Question

A torque applied to a flywheel causes it to accelerate uniformly from a speed of 161 rev/min to a speed of 853 rev/min in 5.0 seconds. determine the number of revolutions n through which the wheel turns during this interval. (suggestion: use revolutions and minutes for units in your calculations.)

Answer 1

First of all, let's convert the time interval into minutes. Since
$60 s: 1 min = 5 s: x$
we find
$\Delta t = \frac{5.0 s}{60 s/min}=0.083 min$

Then we can find the angular acceleration of the flywheel:
$\alpha = \frac{\omega _f - \omega_i}{\Delta t}= \frac{853 rpm-161 rpm}{0.083 min}=8337 rev/min^2$

At this point, we can use the law of motion of an uniformly accelerated rotational motion. The angular displacement after a time $\Delta t$ is given by
$\theta (\Delta t)= \omega_i t + \frac{1}{2} \alpha t^2 =$
$=(161 rpm)(0.083 min)+ \frac{1}{2}(8337 rev/min^2)(0.083 min)^2 =42.1 rev$
So, the flywheel covers 42.1 revolutions.