Rotational motion

So far every object has been treated as a particle, a point with mass. Real machines are full of things that spin: shafts, gears, pulleys, flywheels, turbine rotors, wheels, drill bits and hard-drive platters. To design them you need the rotational versions of the ideas from the last four chapters: angular position, velocity and acceleration; torque in place of force; moment of inertia in place of mass; and rotational kinetic energy and angular momentum.

The good news is that every rotational idea has a direct linear analogue, and the equations look the same once you swap the symbols. This chapter builds that dictionary and uses it on fans, wind turbines, flywheels, pulleys, rolling wheels and shafts.

Angular quantities

Consider a rigid body turning about a fixed axis. Every point moves in a circle, and every point turns through the same angle in the same time. So we describe the motion of the whole body with one angle \(\theta\), measured in radians.

A radian is the angle at which the arc length equals the radius, so an arc of length \(s\) on a circle of radius \(r\) subtends \(\theta = s/r\). One full turn is \(2\pi\) rad = 360°, so 1 rad ≈ 57.3°. Radians are not optional here: the formulas below are only true in radians.

The angular velocity is the rate of turning, and angular acceleration is its rate of change:

Machinery speeds are usually quoted in revolutions per minute (rpm). Convert with \(\omega = \text{rpm} \times 2\pi/60\). A motor at 1800 rpm turns at 188.5 rad/s.

Linking angular and linear motion

A point at distance \(r\) from the axis travels a distance \(s = r\theta\), so its speed and tangential acceleration are

where \(a_c\) is the centripetal acceleration from Lesson 2, directed toward the axis. All points share the same \(\omega\), but points farther out move faster and accelerate more.

Constant angular acceleration

If \(\alpha\) is constant, the angular equations are exactly the constant-acceleration equations from Lesson 1 with \(x \to \theta\), \(v \to \omega\), \(a \to \alpha\):

Torque

Force causes linear acceleration; torque causes angular acceleration. You met torque in Linear Algebra as the moment of a force, \(\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}\). For rotation about a fixed axis its size is

where \(\phi\) is the angle between \(\mathbf{r}\) and \(\mathbf{F}\), and \(d\) is the perpendicular distance from the axis to the force's line of action (the moment arm). Torque has units of N·m. By convention, counterclockwise torques are positive.

The same force produces more torque farther from the axis: this is why door handles are on the side away from the hinges and why a longer wrench loosens a stuck bolt. A force aimed straight through the axis produces no torque at all.

Moment of inertia and the rotational second law

Apply Newton's second law to each small piece of a rotating body. A piece of mass \(m_i\) at radius \(r_i\) has tangential acceleration \(r_i\alpha\), which needs a tangential force \(m_i r_i\alpha\) and hence a torque \(m_i r_i^2\alpha\). Adding over the body:

The moment of inertia \(I\) (units kg·m²) is the rotational counterpart of mass. It depends not just on how much mass there is but on how far it sits from the axis, and because of the \(r^2\) the outer mass dominates. You computed moments of inertia by integration in Calculus II. Some standard results, for total mass \(M\):

• Thin hoop or ring about its centre: \(I = MR^2\).
• Solid disc or cylinder about its axis: \(I = \tfrac12 MR^2\).
• Solid sphere about a diameter: \(I = \tfrac25 MR^2\).
• Thin rod of length \(L\) about its centre: \(I = \tfrac{1}{12}ML^2\); about one end: \(I = \tfrac13 ML^2\).

The parallel-axis theorem gives the moment of inertia about any axis parallel to one through the centre of mass: \(I = I_{\text{cm}} + Md^2\). For instance, the rod about its end is \(\tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2\).

Combining linear and rotational motion

Many machines have parts that translate and parts that rotate, connected by cords or gears. Write \(\sum F = ma\) for each translating part, \(\sum\tau = I\alpha\) for each rotating part, and link them with a constraint such as \(a = R\alpha\) for a cord that does not slip.

Rotational kinetic energy and power

Each piece of a rotating body has kinetic energy \(\tfrac12 m_i(r_i\omega)^2\). Adding them:

Again \(I\) replaces \(m\) and \(\omega\) replaces \(v\). The work done by a torque turning through an angle is \(W = \int\tau\,d\theta\), and the power delivered to a rotating shaft is

This is the relation used to rate every motor, engine and gearbox.

That last remark is the key to gear design. A 3:1 reduction gearbox turns this motor's output into about 150 N·m at 583 rpm, minus a few percent lost in the gears. Power in equals power out plus losses.

Rolling motion

A wheel rolling without slipping both translates and rotates, and the two are locked together: the centre moves at \(v = R\omega\). Its total kinetic energy is the sum of the translational energy of the centre of mass and the rotational energy about it:

Static friction at the contact point makes the wheel roll rather than skid, but does no work, because the contact point is momentarily at rest. So mechanical energy is conserved for an object rolling down a slope. From height \(h\):

Angular momentum

The rotational counterpart of momentum is angular momentum, \(L = I\omega\) for a body spinning about a fixed axis (units kg·m²/s). The rotational second law in its general form is \(\sum\tau = dL/dt\), and it gives another conservation law: if the net external torque on a system is zero, its angular momentum is constant.

If a spinning system pulls mass inward, \(I\) falls and \(\omega\) rises to compensate. A figure skater spins faster by pulling in her arms; a diver tucks to rotate quickly and straightens to slow down before entering the water.

Angular momentum conservation also explains why gyroscopes hold their orientation (used in aircraft instruments and satellite attitude control), why a helicopter needs a tail rotor (to cancel the torque reaction of the main rotor), and why a spinning bicycle wheel resists being tilted.

The linear–rotational dictionary

• Position \(x\) ↔ angle \(\theta\); velocity \(v\) ↔ angular velocity \(\omega\); acceleration \(a\) ↔ angular acceleration \(\alpha\).
• Mass \(m\) ↔ moment of inertia \(I\); force \(F\) ↔ torque \(\tau\).
• \(F = ma\) ↔ \(\tau = I\alpha\); \(K = \tfrac12 mv^2\) ↔ \(K = \tfrac12 I\omega^2\); \(P = Fv\) ↔ \(P = \tau\omega\); \(p = mv\) ↔ \(L = I\omega\).