Answer
a) $v_1=12m/s$, $v_2=9m/s$ and $v_3=18m/s$
b) $KE=1080J$
c) $I=60kg.m^2$
d) $KE_R=1080J$
Work Step by Step
We have $\omega=6rad/s$
a) The tangential speed of each object is:
$v_1=r_1\omega=12m/s$
$v_2=r_2\omega=9m/s$
$v_3=r_3\omega=18m/s$
b) The total kinetic energy of this system is $$KE=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2=1080J$$
c) The moment of inertia of the system is $$I=I_1+I_2+I_3=m_1r_1^2+m_2r_2^2+m_3r_3^2$$ $$I=60kg.m^2$$
d) The rotational kinetic energy of this system is $$KE_R=\frac{1}{2}I\omega^2=1080J$$
