alleryA flywheel of moment of inertia 2 kg m² rotates at 60 rpm. The torque required to bring it to rest in 10 s is:
A0.628 N m
B6.28 N m
C1.257 N m
D12.57 N m
Answer & Solution
Correct answer: C. 1.257 N m
1. Initial ω = 60 rpm = 60 · 2π / 60 = 2π rad s⁻¹.
2. Angular deceleration α = ω / t = 2π / 10 = π/5 rad s⁻².
3. Required torque τ = I α = 2 · π/5 = 2π/5 ≈ 1.257 N m.
4. Distractor 0.628 = π/5 (forgot I); 6.28 = 2π (forgot ÷ t).
_Source: syllabus Physics Ch 6 §6.9 (worked methodology)_
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