## 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)

Question

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?

in progress
0

Physics
4 hours
2021-07-22T04:58:21+00:00
2021-07-22T04:58:21+00:00 2 Answers
0 views
0
## Answers ( )

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final – K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

Answer:

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