Answer:
The angular speed after 6s is [tex]\omega = 1466.67s^{-1}[/tex].
Explanation:
The equation
[tex]I\alpha = Fd[/tex]
relates the moment of inertia [tex]I[/tex] of a rigid body, and its angular acceleration [tex]\alpha[/tex], with the force applied [tex]F[/tex] at a distance [tex]d[/tex] from the axis of rotation.
In our case, the force applied is [tex]F = 22N[/tex], at a distance [tex]d = 6cm =0.06m[/tex], to a ring with the moment of inertia of [tex]I =mr^2[/tex]; therefore, the angular acceleration is
[tex]$\alpha =\frac{Fd}{I} $[/tex]
[tex]$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $[/tex]
[tex]\alpha = 244.44\: s^{-2}[/tex]
Therefore, the angular speed [tex]\omega[/tex] which is
[tex]\omega = \alpha t[/tex]
after 6 seconds is
[tex]\omega = 244.44$\: s^{-2}* 6s[/tex]
[tex]\boxed{\omega = 1466.67s^{-1}}[/tex]
