Physics · Basic concepts of rotational motion, moment of a force, torque, angular momentum, conservation of angular momentum and its applications
A ring type flywheel of mass and diameter is rotating at the rate of Then
A ring type flywheel of mass \( 100 k g \) and diameter \( 2 m \) is rotating at the rate of \( \frac{5}{11} r e v / s . \) Then
- A. the moment of inertia of the wheel is 100 kgm \( ^{-2} \).
- B. the kinetic energy of rotation of the flywheel is \( 5 x \) \( 10^{3} J \)
- C. the angular momentum associated with the flywheel is \( 10^{3} J s \)
- D. the flywheel, if subjected to a retarding torque of \( 250 N m, \) will come to rest in \( 4 s \)
Step-by-step solution
For a ring flywheel, I = m r^2 = 100 kg × (1 m)^2 = 100 kg m^2. Option B: kinetic energy = 1/2 I ω^2 = 0.5 × 100 × (2π × 5/11)^2 ≈ 407.85 J, not 5000 J. Option C: angular momentum = I ω = 100 × (10π/11) ≈ 285.6 J s, not 1000 J s. Option D: angular acceleration α = τ/I = 250/100 = 2.5 rad/s^2, stopping time t = ω/α = (10π/11)/2.5 ≈ 1.14 s, not 4 s. Therefore only A is correct (despite the unit typo).
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