Rotational Kinematics and Going in Circles

The same equations, new letters

When angular acceleration \alpha is constant, the rotational kinematic equations follow line-for-line from the constant-acceleration equations of straight-line motion. Just read \theta for x, \omega for v, and \alpha for a. Nothing new to derive — it is the same calculus.

\begin{aligned} \omega &= \omega_{0} + \alpha t \\ \theta &= \omega_{0}t + \tfrac{1}{2}\alpha t^{2} \\ \omega^{2} &= \omega_{0}^{2} + 2\alpha\,\theta \end{aligned}

Constant-α kinematics — identical in form to v = v₀ + at, x = v₀t + ½at², v² = v₀² + 2ax.

Worked example: spinning up a hard drive

A disk starts from rest and reaches 7200 revolutions per minute (rpm) in 4.0 seconds at constant angular acceleration. Find \alpha, and how many revolutions it makes while spinning up.

Notice the strategy: convert everything to radians first, pick the equation that contains only knowns and the one unknown, then convert back to human units at the end. That discipline solves nearly every kinematics problem.

Constant speed, and yet accelerating

Swing a ball on a string in a circle at steady speed. Its speed never changes, but its velocity does — because velocity is a vector and its direction keeps turning. A changing velocity means acceleration. So uniform circular motion is accelerated motion, even at constant speed. This surprises almost everyone the first time.

The acceleration points straight toward the center — we call it centripetal (center-seeking). Its size follows from the geometry of the turning velocity vector.

a_{c} = \frac{v^{2}}{r} = \omega^{2} r

Centripetal acceleration grows with the square of speed and shrinks with radius — halving the radius doubles it.

This is centripetal acceleration. At v^2/r it explains why a tight, fast turn throws you against a car door: small r and large v both make a_c huge.

Centripetal force is not a new force

By Newton's second law, a center-pointing acceleration needs a center-pointing net force. That inward force is the centripetal force. Its magnitude is:

F_{c} = m\,a_{c} = \frac{mv^{2}}{r} = m\omega^{2} r

The inward net force required to keep mass m moving in a circle of radius r at speed v.

Speeding up while turning

If the object also speeds up around the circle, it has a tangential acceleration a_t = r\alpha along the rim as well as the centripetal a_c toward the center. These two are perpendicular, so the total acceleration is their vector sum.

a = \sqrt{a_{t}^{2} + a_{c}^{2}} = \sqrt{(r\alpha)^{2} + (\omega^{2}r)^{2}}

A car accelerating out of a bend has both components; the total acceleration is their perpendicular sum.