Question

The flywheel of an engine has a moment of inertia 240 kg-m2 about its rotation axis torque is required to bring it up to an angular speed of 440 rev/min in a lime of 7 90s, starting from rest? What N m - Part B

Answer 1

A.We need to determine the angular acceleration of the flywheel to find the torque acting on it.

We have the kinematical law relating the angular velocity and the angular acceleration of the flywheel which is $\omega =\omega _{0}+\alpha t$ .

Since the flywheel starts from rest,$\omega _{0}=0 rad/s$ and final angular velocity $\omega =440 rev/min=440*2\pi /60 rad/s$ as $1 rev=2\pi rad$.

Since t=7.9 sec,$\alpha =\omega -\omega _{0}/t=46-0/7.9=5.8 rad/s^{2}$.

We know that torque $\tau =I\alpha =2.4*5.8=14 Nm$ .

B.The final kinetic energy of the flywheel $K=1/2I\omega ^{2}=1/2*2.4*46^{2}=2539 J$ .