Question

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 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

Step-by-step explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

`\omega =(2\pi * 469)/(60)=49.11 rad/s`

t=16 s

Angular deceleration in 16 s

`\omega =\omega _0+\alpha \cdot t`

`\alpha =(\omega )/(t)=(49.11)/(16)=3.069 rad/s^2`

Moment of Inertia
`I=mr^2=13* 1.8^2=42.12 kg-m^2`

Change in kinetic energy =Work done

Change in kinetic Energy
`=(I\omega ^2)/(2)-(I\omega _0^2)/(2)`

`\Delta KE=(42.12* 49.11^2)/(2)=50,792.34 J`

(a)Work done =50.79 kJ

(b)Average Power

`P_(avg)=(E)/(t)=(50.792)/(16)=3.174 kW`