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The flywheel of a gasoline engine is required to give up 600 j of kinetic energy while its angular velocity decreases from 790 rev/min to 450 rev/min . what moment of inertia is required?

Respuesta :

Answer:

= 0.2594 Kgm²

Explanation:

Using the formula;

K.E = 1.2×I×ω²

Where; K.E is the kinetic energy, I is the moment of inertia and ω is the angular velocity.

Therefore;

ω1 = 790/60 × 2π

     = 82.74 rad/sec

ω2 = 450/60 × 2π

      = 47.13 rad/sec

600J = 1/2 × I × (82.74²-47.13²)

600 J = 0.5 × I × 4,624.67

600 J = 2312.34 I

I = 0.2594 Kgm²

Moment of inertia = 0.2594 Kgm²

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