tolong dibantu caranya

Dinamika Rotasi
▷Titik Berat

Benda 1 :
x₁ = ½ (8) = 4
y₁ = ½ (10) = 5
A₁ = (8)(10) = 80

Benda 2 (bolong) :
x₂ = 4 + (½ (2)) = 5
y₂ = 4 + (½ (4)) = 6
A₂ = (2)(4) = 8

Maka titik berat :
x₀ = (A₁x₁ - A₂x₂) / (A₁ - A₂)
x₀ = (80 × 4 - 8 × 5) / (80 - 8)
x₀ = 3 ⁸/₉

y₀ = (A₁y₁ - A₂y₂) / (A₁ - A₂)
y₀ = (80 × 5 - 8 × 6) / (80 - 8)
y₀ = 4 ⁸/₉

Koordinatnya (x₀ , y₀) = (3 ⁸/₉ , 4 ⁸/₉)