Question

108. A hollow cylinder of mass 2kg and radius 10cm is rotatitng about its axis at 30 rpm.Torque required to stop it after 2 pi rotations in Nm is

Answer 1

Explanation:

Work energy theorem.

W= 21 I(ω f2 −ω i2 )

Here, θ=2π revolution

=2π×2π=4π 2 rad

W i

=3× 602πrad/s

⇒−τθ= 21× 21 mr 2 (0 2−ω i2)

⇒−τ= 4πr221× 2×1×2×(4×10 −2 )(−3× 602π ) 2

⇒τ=2×10 −6 Nm