A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s.
A) How much work must be done to stop it?
B) What is the required average power?

Explanation:

Given that,

Mass of wheel M = 40kg

Radius of thin hoop R= 0.81m

Angular speed ωi = 438 rev/mins

1rev = 2πrad

ωi = 438 × 2π/60 rad/sec

ωi = 45.87 rad/s

Time take to stop t = 21s

Moment of inertial of a hoop around it central axis can be calculated using

I = MR²

I = 40 × 0.81²

I = 26.24 kgm²

Then,

Change in kinetic energy is given as

∆K.E = Kf — Ki

∆K.E = ½Iωf² — ½Iωi²

Then final angular velocity is zero, since the wheel comes to rest

∆K.E = ½Iωf² — ½Iωi²

∆K.E = ½I × 0² — ½ × 26.24 × 45.87²

∆K.E = 0 — 27,609.43

∆K.E = —27,609.43 J

B. Power

Power = Workdone/Time take

The change in kinetic energy of the loop is equal to the net work done on it

∆K•E = W = —27,609.43

Power = |W| / t

P = 27,609.43/21

P = 1314.73 W

The average power to stop the loop is 1314.73 watts

trig 15.3 cos