Newton's Second Law for Rotation — Question 1

Question 1

Net torque rather than motor torque sets angular acceleration

A rigid rotor of inertia $0.50\,\mathrm{kg\,m^2}$ initially turns positively at $2\,\mathrm{rad/s}$. A motor supplies $3.0\,\mathrm{N\,m}$ while bearing resistance supplies $-1.0\,\mathrm{N\,m}$. Both remain constant.

1. Find angular acceleration and angular velocity after $3\,\mathrm s$.

2. Find the resisting torque that would instead maintain constant positive speed.

Question 1 - Solution

Net torque rather than motor torque sets angular acceleration

(a) Derivation and calculation

Sum moments before applying the fixed-axis rotational law: $$\tau_{\mathrm{net}}=3.0-1.0=2.0\ \mathrm{N\,m},\qquad \boxed{\alpha=\frac{\tau_{\mathrm{net}}}I=\frac{2.0}{0.50}=4.0\ \mathrm{rad/s^2}}.$$ $$\boxed{\omega(3)=\omega_0+\alpha t=2+4(3)=14\ \mathrm{rad/s}}.$$

(b) Result and interpretation

Constant positive speed requires $$\alpha=0\Rightarrow \tau_{\mathrm{motor}}+\tau_{\mathrm{resistance}}=0 \Rightarrow\boxed{\tau_{\mathrm{resistance}}=-3.0\ \mathrm{N\,m}}.$$ A nonzero drive torque can coexist with zero acceleration when another torque balances it.

Physical check

The inertia affects the acceleration, while the constant-speed torque balance is independent of inertia.