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- Hubungan roda
2 roda bersinggungan
dx / dy = 4/3
ωx = 2,5 π rad/s
ωy = ___ ?
Pada hubungan 2 roda bersinggungan, kecepatan linear keduanya sama
Vx = Vy
ωx Rx = ωy Ry
ωy = ωx · Rx/Ry
ωy = ωx · dx/dy
ωy = 2,5 π · 4/3
ωy = 3¹/₃π ≈ 3,3 π rad/s ← jawaban