Given,
The length of the rods;
L=0.780 m
l=0.500 m
The angular acceleration, α=0.750 rad/s²
The masses;
m_A=4.00 kg
m_B=3.00 kg
m_C=5.00 kg
m_D=2.00 kg
The moment of inertia of the given system of masses is given by,
[tex]\begin{gathered} I=\Sigma mr^2 \\ =m_A(\frac{L}{2})^2+m_B(\frac{L}{2})^2+m_C(\frac{L}{2})^2+m_D(\frac{L}{2})^2 \\ =(\frac{L}{2})^2(m_A+m_B+m_C+m_D) \end{gathered}[/tex]Where r is the distance between each mass and the axis of rotation.
On substituting the known values,
[tex]\begin{gathered} I=(\frac{0.780}{2})^2(4.00+3.00+5.00+2.00) \\ =2.13\text{ kg}\cdot\text{m}^2 \end{gathered}[/tex]The torque required is given by,
[tex]\tau=I\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=2.13\times0.750 \\ =1.6\text{ Nm} \end{gathered}[/tex]Thus the torque that must be applied to cause the required acceleration is 1.6 Nm