Question

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration

Answer 1

Answer:

The time of motion is 333.3 s

The angular acceleration is -0.0045 rad/s²

Explanation:

Given;

angular distance of the flywheel, θ = 40 rev

initial angular speed, $\omega_i$ = 1.5 rad/s

When the wheel comes to rest, the final angular speed, $\omega_f$ = 0

The angular acceleration is calculated as follows;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta \\\\0 = (1.5 \ rad/s)^2 + 2\alpha (40 \ rev\times \frac{2\pi \ rad}{1 \ rev} )\\\\0 = 2.25 + 160\pi \alpha\\\\160\pi \alpha = - 2.25\\\\\alpha = -\frac{2.25 }{160\pi} \\\\\alpha = -0.0045 \ rad/s^2$

The time of motion is calculated as;

$\omega_f = \omega _i + \alpha t\\\\0 = 1.5 + (-0.0045t)\\\\0 = 1.5 - 0.0045t\\\\0.0045t = 1.5\\\\t = \frac{1.5}{0.0045} = 333.3 \ s$