Question

A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 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

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, $\omega_i=114\ rev/min = 11.93\ rad/s$

Final angular speed of the wheel, $\omega_f=0$

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

$I=mr^2$

$I=49\times (0.73)^2$

$I=26.11\ kg-m^2$

We know that the work done is equal to change in kinetic energy.

$W=\Delta E$

$W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

$W=-\dfrac{1}{2}\times 26.11\times (11.93^2)$

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{1858.05\ J}{22\ s}$

P =84.45 watts

Hence, this is the required solution.