Question

A 12.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 104 rev/min. It must be brought to a stop in 30.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.

Answer 1

Solution)

Part a)

The moment of inertia I'
I = m * r^2
E = (1/2) * I * w^2, where E = angular kinetic energy

w is the angular speed in rads/s

We know, w = 2π * f

= 2π * 104 [rev/min] / 60s/min = 10.88 rad/s

Now,

E = (1/2) * I * w^2

= (1/2) * (m * r^2) * 10.88^2

= (1/2) * (12kg * .810m^2) * 10.88^2 = 575.29 J of work to stop (Ans)

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B)

Power = Work / t

= 575.29J / 30s = 19.17 W (Ans)

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Good luck!:)