Steady Rotation Versus Speeding Up and Slowing Down

Steady Rotation Versus Speeding Up and Slowing Down

Rotation is steady when angular velocity does not change. It speeds up or slows down when angular velocity changes with time. That difference decides whether acceleration is zero and whether a net torque is required.

Angular displacement $\theta$θ measures how far a body has turned. Angular velocity $\omega$ω is the rate of that turn. Angular acceleration $\alpha$α is the rate of change of $\omega$ω. In steady rotation, $\omega$ω is constant, so $\alpha = 0$α=0. A ceiling fan at fixed speed and a wheel coasting at constant spin are examples. The angle still increases. The speed of rotation does not.

Speeding up means $\omega$ω and $\alpha$α point in the same sense. Slowing down means they point in opposite senses. Therefore, a sign chart matters. If counterclockwise is positive, a positive $\alpha$α increases a positive $\omega$ω and decreases a negative $\omega$ω. The words “faster” and “slower” refer to the size of $\omega$ω, not only to its sign.

With constant $\alpha$α, the kinematic links are

$$\omega = \omega_0 + \alpha t$$ω=ω0​+αt

$$\theta = \omega_0 t + \frac{1}{2}\alpha t^{2}$$θ=ω0​t+21​αt2

$$\omega^{2} = \omega_0^{2} + 2\alpha\theta$$ω2=ω02​+2αθ

If $\alpha = 0$α=0, these collapse to $\omega = \omega_0$ω=ω0​ and $\theta = \omega t$θ=ωt. A zero acceleration does not mean zero motion.

Dynamics supplies the cause. Newton’s second law for rotation is $\tau_{\mathrm{net}} = I\alpha$τnet​=Iα. Steady rotation needs zero net torque about the axis, even if individual torques cancel. Speeding up or slowing down needs a nonzero net torque. Friction at an axle, a brake, or a driving belt can supply it. A balanced motor torque can also hold $\alpha$α at zero once the target speed is reached.

Linear motion at the rim follows the same split. Tangential speed is $v = r\omega$v=rω. Tangential acceleration is $a_t = r\alpha$at​=rα. In steady rotation, $a_t = 0$at​=0, but a point on a circular path still has centripetal acceleration $a_r = r\omega^{2}$ar​=rω2 toward the center. Therefore, constant $\omega$ω is not force-free motion for a particle on the rim. It is torque-free motion of the rigid body about its axis.

The practical test is simple. Check whether $\omega$ω changes. If it does not, $\alpha = 0$α=0 and net torque is zero. If it does, use the kinematic equations and match the sign of $\alpha$α to speeding up or slowing down.

Disclaimer: This article explains standard rotational kinematics for study only. Sign conventions should follow the textbook in use. It is not a graded solution.