Rotational Mechanics

A bucket drops into a well and the pulley above it whirls faster and faster. The pulley’s centre never moves, so the net force on it is zero, yet clearly something is changing. Welcome to rotation.

What “rotating about an axis” means#

Pinch a book between two fingertips and twist it. Whatever you do, the two pinched points stay put and every other bit of the book swings round in a circle. The circles all lie in parallel planes, each centred on the line joining your fingertips, and every particle completes its circle in the same time. Particles on that line don’t move at all; those far from it sweep out big circles.

Hinged doors, ceiling fans and the knob on a gas stove all rotate this way. The axis need not even pass through the body: a fly sitting near the rim of a spinning turntable rotates about the turntable’s axis.

One angle describes the whole body#

Because the body is rigid, when one particle’s radius turns through an angle θ, every particle’s radius turns through the same θ. So a single number, the angular position θ of the body, tells us where everything is. From it we build the same ladder of quantities we used for a particle on a line:

Angular velocity is measured in rad/s (often quoted in rev/s or rpm; 1 rev = 2π rad). It is given a direction along the axis by the right-hand rule: curl your fingers the way the body turns and your thumb points along $\vec\omega$. Choose one sense of rotation as positive and stick with it; θ, ω and α can then be positive or negative, just like x, v and a.

From the body back to its particles#

A particle at perpendicular distance r from the axis moves along an arc of length $s = r\theta$. Differentiate:

Every particle also has its centripetal acceleration $\omega^2 r$ toward the axis, as in circular motion.

A wheelchair doesn’t start spinning just because you wish it to; you have to push. But a ceiling fan spins up while its centre stays still, so the net force on it is zero. Something other than net force decides whether a body starts to turn. That something is torque.

Torque about a point#

Being a cross product, $\vec\tau$ is perpendicular to both $\vec r$ and $\vec F$, and its size is $rF\sin\theta$. Its unit is N·m (not written as joule, though dimensionally the same).

Torque about an axis#

For a body turning about a fixed axis, what matters is the torque about that axis: pick any point O on the axis and take the component of $\vec r \times \vec F$ along the axis. Shifting O along the axis adds a vector perpendicular to the axis, so the answer doesn’t depend on which point you pick. Four situations cover everything:

Case 1: force parallel to the axis

$\vec r\times\vec F$ is perpendicular to $\vec F$, hence to the axis. Zero torque about the axis. Pushing a door upward along its hinges doesn’t open it.

Case 2: force whose line passes through the axis

Choose O where the line meets the axis; then r and F are parallel. Zero torque. Pushing a door straight at its hinges does nothing.

Case 3: force perpendicular to the axis but not meeting it

This is the everyday case. The torque about the axis is the force times the length of the common perpendicular between the axis and the force’s line of action: the lever arm (or moment arm).

Case 4: anything else

Split the force into a part parallel to the axis (no torque, case 1) and a part perpendicular to it (case 3).

By convention a torque that would turn the body anticlockwise, as seen looking back along the axis toward you, is positive.

Adding torques#

The total torque on a body is the sum of the individual torques:

Net force decides linear acceleration. We now show that net torque decides angular acceleration, and meet the quantity that plays the part of mass: the moment of inertia.

Following one particle#

Take a rigid body turning about a fixed axis with angular velocity ω and angular acceleration α. A particle of mass m at distance r from the axis moves on a circle, so the net force on it has two parts:

• a radial part $m\omega^2 r$ toward the axis. Its line passes through the axis, so its torque about the axis is zero;
• a tangential part $m\,\dfrac{dv}{dt} = m r\alpha$, perpendicular to the radius. Its lever arm is r, so its torque is $mr^2\alpha$.

The torque of the net force on that particle is therefore $m r^2 \alpha$. Add this up over all particles (they all share α):

Internal torques cancel#

The total includes the forces particles exert on each other. By the third law these come in equal, opposite pairs, and each pair acts along the line joining the two particles. The two torques $\vec r_i\times\vec F_{ij} + \vec r_j\times\vec F_{ji} = (\vec r_i - \vec r_j)\times\vec F_{ij}$ vanish because $\vec r_i - \vec r_j$ is parallel to $\vec F_{ij}$. Only external torques survive.

A bridge, a bookshelf, a ladder against a wall: engineering is mostly about things that must not move. For that, two balances must hold at once.

Two conditions#

$\vec F_{\text{ext}} = M\vec a_{cm}$ says the centre of mass stays at rest only if the external forces add to zero. $\tau = I\alpha$ says the body won’t start turning only if the external torques add to zero. A body at rest in an inertial frame needs both: zero net force in every direction and zero net torque about every line.

