Answer:
[tex]K_r=5992J[/tex]
Explanation:
The rotational kinetic energy is given by:
[tex]K_r=\frac{1}{2}I*\omega^2[/tex]
the moment of inertia for this case is given by:
[tex]I=\frac{1}{3}*m*l^2\\I=0.460kg.m^2[/tex]
[tex]\omega=\frac{V_{rpm}}{60}*2\pi\\\\\omega=\frac{577}{60}*2\pi\\\\\omega=60.4rad/s[/tex]
So the total kinetic energy is:
[tex]K_r=5*\frac{1}{2}*0.460kg.m^2*(60.4rad/s)^2\\K_r=4195J[/tex]
