Work and Rotational Kinetic Energy — Question 2

Question 2

The area under a torque-angle graph sets the speed

A rotor has inertia $2\,\mathrm{kg\,m^2}$ and starts at $\theta=0$ with $\omega=+1\,\mathrm{rad/s}$. Its net torque is $\tau(\theta)=6-2\theta\,\mathrm{N\,m}$ with numerical angle in radians.

1. Find its angular speed when it first reaches $\theta=2\,\mathrm{rad}$.

2. Find the first positive turning angle and explain why negative torque does not immediately imply reversal.

Question 2 - Solution

The area under a torque-angle graph sets the speed

(a) Derivation and calculation

Integrate torque over angle and apply the work-energy theorem: $$W_{0\to2}=\int_0^2(6-2\theta)\,d\theta =[6\theta-\theta^2]_0^2=8\ \mathrm J.$$ $$K_i=\tfrac12(2)(1)^2=1\ \mathrm J,\quad K_f=9\ \mathrm J, \qquad\boxed{\omega(2)=+\sqrt{\frac{2(9)}2}=3\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The energy along the forward branch is $$K(\theta)=1+6\theta-\theta^2.$$ $$K(\theta_t)=0\Rightarrow\theta_t=3\pm\sqrt{10} \Rightarrow\boxed{\theta_t=3+\sqrt{10}=6.162\ \mathrm{rad}}.$$ The negative root is outside the first forward passage. Torque changes sign at 3 rad, but reversal occurs only when the positive kinetic energy has been exhausted.

Physical check

Energy is maximal at the zero of torque, then decreases until the turning point.