Answer to Question #278079 in Physics for Llll

Question

A flywheel initially rotating at 600 rpm is bought to a stop by a constant torque in 15 a. How many revolutions does the flywheel make before coming to stop

Answer 1

Let's first convert rev/min to rad/s:

$\omega_i=600\ \dfrac{rev}{min}\times\dfrac{1\ min}{60\ s}\times\dfrac{2\pi\ rad}{1\ rev}=62.8\ \dfrac{rad}{s}.$

Then, we can find the angular deceleration of the flywheel:

$\alpha=\dfrac{\omega_f-\omega_i}{t}=\dfrac{0-62.8\ \dfrac{rad}{s}}{15\ s}=-4.19\ \dfrac{rad}{s^2}.$

Let's find the angular displacement of the flywheel:

$\omega_f^2=\omega_i^2+2\alpha \theta,$ $0=\omega_i^2+2\alpha \theta,$ $\theta=-\dfrac{\omega_i^2}{2\alpha}=-\dfrac{(62.8\ \dfrac{rad}{s})^2}{2\times(-4.19\ \dfrac{rad}{s^2})}=471\ rad.$

Finally, we can find how many revolutions does the flywheel make before coming to stop:

$n=\dfrac{\theta}{2\pi\ \dfrac{rad}{rev}}=\dfrac{471\ rad}{2\pi\ \dfrac{rad}{rev}}=75\ rev.$