Question

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 72.0 cm. The power is off for 35.0 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 180 complete revolutions
A.) At what rate is the flywheel spinning when the power comes back on?
= ? rad/s

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?
= ? s

C.) How many revolutions would the wheel have made during this time?
= ? rev

Answer 1

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

• mass of flywheel, $m=40\ kg$

• diameter of flywheel, $d=0.72\ m$

• rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

• duration for which the power is off, $t_0=35\ s$

• no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

Using equation of motion:

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

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

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

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.