tolongin no 30 .... pake caranya... tolongin ya please....

Dinamika Rotasi
- Momen Inersia

[tex] m_A = 2 \ kg \\
m_B = 4 \ kg \\
m_C = 2 \ kg \\
R_A = 0 \ m \\
R_B = 0,8 \ m \\
R_C = 0,6 \ m [/tex]

[tex] \sum{I_A} [/tex] = __ ?

[tex] \sum{I_A} = m_AR_{A}^{2} + m_BR_{B}^{2} + m_CR_{C}^{2} \\
\sum{I_A} = (0) + (4 \cdot 0,8^2) + (2 \cdot 0,6^2) \\
\sum{I_A} = 3,28 \ kgm^2 [/tex]

Jawaban : D