A26.0 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.

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

Average power P = 924.7456 watt

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