When all forces lie in one plane (the x–y plane), this reduces to three equations:

The torques about the x- and y-axes vanish automatically for forces in that plane. And if the net force is zero, the net torque is the same about every perpendicular axis, so you may take torques about whichever point is most convenient.

Standing, tipping and stability#

An object resting on a surface feels its weight at the centre of mass and contact forces spread over its base. The contact forces can only push, so they can only cancel gravity’s torque if the vertical line through the centre of mass lands inside the base (or inside the region bounded by the contact points). That is why you lean the other way when carrying a heavy bag in one hand.

Disturb a body slightly and watch what its centre of mass does:

• Stable equilibrium: the centre of mass rises; gravity brings the body back (a box sitting flat).
• Unstable equilibrium: the centre of mass falls; the body moves further away (a box balanced on an edge, a pencil on its tip).
• Neutral equilibrium: the centre of mass stays at the same height; the body stays where you leave it (a ball on a level floor).

The leaning cyclist#

A cyclist rounding an unbanked turn of radius r at speed v is not in equilibrium in the ground frame (the centre of mass is accelerating). Work instead in a frame rotating with the cyclist at ω = v/r. In that frame we add the centrifugal pseudo-force $Mv^2/r$ at the centre of mass, and then the cyclist is at rest, so the torques about the contact point A must cancel. Gravity tries to topple the cycle inward and the centrifugal force outward. With the cycle at angle θ from the vertical and the centre of mass at distance l from A:

A figure skater begins a spin with arms wide, then draws them in and blurs into a whirl. No one pushed her. What stayed the same was her angular momentum.

Angular momentum of a particle#

Its size is

where $r_\perp$ is the perpendicular distance from O to the particle’s line of motion. Even a particle moving in a straight line has angular momentum about a point off that line. As with torque, the angular momentum about an axis is the component of $\vec r\times\vec p$ along it, for any O on the axis.

A rigid body: L = Iω#

For a particle going round a circle of radius r, pick O at the centre: r ⊥ v, so $l = mvr = mr^2\omega$, along the axis. For a rigid body each particle contributes $m_i r_i^2\omega$, and they share ω:

Torque changes angular momentum#

Differentiate the definition:

The first term vanishes (a vector crossed with itself) and internal torques cancel in pairs, leaving

Angular impulse#

Integrating $\tau = dL/dt$ over a time interval gives the rotational version of impulse:

A potter’s wheel keeps turning after the foot stops pushing; a flywheel smooths an engine’s jerks. Spinning things store energy, and we can count it.

Kinetic energy of a spinning body#

A body turning about a fixed axis at angular speed ω has its i-th particle moving at $v_i = r_i\omega$. Add up the ordinary kinetic energies:

Energy bookkeeping with wheels#

When a block on a string unwinds a wheel, the string doesn’t slip, so the rim moves as fast as the block: ω = v/r. As the block falls a height h, gravity’s lost potential energy goes into both bodies:

Work and power from a torque#

A torque that speeds a body up raises its kinetic energy at a rate

Mass tells you how hard a body is to push. Moment of inertia tells you how hard it is to spin, and it depends on where the mass sits. Here is how to calculate it.

Collections of particles#

For separate particles, use the definition directly: $I = \sum m_i r_i^2$, where each $r_i$ is the perpendicular distance from the axis. Two heavy balls on a light rod, at distances r₁ and r₂ from an axis, give $I = m_1 r_1^2 + m_2 r_2^2$; the rod adds nothing.

Continuous bodies#

For a smooth distribution of matter, slice the body into elements of mass dm chosen so that every point of an element is practically the same distance r from the axis. Then

The integration limits must make the elements cover the whole body exactly once.

Uniform rod about its perpendicular bisector

Rod of mass M, length L, origin at its middle. An element between x and x + dx has mass $(M/L)\,dx$ and distance |x|:

Rectangular plate about a central line parallel to one edge

Plate of sides l and b, axis through the centre parallel to the side of length b. Slice it into strips parallel to the axis: a strip at distance x of width dx has mass $\frac{M}{lb}\,b\,dx = \frac Ml\,dx$, exactly like the rod. So $I = Ml^2/12$; the side b parallel to the axis doesn’t matter.

Ring about its axis

Every bit of a thin ring of radius R is at distance R from its axis: $I = \int R^2\,dm = MR^2$. A thin-walled hollow cylinder gives the same result for the same reason.

