ANSWER:
2167.68 kg*m^2
STEP-BY-STEP EXPLANATION:
Given:
mr = 8 kg
L = 6 m
ms = 36.25 kg
R = 1.5 m
Moment of inertia of sphere about its center is:
[tex]I_{CM}=\frac{2}{5}m_s\cdot R^2[/tex]
Using paraller theorem, moment of inertia of shpere about end of rop is:
[tex]I_{\text{rod}}=m_s\cdot(R+L)^2+\frac{1}{3}m_r\cdot L^2[/tex]
Therefore:
[tex]\begin{gathered} I=I_{cm}+I_{\text{rod}}_{} \\ I=\frac{2}{5}\cdot m_s\cdot R^2+m_s\cdot(R+L)^2+\frac{1}{3}\cdot m_r\cdot L^2 \end{gathered}[/tex]
Replacing:
[tex]\begin{gathered} I=\frac{2}{5}\cdot36.25\cdot1.5^2+36.25\cdot(1.5+6)^2+\frac{1}{3}\cdot8\cdot6^2 \\ I=2167.69\text{ kg}\cdot m^2 \end{gathered}[/tex]
The moment of inertia is 2167.68 kg*m^2
