Question

Tolong di bantu buat besok

Answer 1

Verified answer

Dinamika Rotasi
• Titik Berat

Benda I : Persegi panjang Besar
x₁ = ½(6) = 3 cm
y₁ = ½(8) = 4 cm
A₁ = (6)(8) = 48 cm²

Benda II : Persegi panjang bolong

x₂ = ½(2) + 2 = 3
y₂ = ½(6) + 2 = 5
A₂ = (2)(6) = 12 cm²

MAKA :
x₀ = (A₁x₁ - A₂x₂) / (A₁ - A₂)
x₀ = (48 × 3 - 12 × 3) / (48 - 12)
x₀ = 3 cm

y₀ = (A₁y₁ - A₂y₂) / (A₁ - A₂)
y₀ = (48 × 4 - 12 × 5) / (48 - 12)
y₀ = 3 ⅔ cm

Jadi , titik beratnya (3 , 3⅔)
Opsi : E