Respuesta :
Explanation:
Mass of the wheel, m = 49 kg
Radius of the hoop, r = 0.73 m
Initial angular speed of the wheel, [tex]\omega_i=114\ rev/min = 11.93\ rad/s[/tex]
Final angular speed of the wheel, [tex]\omega_f=0[/tex]
Time, t = 22 s
(a) If I is the moment of inertia of the hoop. It is equal to,
[tex]I=mr^2[/tex]
[tex]I=49\times (0.73)^2[/tex]
[tex]I=26.11\ kg-m^2[/tex]
We know that the work done is equal to change in kinetic energy.
[tex]W=\Delta E[/tex]
[tex]W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)[/tex]
[tex]W=-\dfrac{1}{2}\times 26.11\times (11.93^2)[/tex]
W = -1858.05 Joules
(b) Let P is the average power. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
[tex]P=\dfrac{1858.05\ J}{22\ s}[/tex]
P =84.45 watts
Hence, this is the required solution.
