Answer
a) $I_1=229kg.m^2$ and $I_2=321kg.m^2$
b) $\tau_1=-1272N.m$ and $\tau_2=0$
c) $\omega_1=-27.77rad/s$ and $\omega_2=0$
Work Step by Step
a) System 1: $$I_1=m_2(3m)^2+m_3(5m)^2=6\times9+7\times25=229kg.m^2$$
System 2: $$I_2=m_2(4m)^2+m_1(5m)^2=6\times16+9\times25=321kg.m^2$$
b) System 1:
Force $F$ produces a clockwise (negative) torque with a lever arm of $3m$, so $$\tau_1=-(424N)\times(3m)=-1272N.m$$
System 2:
Force F passes through the axis of rotation, so it produces no torque. In other words, $\tau_2=0$
c) System 1:
The angular acceleration of the system is $$\alpha_1=\frac{\tau_1}{I_1}=-5.55rad/s^2$$
We have $\omega_0=0$ and $t=5s$, so $$\omega_1=\omega_0+\alpha_1t=-27.77rad/s$$
System 2:
Because there is no torque, the system does not accelerate, and the velocity does not change. Therefore, the angular velocity of the system after $5s$ is $\omega_2=0$
