Answer to Question #274111 in Physics for thea

Question

The force due to weight of a certain flywheel is 5000 lb, having a radius of gyration of 3 ft. If there is a a driving torque of 600 lb - ft and a resisting torque of 250 lb - ft, find the time to increase its speed from 20 rev/min to 80 rev/min. (Hint : The moment of inertia is directly proportional to the square of radius of gyration, having the constant of proportionality to be the mass of the system.)

Answer 1

1) Determine the moment of inertia:

$$I=mr^2.$$

2) Determine the total torque:

$$T=T_d-T_r.$$

3) Determine the angular acceleration:

$$\alpha=\frac{T}{I}=\frac{T_d-T_f}{mr^2}.$$

4) The time is

$$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{mr^2(\omega_f-\omega_i)}{T_d-T_f},\\\space\\ t=\frac{5000·3^2(80-20)}{600-250}=7714\text{ s}.$$