Disc about its axis

Slice the disc into thin rings of radius x and width dx. Area $2\pi x\,dx$, mass $dm = \dfrac{M}{\pi R^2}2\pi x\,dx = \dfrac{2M}{R^2}x\,dx$, and each ring contributes $x^2\,dm$:

A solid cylinder is a stack of such discs, so it too has $I = MR^2/2$ about its axis, whatever its length.

Hollow sphere about a diameter

Cut the shell into rings by planes perpendicular to the axis. The ring at polar angle θ has radius $R\sin\theta$ and width $R\,d\theta$, so its mass is $\dfrac{M}{4\pi R^2}(2\pi R\sin\theta)(R\,d\theta) = \tfrac M2\sin\theta\,d\theta$:

Solid sphere about a diameter

Build it from thin hollow shells of radius x and thickness dx. Each has mass $\dfrac{3M}{4\pi R^3}4\pi x^2\,dx = \dfrac{3M}{R^3}x^2\,dx$ and moment of inertia $\tfrac23(dm)x^2$:

Integrating for every new axis would be exhausting. Two theorems let one known moment of inertia generate many others.

The parallel axis theorem#

Why it works. Put the centre of mass C at the origin with the z-axis through it, and let the given axis be the parallel line through (d, 0, 0). A particle at (x, y, z) has squared distance $x^2 + y^2$ from the z-axis and $(x - d)^2 + y^2$ from the given axis. So

The first sum is $I_{cm}$. The last sum is M times the x-coordinate of the centre of mass, which is zero because the centre of mass is at the origin. Done.

The perpendicular axis theorem#

This one works only for flat (plane) bodies. Put the body in the x–y plane with the z-axis perpendicular to it, all three axes through one point:

Why. For a particle in the plane, its squared distance from the z-axis is $x^2 + y^2$, the sum of its squared distances from the y-axis (x²) and the x-axis (y²). Multiply by mass and add over all particles.

Two more results that follow from the parallel axis theorem:

• Solid cylinder about a line on its curved surface, parallel to its axis: $\tfrac12MR^2 + MR^2 = \tfrac32MR^2$.
• Solid sphere about a tangent: $\tfrac25MR^2 + MR^2 = \tfrac75MR^2$.

Radius of gyration#

For a disc about its axis, $Mk^2 = MR^2/2$, so $k = R/\sqrt2 \approx 0.71R$. For a solid sphere about a diameter, $k = \sqrt{2/5}\,R \approx 0.63R$.

A bicycle wheel neither simply slides nor simply spins. It does both at once, and the combination has some surprises: the bit touching the road is standing still.

Translation plus rotation#

When a body both moves and turns, describe it in two parts: the motion of its centre of mass, and its rotation about an axis through the centre of mass. The first obeys $\vec F_{\text{ext}} = M\vec a_{cm}$ as always. For the second there is a pleasant fact: although the centre-of-mass frame may be accelerating, the pseudo-forces −mᵢa all act like a single force through the centre of mass, so they have no torque about it. Hence τ = Iα still holds there.

Pure rolling#

Paint one spoke of a bicycle wheel. Each time the spoke returns to pointing straight down, the wheel has turned once and the bike has moved one circumference, 2πR. Over a short time the centre moves Δx while the wheel turns Δθ, with Δx = RΔθ:

Now add velocities. Relative to the centre, every rim point moves at Rω along the rim. At the bottom that is backward, at the top forward. Adding the centre’s forward velocity v:

If the condition fails, the wheel slips:

• $v_{cm} > R\omega$: forward slipping. The contact point rubs forward, as when you brake hard on a wet road and the wheel barely turns.
• $v_{cm} < R\omega$: backward slipping. The contact point rubs backward, as when a wheel spins in mud.

Kinetic energy of a rolling body#

Write each particle’s velocity as the centre’s velocity plus its velocity relative to the centre, and expand $\sum\tfrac12 m_i v_i^2$. The cross term is proportional to the total momentum in the centre-of-mass frame, which is zero, leaving

For pure rolling, put $v_{cm} = R\omega$: $K = \tfrac12(I_{cm} + MR^2)\omega^2 = \tfrac12 I_P\omega^2$, where $I_P$ is the moment of inertia about the contact line (parallel axis theorem). At each instant a rolling body is in pure rotation about its contact point.

Down a slope#

Release a body from rest on a slope of angle θ. If it rolls without slipping, static friction does no work (the contact point doesn’t move), so after a drop h, with $I_{cm} = kMR^2$:

Angular momentum of a rolling body#

About a fixed point O, splitting each position and velocity into centre-of-mass and relative parts gives two terms (the cross terms vanish just as for energy):

the spin about the centre of mass, plus the angular momentum of the whole mass imagined concentrated at the centre of mass.