Work and Rotational Kinetic Energy — Question 3

Question 3

Motor input splits between kinetic energy and bearing heating

A rotor with $I=0.40\,\mathrm{kg\,m^2}$ starts at $2\,\mathrm{rad/s}$. A motor applies $4.0\,\mathrm{N\,m}$ while a constant bearing torque of magnitude $1.5\,\mathrm{N\,m}$ opposes its positive motion through $8\,\mathrm{rad}$.

1. Find motor work, bearing dissipation and final angular speed.

2. Check the result using angular acceleration and the kinematic speed-angle relation.

Question 3 - Solution

Motor input splits between kinetic energy and bearing heating

(a) Derivation and calculation

Keep supplied work and dissipated energy distinct: $$\boxed{W_m=4.0(8)=32\ \mathrm J},\qquad \boxed{Q_b=1.5(8)=12\ \mathrm J},\qquad W_b=-12\ \mathrm J.$$ $$K_f=\tfrac12(0.40)(2)^2+32-12=20.8\ \mathrm J.$$ $$\boxed{\omega_f=\sqrt{\frac{2K_f}{I}}=\sqrt{104}=10.198\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The independent constant-acceleration calculation gives $$\alpha=\frac{4.0-1.5}{0.40}=6.25\ \mathrm{rad/s^2}.$$ $$\omega_f^2=\omega_i^2+2\alpha\Delta\theta=4+2(6.25)(8)=104.$$ Both methods use the same net torque and produce the same speed.

Physical check

The 32 J motor input exceeds the 20 J increase in kinetic energy by the 12 J dissipated.