Angular Kinematics

Why rotation gets its own variables

Put a strip of tape near the axle of a bicycle wheel and another strip out on the rim, then spin the wheel. In one second the outer strip races through something like a metre of arc; the inner one barely moves. If you tried to describe the wheel by "where its parts are", you would need a different answer for every point on it, and none of those answers would be the same two moments running.

But something is shared. In the time the outer strip sweeps a quarter turn, so does the inner strip. Every particle of a rigid spinning body turns through the same angle in the same time. That single number describes the whole object, and it is why rotation gets its own set of variables. In the last lesson you replaced a swarm of particles by one point, the centre of mass, to talk about how a system moves as a whole. Now you need the other half: how it turns about that point.

Angular position $\theta$ is the angle between some line painted on the body and a fixed reference direction. Positive is usually counterclockwise; that is a convention, and you are free to pick the other one as long as you keep it.

The unit matters more than it looks. Define the radian by the rule that an angle of $\theta$ radians subtends an arc of length

$s = r\theta$

on a circle of radius $r$. A full turn wraps the whole circumference, $s = 2\pi r$, so a full turn is $2\pi$ radians. That is the entire content of "one revolution equals $2\pi$ rad". Three and a half revolutions is $3.5 \times 2\pi = 21.99$ rad.

Every formula below comes out of $s = r\theta$, and $s = r\theta$ is only true in radians. Degrees and revolutions are data formats: convert them the moment they appear, and never carry them into a calculation.

Rpm (revolutions per minute) is the commonest one you will meet. A motor at 1800 rpm turns 1800 revolutions in 60 seconds, so

$\omega = \frac{1800 \times 2\pi\ \text{rad}}{60\ \text{s}} = 188.5\ \text{rad/s}.$

Going the other way, the period $T$ is the time for one full turn, so $\omega T = 2\pi$ and $T = 2\pi/\omega$. A carousel at 0.40 rad/s has $T = 2\pi/0.40 = 15.7$ s.

Rates: angular velocity and angular acceleration

Angular velocity is how fast the angle changes,

$\omega = \frac{d\theta}{dt},$

measured in rad/s, and angular acceleration is how fast that changes,

$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2},$

in rad/s². These are the same two derivatives that took you from $x$ to $v$ to $a$ in a straight line. Nothing new is being claimed. If you are handed $\theta(t)$ you differentiate; if you are handed $\alpha(t)$ you integrate.

Sign works the way it does on a line. Counterclockwise positive means a wheel spinning counterclockwise and slowing has $\omega > 0$ and $\alpha < 0$, exactly as a car moving forward and braking has $v > 0$ and $a < 0$. A negative $\alpha$ does not mean "slowing"; it means "turning more clockwise per second". Those coincide only when $\omega$ is positive.

Worked example 1. A disk turns so that $\theta(t) = 2t^3 - 5t$, with $\theta$ in radians and $t$ in seconds. Find $\omega$ at $t = 2.0$ s and $\alpha$ at $t = 3.0$ s.

Differentiate once: $\omega = d\theta/dt = 6t^2 - 5$. At $t = 2.0$, $\omega = 6(4) - 5 = 24 - 5 = 19$ rad/s.

Differentiate again: $\alpha = d\omega/dt = 12t$. At $t = 3.0$, $\alpha = 12(3.0) = 36$ rad/s². Note that $\alpha$ here is not constant, so the constant-acceleration formulas in the next section would give nonsense. Check that before you reach for them.

When alpha is constant

If $\alpha$ is a constant, the integrals are the ones you already did for straight-line motion. Integrating $d\omega/dt = \alpha$ from $0$ to $t$ gives $\omega = \omega_0 + \alpha t$, where $\omega_0$ is the angular velocity at $t = 0$. Integrating that again gives the angle turned:

and eliminating $t$ between them gives the timeless one,

$\omega^2 = \omega_0^2 + 2\alpha\theta.$

Here $\theta$ means the angle turned since the start, not an absolute position. Every technique you used on the linear versions carries over untouched: pick the equation missing the variable you neither know nor want.

A flywheel at 60 rad/s stopped at 5.0 rad/s² takes $t = (0 - 60)/(-5.0) = 12$ s. A disk from rest at 4.0 rad/s² turns $\theta = \tfrac12 (4.0)(5.0)^2 = 50$ rad in 5.0 s. That is all these are.

