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

(a) How much work must be done to stop it?

(b) What is the required average power?

Give absolute values for both parts.

Answer :

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.

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