Work and Rotational Kinetic Energy — Question 10

Question 10

Storage capacity limits how far regenerative braking can proceed

A flywheel has inertia $2.0\,\mathrm{kg\,m^2}$ and initial angular speed $30\,\mathrm{rad/s}$. A regenerative brake stores $75\%$ of the removed kinetic energy; its storage device has $500\,\mathrm J$ of free capacity. Assume this is the only braking mechanism.

1. Can it slow the flywheel to $10\,\mathrm{rad/s}$ without filling the device?

2. Find the angular speed when capacity is first reached and the energy dissipated by then.

Question 10 - Solution

Storage capacity limits how far regenerative braking can proceed

(a) Derivation and calculation

The desired kinetic-energy reduction and stored share are $$\Delta K_{\mathrm{removed}}=\tfrac12(2)(30^2-10^2)=800\ \mathrm J.$$ $$\boxed{E_{\mathrm{stored}}=0.75(800)=600\ \mathrm J}>500\ \mathrm J.$$ The target speed cannot be reached with uninterrupted regeneration into this storage device alone.

(b) Result and interpretation

At capacity, the removed and remaining energies are $$\Delta K_{\mathrm{cap}}=\frac{500}{0.75}=\frac{2000}{3}\ \mathrm J,\qquad K_f=900-\frac{2000}{3}=\frac{700}{3}\ \mathrm J.$$ $$\boxed{\omega_{\mathrm{cap}}=\sqrt{\frac{2K_f}{I}}=\sqrt{700/3}=15.275\ \mathrm{rad/s}}.$$ $$\boxed{Q=(1-0.75)\frac{2000}{3}=\frac{500}{3}=166.67\ \mathrm J}.$$

Physical check

The final kinetic energy, stored energy, and heat sum to $233.33+500+166.67=900$ J.