Answer
See answers.
Work Step by Step
a. The kinetic energy of the system is the total kinetic energy of the two masses, because the rod is “very light”, i.e., massless.
Let 1 represent the heavier mass and 2 the lighter mass.
$$KE=\frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2$$
$$KE=\frac{1}{2}m_1r_1^2\omega_1^2+\frac{1}{2}m_2r_2^2\omega_2^2$$
$$KE=\frac{1}{2}(m_1+m_2)r^2\omega^2=\frac{1}{2}(7.00kg)(0.210m)^2(5.60rad/s)^2=4.84J$$
b. The net force on each object is its mass multiplied by its acceleration, which is centripetal acceleration $\omega^2 r$.
$$F_{net,1}=m_1\omega_1^2r_1=(4.00kg)(5.60rad/s)^2(0.210m)=26.3N$$
$$F_{net,2}=m_2\omega_2^2r_2=(3.00kg)(5.60rad/s)^2(0.210m)=19.8N$$
