Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved
[tex]L_i = L_f[/tex]
[tex]mv \frac{L}{2} = (I_1 + I_2)\omega[/tex]
here we will have
L = 0.250 m
v = 385 m/s
m = 1.90 gram
now moment of inertia of the plate will be
[tex]I_1 = \frac{ML^2}{3}[/tex]
[tex]I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2[/tex]
[tex]I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2[/tex]
now from above equation
[tex]0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega[/tex]
[tex]\omega = 5.85 rad/s[/tex]
