Answer to Question #123780 in Mechanics | Relativity for Ojugbele Daniel

Question

The flywheel of motor has a mass of 300kg and a radius of gyration of 1.5m. the motor develops a constant torque of 2000Nm and the flywheel starts from rest.
(i) what is the angular acceleration of the flywheel?
(ii) what will be it's angular velocity after making four revolutions?
(iii) how much work is done by the motor during the first four revolutions?

Answer 1

Solution: the moment of inertia of a flywheel is equal to

$J=\frac{1}{2}\times m\times R^2,then\;\epsilon=\frac{M}{J},\;or\;\epsilon=\frac{2\times M}{m\times R^2};\\\epsilon=\frac{2\times2000}{300\times2,25}=5,926s^{-2};\\\phi=\frac{\omega^2}{\epsilon};\phi=2\times\pi\times N;\omega=\sqrt{2\times\pi\times N\times\epsilon}=\\=\sqrt{2\times3,14\times 4\times5,926}=12,201s^{-1};\\A=E_k=\frac{J\times\omega^2}{2}=\frac{\frac{1}{2}\times m\times R^2\times\omega^2}{2}=\\=\frac{\frac{1}{2}\times 300\times 2,25\times148,861}{2}=25120,294J.\\Answer:\;\epsilon=5,926s^{-2};\\\omega=12,201s^{-1};A=25120,294J$