Circular Motion

A fan blade, a merry-go-round, a satellite: for anything going round a circle, one number, an angle, tells you where it is.

Angular position and angular velocity#

Let a particle P move on a circle of radius $r$ with centre O. Put the origin at O and draw an x-axis OX. The angle $\theta$ between OP and OX pins down where P is; call it the angular position. As P moves to a nearby point in time $\Delta t$, the angle grows by $\Delta\theta$. The rate of change is the angular velocity, and its rate of change is the angular acceleration:

Angles are measured in radians, so ω is in rad/s and α in rad/s².

Constant angular acceleration#

These definitions have exactly the same form as $v = dx/dt$ and $a = dv/dt$. So when α is constant, integrating gives three equations that mirror the equations of motion, with θ in place of x, ω for v and α for a (θ = 0 at t = 0):

From angles back to metres#

In time $\Delta t$ the particle moves along the arc a distance $\Delta s = r\,\Delta\theta$ (that is the definition of a radian). Divide by $\Delta t$ and take the limit:

Differentiate once more to see how fast the speed itself changes:

A car taking a roundabout at a steady 30 km/h is accelerating the whole time. The speed never changes; the direction never stops changing.

Two unit vectors that ride along#

Put the centre of the circle at the origin, and let the particle be at angular position θ. Two unit vectors are handy:

The position vector is $\vec r = r(\hat\imath\cos\theta + \hat\jmath\sin\theta)$. Differentiate with r fixed and θ changing:

So the velocity is always along the tangent, with size $r\omega$. Differentiate again (now ω may also change):

Uniform circular motion#

If the speed is constant, $dv/dt = 0$ and only the first term survives. It points along $-\hat e_r$, straight at the centre:

You saw the same thing from the velocity arrows in chapter 3: turn the velocity through a small angle and the change $\Delta\vec v$ points inward.

When the speed changes too#

If the speed changes, the acceleration has both parts:

where α is the angle between the acceleration and the inward radius.

Every circling object is accelerating toward the centre, so by Newton’s second law something must be pulling it there. The only question is: what?

The force circular motion requires#

Seen from an inertial frame, a particle moving in a circle is accelerating, and acceleration needs a net force. For uniform circular motion that acceleration is $v^2/r$ toward the centre, so the net force must point toward the centre too, with

Because this net force points to the centre, it is called the centripetal force (“centre-seeking”).

Take a bend too fast on a wet road and the car slides wide. Racetracks and highway ramps are tilted so that the road itself helps you round the corner.

A level turn: friction does the steering#

A car of mass $M$ takes a level bend of radius $r$ at speed $v$. The forces on it are its weight $Mg$, the road’s normal force $N$ (vertical on a flat road) and friction. The only horizontal force available to point toward the centre is static friction: the tyres tend to skid outward, and friction opposes that.

Banking: let the normal force help#

Friction is unreliable at high speeds, on sharp bends and in rain. So roads are banked: the outer edge is raised, tilting the road surface at angle θ. The normal force now tilts toward the centre, and its horizontal component can do the steering. At the right speed, friction isn’t needed at all:

A road is banked for the typical expected speed. A car going a little slower or faster stays put because static friction adjusts itself, pointing down the slope for a fast car and up the slope for a slow one. Too far from the design speed, and even maximum friction cannot prevent a slide.

On a spinning ride you feel flung outward, yet nothing outside you is pushing. That feeling has a name, and a precise place in physics.

A box on a turntable#

Put a heavy box on the rough floor of a closed cabin rotating steadily about a vertical axis with angular velocity ω. The box sits at distance $r$ from the axis and doesn’t slip.

From the ground (inertial): the box goes round a circle, so it needs $m\omega^2 r$ toward the axis. The forces on it are its weight, the floor’s normal force and friction; only friction is horizontal, so $f = m\omega^2 r$, pointing inward.

From inside the cabin: the box is at rest. Newton’s first law says the forces must cancel, but friction still points inward. To make the books balance we must add a pseudo force of size $m\omega^2 r$, pointing radially outward. Then $f = m\omega^2 r$ again: the same physics, described from a different seat.

Centrifugal force is enough for bodies at rest in the rotating frame. A body that moves within the rotating frame also needs the Coriolis force, which is perpendicular both to its velocity (in that frame) and to the rotation axis. It explains why large weather systems swirl, and why a ball rolled across a spinning platform appears to curve.

Earth is a rotating frame#

Earth spins once a day, $\omega = 2\pi/(24\ \text{h}) = 7.27\times10^{-5}$ rad/s. A point at colatitude θ (angle from the north pole) circles the axis on a radius $R\sin\theta$. Working from Earth’s frame, a mass hanging on a string feels gravity $mg$ toward Earth’s centre and centrifugal force $m\omega^2R\sin\theta$ away from the axis. Their resultant is the apparent weight $mg'$:

Since $2g > \omega^2R$, $g' < g$ everywhere except at the poles. The direction of $g'$ is the apparent vertical: plumb lines hang along it, walls are built along it, and still water lies perpendicular to it. At the equator the two forces are exactly opposite: