Question

When a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 12.0 rad/s to 7.0 rad/s in 5.00 s. if its angular acceleration does not change, how long after it is turned off does it take the fan to stop?

Answer 1

Answer: 13.5

Explanation: Obtained from this

C) ω = 0 = ωo + αt = 6.3rad/s - 0.74rad/s² * t

t = 8.5 s ← after the given 5 s, or 13.5s total

-- alternate: --

ω = 0 = ωo + αt = 10rad/s - 0.74rad/s² * t

t = 13.5 s

Answer 2

The time taken for the fan to stop after it has been turned off is 7 s.

The given parameters;

• mass of the fan, m = 2.75 kg
• length of the blade, r = 18.5 cm = 0.185 m
• initial angular speed, ω1 = 12 rad/s
• final angular speed, ω2 = 7 rad/s
• time of motion, t = 5 s

The angular acceleration of the fan is calculated as;

$\alpha = \frac{\Delta \omega }{t} = \frac{\omega _2 - \omega _1}{t} \\\\\alpha = \frac{7-12}{5} \\\\\alpha = -1 \ rad/s^2$

The angular displacement of the blades  before stopping is calculated as;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta \\\\\0 = \omega_i^2 + 2\alpha \theta\\\\-2\alpha \theta = \omega_i^2 \\\\\theta = \frac{\omega_i^2}{-2\alpha } \\\\\theta = \frac{(7)^2}{-2(-1)} \\\\\theta = 24.5 \ rad$

The time taken for the blade to make this angular displacement is calculated as;

$\theta = \omega_it + \frac{1}{2} \alpha t^2\\\\24.5 = 7t + (0.5)(-1)(t)^2\\\\24.5 = 7t - 0.5t^2\\\\0.5t^2 -7t + 24.5 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method\\\\a = 0.5, \ \ b = -7, \ \ c = 24.5 \\\\t = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-7) \ \ +/- \ \sqrt{(-7)^2 - 4(0.5\times 24.5)} }{2(0.5)}\\\\t = 7 \ s$

Thus, the time taken for the fan to stop after it has been turned off is 7 s.

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