Answer:
[tex]I = 0.23 kg m^2[/tex]
Explanation:
As we know that there is no external torque on the system
So here we can say that the angular momentum of the system will remain conserved
So we can say
[tex]m_b v_i L = I\omega + m_b v_f L[/tex]
here we know that
[tex]m_b = 0.628 kg[/tex]
[tex]v_i = 12.1 m/s[/tex]
[tex]v_f = 5.10 m/s[/tex]
[tex]\omega = 2.03 rad/s[/tex]
L = 10.7 cm = 0.107 m
now we will have
[tex]0.628(12.1)(0.107) = I(2.03) + (0.628)(5.10)(0.107)[/tex]
[tex]I = 0.23 kg m^2[/tex]