Worked example 2. A centrifuge starts from rest and reaches 8000 rpm after 220 revolutions. Find its angular acceleration, assumed constant.

First convert, before anything else. The final angular speed is

$\omega = \frac{8000 \times 2\pi}{60} = \frac{50265}{60} = 837.76\ \text{rad/s},$

and the angle turned is $\theta = 220 \times 2\pi = 1382.3$ rad.

You know $\omega_0 = 0$, $\omega$, and $\theta$; you want $\alpha$ and you do not know or care about $t$. So use the equation without $t$:

If you had converted 8000 rpm to anything else, or left $\theta$ in revolutions, the number would have been wrong by a factor of $2\pi$ or 60 and nothing in the algebra would have warned you.

From the axle back to the rim

The whole point of $\theta$ was that it describes every point at once. To get back to what an individual point is doing, multiply by how far out it sits.

Take a point at radius $r$. In time $dt$ the body turns $d\theta$, so the point travels an arc $ds = r\, d\theta$. Divide by $dt$:

$v = \frac{ds}{dt} = r\frac{d\theta}{dt} = \omega r.$

That is its speed along its circular path. A rim point 0.35 m out on a wheel at 20 rad/s moves at $v = 20 \times 0.35 = 7.0$ m/s. Differentiate again, holding $r$ fixed, and you get the rate at which that speed changes:

$a_t = \frac{dv}{dt} = r\frac{d\omega}{dt} = \alpha r.$

This is the tangential acceleration: tangential because it points along the direction of motion. It is the piece that makes the point go faster or slower.

But a point going in a circle is accelerating even at constant speed, because its direction keeps turning. That is the centripetal acceleration you met with circular motion, $a_c = v^2/r$, which in angular language is

$a_c = \frac{v^2}{r} = \frac{(\omega r)^2}{r} = \omega^2 r,$

pointing straight at the axis. A point 0.20 m out on a wheel at 30 rad/s has $a_c = (30)^2(0.20) = 900 \times 0.20 = 180$ m/s².

These two pieces are perpendicular: one along the motion, one across it. So the total acceleration of the point has magnitude

$a = \sqrt{a_t^2 + a_c^2} = r\sqrt{\alpha^2 + \omega^4},$

and it is tilted away from the velocity by an angle $\phi$ with

$\tan\phi = \frac{a_c}{a_t} = \frac{\omega^2}{\alpha}.$

The radius cancels out of that angle. Every point on the body, near the axle or out at the rim, has its acceleration tilted by the same $\phi$.

Worked example 3. A point on a wheel of radius 0.25 m has $\omega = 4.0$ rad/s and $\alpha = 8.0$ rad/s². What angle does its total acceleration make with its velocity?

The velocity is tangential, so the angle you want is the angle between the total acceleration and the tangential direction. The two components are

The centripetal part is across the velocity, the tangential part along it, so

The radius did cancel, as promised: $\tan\phi = \omega^2/\alpha = 16/8 = 2$ without ever using 0.25 m.

Ropes, spools and rolling

Whenever something flexible runs over a wheel without slipping, the contact forces its speed to match the rim's. A cord unwinding from a spool of radius $r$ pays out length at exactly the rate the rim moves, so $v_{\text{cord}} = \omega r$ and $a_{\text{cord}} = \alpha r$. Pull a cord off a 0.12 m spool from rest at a steady 2.0 m/s²: then $\alpha = a/r = 2.0/0.12 = 16.67$ rad/s², and after 4.0 s the spool turns at $\omega = \alpha t = 16.67 \times 4.0 = 66.7$ rad/s.

A wheel rolling on the ground is the same condition in disguise. "Rolling without slipping" means the patch of rim touching the ground is not sliding along it: at that instant, the contact point is at rest. Work out what that forces. The axle moves forward at $v$; relative to the axle, every rim point moves at $\omega r$. At the bottom of the wheel the rim's motion relative to the axle points backwards, so the ground-frame speed there is $v - \omega r$. Setting that to zero gives the rolling condition:

$v = \omega r, \qquad a = \alpha r.$

Now add the two motions everywhere else. At the top, the rim's motion relative to the axle points forwards, in the same direction as $v$, so

$v_{\text{top}} = v + \omega r = 2v.$

A wheel of radius 0.30 m rolling at 4.0 m/s has a top point moving at 8.0 m/s. This is not a trick: the top of a rolling wheel really does move twice as fast as the car it carries, and the bottom is momentarily stationary.

