Given
Mass of the particles,
[tex]\begin{gathered} m_1=10.3\text{ kg} \\ m_2=2.5\text{ kg} \\ m_3=2.7\text{ kg} \end{gathered}[/tex]
Their respective coordinates are
[tex]\begin{gathered} (x_1,y_1)=(-12,-5) \\ (x_2,y_2)=(4,3) \\ (x_3,y_3)=(14,1) \end{gathered}[/tex]
The angular velocity is
[tex]\omega=15\text{ rad/s}[/tex]
To find
Rotational kinetic energy of the system
Explanation
The moment of inertia about the y axis is
[tex]\begin{gathered} I=m_1x_1^2+m_2x_2^2+m_3x_3^2 \\ \Rightarrow I=10.3\times12^2+2.5\times4^2+2.7\times^14^2 \\ \Rightarrow I=2052.4\text{ kgm}^2 \end{gathered}[/tex]
The rotational kinetic energy is
[tex]\begin{gathered} K=\frac{1}{2}I\omega^2 \\ \Rightarrow K=\frac{1}{2}\times2052.4\times15^2 \\ \Rightarrow K=230895J \end{gathered}[/tex]
Conclusion
The rotational kinetic energy is A.230895 J
