Answer:1.6 rad/s
Explanation:
Given
moment of Inertia of disk [tex]I=5 kg-m^2[/tex]
radius of disc [tex]r=0.25 m[/tex]
Force [tex]F=8 N[/tex]
Torque [tex]T=I\alpha =F\cdot r[/tex]
[tex]5\times \alpha =8\times 0.25[/tex]
[tex]\alpha =0.4 rad/s^2[/tex]
using
[tex]\theta =\omega _0\times t+\frac{\alpha t^2}{2}[/tex]
[tex]\pi =0+\frac{0.4t^2}{2}[/tex]
[tex]2\pi =0.4t^2[/tex]
[tex]t^2=5\pi [/tex]
[tex]t=\sqrt{5\pi }[/tex]
[tex]t=3.96 s[/tex]
[tex]\omega =\omega _0+\alpha t[/tex]
[tex]\omega =0+0.4\times 3.96[/tex]
[tex]\omega =1.58 rad/s\approx 1.6 rad/s[/tex]
