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As per your syllabus, the rotational analogue of v² = v₀² + 2 a x is which?
Aω = ω₀ + α (θ − θ₀)
Bω² = ω₀² + α θ
Cθ = ω₀ t + (1/2) α t² only
Dω² = ω₀² + 2 α (θ − θ₀)
Answer & Solution
Correct answer: D. ω² = ω₀² + 2 α (θ − θ₀)
syllabus eqn 6.38: ω² = ω₀² + 2 α (θ − θ₀).
Useful when time is unknown.
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