Answer:
Explanation:
Given
Initial linear speed [tex]v_1=8 m/s[/tex]
initial angular velocity [tex]\omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s[/tex]
Speed after 4.1 s is [tex]v_2=2.2 m/s[/tex]
[tex]\omega _2=\frac{2.2}{0.6}=3.66 rad/s[/tex]
using [tex]\omega _2=\omega _1+\alpha t[/tex]
where [tex]\alpha [/tex]is angular acceleration
[tex]3.66=13.33+\alpha \cdot 4.1[/tex]
[tex]\alpha =-2.37 rad/s^2[/tex] i.e. clockwise
(b)angular displacement
[tex]\theta =\omega _1t+\frac{\alpha t^2}{2}[/tex]
[tex]\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}[/tex]
[tex]\theta =54.66-19.75[/tex]
[tex]\theta =34.91 rad[/tex]
