Answer:
4.62 s
Explanation:
We are given that
Initial angular speed,[tex]\omega=3.4 rad/s[/tex]
[tex]\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad[/tex]
[tex]\omega'=0[/tex]
[tex]\omega'^2-\omega^2=2\alpha \theta[/tex]
Substitute the values
[tex]0-(3.4)^2=2\times 2.5\pi \alpha[/tex]
[tex]\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2[/tex]
[tex]\omega'=\omega+\alpha t[/tex]
[tex]0=3.4-0.736 t[/tex]
[tex]-0.736t=-3.4[/tex]
[tex]t=\frac{-3.4}{-0.736}=4.62 s[/tex]
Hence, the wheel takes 4.62 s to come to rest.
