Question

A clindrical rod of uniform density is located with its center at the origin, and its axis along the axis. It rotates about its center in the plane, making one revolution every 0.02 s. The rod has a radius of 0.06 m, length of 0.6 m, and mass of 4 kg. What is the rotational kinetic energy of the rod?

Answer 1

Answer:

6093.2328 J

Explanation:

For cylindrical rod moment of inertia will be

$I_X=I_Y=\frac{1}{12}m(3r^2+h^2)$

$I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236$ $kg -m^2$

we have given time =0.02 sec

Angular speed =$\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec$

Rotational KE = $\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J$