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

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

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

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy$=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$