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

The time of motion is 333.3 s

The angular acceleration is -0.0045 rad/s²

Step-by-step 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* (2\pi \ rad)/(1 \ rev) )\\\\0 = 2.25 + 160\pi \alpha\\\\160\pi \alpha = - 2.25\\\\\alpha = -(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 = (1.5)/(0.0045) = 333.3 \ s`