At the side, level with the axle, the relative motion is vertical while the axle's motion is horizontal. They are perpendicular, so you add them as vectors:

In general, a point whose angle from the contact point is $\psi$ has speed $2v\sin(\psi/2)$, which gives 0 at the bottom, $v\sqrt2$ at the side and $2v$ at the top. Rolling also links turns to distance travelled: one revolution rolls out one circumference, so 40 turns of a 0.34 m wheel covers $40 \times 2\pi \times 0.34 = 85.5$ m.

When alpha is not constant

Two kinds of problem break the constant-$\alpha$ formulas, and each has a standard escape.

Alpha given as a function of time. Then you integrate. If a wheel starts from rest with $\alpha(t) = 6t$, integrate once for the angular velocity:

$\omega(t) = \int_0^t 6t'\,dt' = 3t^2,$

with no constant because $\omega(0) = 0$. Integrate again for the angle:

$\theta(t) = \int_0^t 3t'^2\,dt' = t^3.$

At $t = 4.0$ s, $\theta = 4.0^3 = 64$ rad. Had you wrongly used $\theta = \tfrac12\alpha t^2$ with the $\alpha$ at $t=4$, you would have got $\tfrac12(24)(16) = 192$ rad, three times too big.

Alpha given in terms of omega. A rotor coasting against a drag torque proportional to speed obeys $\alpha = -k\omega$. You cannot integrate that in $t$ directly, because the right side keeps changing. Separate the variables:

With $\omega_0 = 12$ rad/s and $k = 0.25$ s⁻¹, after 6.0 s

It never quite reaches zero, but the total angle is still finite. To get it, switch the variable you differentiate against. Use the chain rule to trade $t$ for $\theta$:

That identity is worth memorising; it is the rotational twin of $a = v\,dv/dx$ and it is the tool for anything that depends on position rather than time. Putting $\alpha = -k\omega$ into it,

so $\omega$ falls off linearly in angle, even though it falls off exponentially in time. Starting at $\omega_0$ and ending at $\omega = 0$:

$\theta_{\text{total}} = \frac{\omega_0}{k} = \frac{12}{0.25} = 48\ \text{rad}.$

Two things going round at once

When two objects circle the same axis, subtract their angles the way you subtract positions in a relative-velocity problem. The angle by which the faster one leads is $\Delta\theta = (\omega_1 - \omega_2)t$, and it laps the slower one when that gap reaches a full turn:

Two runners on a track at 0.16 and 0.10 rad/s have a relative angular speed of 0.06 rad/s, so the lap happens at $t = 2\pi/0.06 = 104.7$ s. The radius of the track, 50 m, never enters: lapping is about angle, not distance. That is a good habit to notice: if the question is about who is where, work in angle and leave the radius alone until the very end.

Recognising the technique

You are in an angular kinematics problem when the quantities given are about turning rather than travelling: revolutions, rpm, a period, "how many turns", an angle in radians, a radius paired with a rate. Then:

• Everything constant, a wheel spinning up or down uniformly. Convert to radians and rad/s, then use the three constant-$\alpha$ equations, choosing the one that omits the variable you neither have nor want. If the problem gives you revolutions and asks for time, that is $\omega = \omega_0 + \alpha t$ after converting.
• The answer is wanted in metres or m/s, but the data is angular (or the reverse). You need $s = r\theta$, $v = \omega r$ or $a_t = \alpha r$: one multiplication by the radius, at the end.
• A question about the acceleration of a point, not of the body. Two perpendicular pieces: $a_t = \alpha r$ along the motion and $a_c = \omega^2 r$ toward the axis. Combine with Pythagoras, and use $\tan\phi = \omega^2/\alpha$ for the direction.
• A cord, belt, chain or a wheel on the ground, with "does not slip". That phrase is the equation $v = \omega r$. Write it down first.
• $\alpha$ given as a formula. If it depends on $t$, integrate in $t$. If it depends on $\omega$ or $\theta$, separate variables, and reach for $\alpha = \omega\, d\omega/d\theta$ the moment the question asks for a total angle rather than a time.
• Two things going round together. Work with the difference of the angles, and set it to $2\pi$ for a lap.

The one mistake that costs more marks than all the others is a unit left unconverted. Write the radian conversion on the first line, every time, even when it feels unnecessary.

Next you will stop describing the spin and start asking what causes it: forces applied off-axis, and the quantity that measures their turning effect, in Torque & Static Equilibrium.