Answer:
The final angular speed becomes [tex]264rad/sec[/tex]
Explanation:
The work done by the force shall appear as the rotational kinetic energy of the system.
We know that work done by force is given by
[tex]W=Force\times displacement\\\\W=44\times 0.9=39.6Joules[/tex]
Now the rotational kinetic energy of the disc equals
[tex]K.E=\frac{1}{2}I\omega ^{2}[/tex]
We know that moment of inertia of disc is given by
[tex]I=\frac{1}{2}mr^{2}[/tex]
Thus applying values we get
[tex]39.6=\frac{1}{2}\times 7\times.21^{2}\times \omega _{f}\\\\\therefore \omega _{f}=\frac{39.6}{0.15}=264rad/sec[/tex]
