Work and Rotational Kinetic Energy — Question 9

Question 9

Torsional spring work reveals a reduced opposite turning angle

A rotor of inertia $0.50\,\mathrm{kg\,m^2}$ is attached to a torsional spring exerting $\tau_s=-k\theta$, with $k=4\,\mathrm{N\,m/rad}$. It is released from rest at $\theta=1\,\mathrm{rad}$. A constant resisting torque of magnitude $0.50\,\mathrm{N\,m}$ opposes motion; static capacity is no larger than this value.

1. Derive the spring energy and find angular speed at $\theta=0$.

2. Find the first negative turning angle, accounting for total angular travel.

Question 9 - Solution

Torsional spring work reveals a reduced opposite turning angle

(a) Derivation and calculation

Define spring energy by the negative work of its restoring torque: $$U(\theta)-U(0)=-\int_0^\theta(-kq)\,dq=\boxed{\frac12k\theta^2}.$$ The spring torque initially exceeds static resistance, so motion begins toward decreasing angle. At zero angle, $$K(0)=\tfrac12(4)(1)^2-(0.50)(1)=1.50\ \mathrm J.$$ $$\boxed{|\omega(0)|=\sqrt{2K/I}=\sqrt6=2.449\ \mathrm{rad/s}},\qquad\omega(0)<0.$$

(b) Result and interpretation

Write the first negative turning angle as $-a$ with $a>0$. The travel from $+1$ to $-a$ is $1+a$, so $$2=\frac12(4)a^2+0.50(1+a)\Rightarrow 2a^2+0.50a-1.50=0.$$ $$(a+1)(2a-1.50)=0\Rightarrow\boxed{a=0.75},\qquad \boxed{\theta_t=-0.75\ \mathrm{rad}}.$$ The root $a=-1$ is not a negative-side amplitude.

Physical check

At the turn, spring energy is 1.125 J and dissipation is 0.875 J, totaling the initial 2 J.