From rotational kinematic equations,
ωf2−ωi2=2αθ
02−(3602π)2=2α2π(2π)
α=−8001rads−2
∴ Torque required τ=∣Iα∣
=22(4×10−2)2×8001
=80016×10−4
=2×10−6Nm
NEET UG 2019 — Physics Mechanics
A solid cylinder of mass 2kg and radius 4cm is rotating about its axis at the rate of 3rpm. The torque required to stop after 2π revolutions is
Held on 30 Apr 2019 · Verified 9 Jul 2026.
2×10−6Nm
2×10−3Nm
12×10−4Nm
2×106Nm
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