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Q.The moment of inertia of a flywheel making 300 revolutions per minute is 0.3 kgm^2. Find the torque required to bring it to rest in 20 s.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2020Subjective· 4mImportance★★★★★
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Convert the flywheel's rotational speed to angular velocity, use ω = 2πN/60, find the angular retardation from ω_f = ω_i + αt, and then find the torque from τ = Iα, giving about 0.47 N·m.

Given:

  • Moment of inertia, I = 0.3 kg m²
  • Initial rotational speed, N = 300 revolutions per minute (rpm)
  • Final angular velocity, ω_f = 0 (brought to rest)
  • Time taken, t = 20 s

Step 1 — Convert rpm to initial angular velocity (rad/s):

ω_i = 2πN / 60 = 2π × 300 / 60 = 2π × 5 = 10π rad/s ≈ 31.42 rad/s

Step 2 — Find the angular retardation using ω_f = ω_i + αt:

0 = 31.42 + α × 20

α = −31.42 / 20 = −1.571 rad/s²

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