Answer:
6093.2328 J
Explanation:
For cylindrical rod moment of inertia will be
[tex]I_X=I_Y=\frac{1}{12}m(3r^2+h^2)[/tex]
[tex]I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236[/tex] [tex]kg -m^2[/tex]
we have given time =0.02 sec
Angular speed =[tex]\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec[/tex]
Rotational KE = [tex]\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J[/tex]
