Newton's Second Law for Rotation — Question 8

Question 8

Breakaway torque and running resistance define different regimes

A rotor of inertia $0.20\,\mathrm{kg\,m^2}$ starts at rest. Its bearing can supply any opposing static torque up to $1.0\,\mathrm{N\,m}$; while rotating, the resisting magnitude is $0.60\,\mathrm{N\,m}$.

1. Find the response to a constant applied torque of $0.80\,\mathrm{N\,m}$.

2. Find the initial moving acceleration if the applied torque is instead $1.20\,\mathrm{N\,m}$, and state what happens if it is then reduced to $0.80$ while motion remains positive.

Question 8 - Solution

Breakaway torque and running resistance define different regimes

(a) Derivation and calculation

At rest, the required balancing moment is within static capacity: $$|\tau_s|=0.80\le1.0\ \mathrm{N\,m},\qquad \tau_s=-0.80\ \mathrm{N\,m}.$$ $$\boxed{\alpha=\frac{0.80-0.80}{0.20}=0};\qquad\omega=0.$$ The running resistance is not applicable before breakaway.

(b) Result and interpretation

For the larger drive, $1.20>1.0$, so rest cannot persist. Once moving positively, $$\boxed{\alpha=\frac{1.20-0.60}{0.20}=3.0\ \mathrm{rad/s^2}}.$$ Reducing the drive to 0.80 during positive motion gives $$\boxed{\alpha=\frac{0.80-0.60}{0.20}=1.0\ \mathrm{rad/s^2}}.$$ It continues to speed up despite that drive being insufficient to start from rest.

Physical check

The different responses at 0.80 N m follow from the different static and moving states.