velocity of yo-yo at the bottom:
[tex]v^2=u^2+2g.h[/tex]
[tex]v^2=0^2+2\times 9.8\times 1[/tex]
[tex]v=4.4272\ m.s^{-1}[/tex]
Now using the relation between angular and linear velocity:
[tex]\omega=\frac{v}{r}[/tex]
[tex]\omega=\frac{4.4272}{0.025}[/tex]
[tex]\omega=177.09\ rad.s^{-1}[/tex]