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A 26.0 kg wheel, essentially a thin hoop with radius 1.30 m, is rotating at 297 rev/min. It must be brought to a stop in 23.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.

User Fhollste
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1 Answer

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Answer:

Work done on the loop to stop it = 21,269.1496 J

Average power P = 924.7456 watt

Step-by-step explanation:

Mass of the wheel M = 26.0 Kg

Radius r = 1.30 m

The wheel is rotating at a speed of 297 rev/min

1 minute = 60 seconds,

So,

The wheel is rotating at a speed of 297/60 rev/sec

Initial angular speed (ω₁) = 2π(297/60) = 31.1143 rad/s

Final angular speed (ω₂) = 0 rad/s

Time taken to stop , (t) = 23 s econds

Moment of inertia I of a circular hoop around its central axis = Mr²

Where m is the mass of the wheel and r is the radius of the wheel

Thus, I = 26.0×(1.30)² Kgm² = 43.94 Kgm²

(a)

Work done to stop it is the difference in the kinetic energy of the initial and the final system. So,

Work done W = (1/2)I(ω₂² - ω₁²) = 0.5*43.94(0 - (31.1143)²) = -21,269.1496 J

Thus, work done on the loop to stop it = - 21,269.1496 J

The answer has to be answered in absolute value so, Work = |-21,269.1496 J| = 21,269.1496 J

(b)

Average power P = |W|/t = 21,269.1496 J/23.0 s = 924.7456 watt

User Bhavinjr
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