Question

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions.
(a) At what rate is the flywheel spinning when the power comes back on?
(b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer 1

Answer:

Part a)

$\omega_f = 31.4 rad/s$

$\omega_f = 300 rpm$

Part b)

$t = 75 s$

$N = 311.5 revolutions$

Explanation:

Initial speed of the flywheel is given as

$f_1 = 500 rpm$

initial angular speed is given as

$\omega_i = 2\pi(\frac{500}{60})$

$\omega_i = 52.35 rad/s$

Number of revolutions

$N = 200$

angular displacement is given as

$\theta = 2N\pi$

$\theta = 2(200)\pi = 1256.6 rad$

now we have

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

$1256.6 = 52.35(30) + \frac{1}{2}\alpha (30)^2$

$\alpha = -0.7 rad/s^2$

Part a)

final angular speed of the wheel is given as

$\omega_f = \omega_i + \alpha t$

$\omega_f = 52.35 - 0.7(30)$

$\omega_f = 31.4 rad/s$

$\omega_f = 300 rpm$

Part b)

If the final angular speed becomes zero then time taken by the wheel is given as

$\omega_f - \omega_i = \alpha t$

$0 - 52.35 = -0.7 t$

$t = 75 s$

Now total number of revolution that it makes is given as

$N = \frac{\omega_f + \omega_i}{4\pi} t$

$N = \frac{52.35 + 0}{4\pi} (75)$

$N = 311.5 revolutions$

Answer 2

The final angular speed of the flywheel before the power comes back on is 31.43 rad/s.

The time taken for the flywheel to stop if the power had not come back is 75 s.

The number of revolutions the wheel would have made during this time is 312.5 revolutions.

The given parameters;

• initial angular speed of the wheel, $\omega _i$ = 500 rpm
• mass of the flywheel = 40 kg
• diameter of the flywheel, d = 75 cm
• time, t = 30 s
• number of revolution of the flywheel, N = 200

The angular acceleration of the wheel is calculated as follows;

$\theta = \omega _i t + \frac{1}{2} \alpha t^2\\\\(200\ rev \times\frac{2 \pi \ rad}{1 \ rev} ) = (500 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s})(30 \ s) \ + \ \frac{1}{2} (30 \ s)^2 (\alpha )\\\\1256.8 = 1571 + 450 \alpha \\\\450 \alpha = 1256.8 - 1571\\\\450 \alpha = -314.2\\\\\alpha = \frac{-314.2}{450} \\\\\alpha = -0.698 \ rad/s^2$

(a) The final angular speed of the flywheel is calculated as follows;

$\omega_f = \omega _i \ + \ \alpha t\\\\\omega _f = (500 \ \frac{rev}{1\min} \times \frac{2\pi \ rad }{1 \ rev} \times \frac{1 \min}{60 \ s} ) \ + \ (-0.698\times 30)\\\\\omega_f = 52.37 \ - \ 20.94\\\\\omega _f = 31.43 \ rad/s$

(b) The time taken for the flywheel to stop if the power had not come back, is calculated as follows;

$\omega_f = \omega _i + \alpha t\\\\0 = 52.37 - (0.698t)\\\\0.698t = 52.37\\\\t = \frac{52.37}{0.698} \\\\t = 75 \ s$

The number of revolutions the wheel would have made during this time is calculated as follows;

$\theta = (\frac{\omega _i + \omega _f}{2} )t\\\\\theta = (\frac{52.37 + 0}{2} )75\\\\\theta = 1963.875 \ rad\\\\\theta = \frac{1 \ rev}{2\pi \ rad} \times 1963.875 \ rad\\\\\theta = 312.5 \ rev$

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