Answer :
Answer:
[tex]\Delta K = 315.667\,J[/tex]
Explanation:
The final angular speed of the system is given by the help of the Principle of the Angular Momentum Conservation:
[tex][2\cdot (50\,kg)\cdot (3.5\,m)^{2}] \cdot (0.5\,\frac{rad}{s} ) = [2\cdot (50\,kg)\cdot (2\,m)^{2}] \cdot \omega[/tex]
The final angular speed is:
[tex]\omega = 1.531\,\frac{rad}{s}[/tex]
Finally, the change in the rotational kinetic energy is:
[tex]\Delta K = \frac{1}{2}\cdot \left[2\cdot (50\,kg)\cdot (2\,m)^{2}\cdot \left(1.531\,\frac{rad}{s} \right)^{2}-2\cdot (50\,kg)\cdot (3.5\,m)^{2}\cdot \left(0.5\,\frac{rad}{s} \right)^{2}\right][/tex]
[tex]\Delta K = 315.667\,J[/tex]