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Skill Check 0 / 10 complete

Rotational kinematics

A spinning disk. A record on a turntable. A wheel on a fixed axle. Same physics: every point on the rigid body sweeps through one angle in the same time, with one angular velocity $\omega$ and one angular acceleration $\alpha$. This lesson sets up the rotational analogs of $x$, $v$, and $a$, reads them off three kinematic graphs, and pulls the sign conventions through with discipline. The traps cluster around three errors: thinking the rim spins faster than the hub, swapping $\omega$ for $\alpha$, and losing track of the CW/CCW sign convention mid-problem.

§1

One disk. One omega.

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Picture a record on a turntable. Or a wheel on a fixed axle. Same physics either way. The disk is rigid, so every point on it sweeps the same angle in the same time. Center, edge, anywhere in between: they all turn together.

To describe that motion, we need three quantities, one for each kinematic variable in translation:

Angular position $\theta$. The angle the disk has rotated, in radians, measured from some reference line. Positive by convention means counterclockwise (CCW).
Angular velocity $\omega$. How fast the angle is changing, in rad/s. The slope of $\theta$-vs-$t$. Positive means CCW.
Angular acceleration $\alpha$. How fast $\omega$ is changing, in rad/s$^2$. The slope of $\omega$-vs-$t$. Positive means $\omega$ is becoming more CCW.

Two questions to ask of any rotating body:

What is the angular velocity right now? Read $\omega(t)$.
Is that angular velocity changing? Read $\alpha$. If $\alpha = 0$, then $\omega$ holds steady. If $\alpha \ne 0$, then $\omega$ is changing at that rate every second.

These are two different quantities, not one. A wheel can have a big $\omega$ and zero $\alpha$ (steady spin). A wheel can have zero $\omega$ and big $\alpha$ (just starting to spin from rest, or at the moment it reverses direction).

§2

The rotational kinematic equations.

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When $\alpha$ is constant, the angular kinematic equations have exactly the same shape as the linear ones. The rotational variables replace the translational ones, slot for slot:

Linear (constant $a$)

$v = v_0 + at$
$\Delta x = v_0 t + \tfrac{1}{2}a t^2$
$v^2 = v_0^2 + 2 a \Delta x$

Rotational (constant $\alpha$)

$\omega = \omega_0 + \alpha t$
$\Delta \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$
$\omega^2 = \omega_0^2 + 2 \alpha \Delta\theta$

If you know the linear equations, you already know the rotational ones. The substitutions: $x \to \theta$, $v \to \omega$, $a \to \alpha$. Mass and force do not appear in these equations because kinematics, by itself, does not care about what causes the motion. That is for later topics in this unit.

Setup

A wheel starts at $\omega_0 = +6$ rad/s and has constant $\alpha = -2$ rad/s$^2$. Find $\omega$ at $t = 5$ s, find $\Delta\theta$ at $t = 5$ s, and find the time at which $\omega = 0$. CCW positive.

Angular velocity at $t = 5$ s

$$\omega = \omega_0 + \alpha t = (+6) + (-2)(5) = -4 \text{ rad/s}.$$

The wheel has reversed. It is now rotating CW at $4$ rad/s. The minus sign is direction, not size.

Angular displacement at $t = 5$ s

$$\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = (+6)(5) + \tfrac{1}{2}(-2)(25) = 30 - 25 = +5 \text{ rad}.$$

The wheel ends up $5$ rad CCW of where it started, even though it is currently spinning CW. Total angle traveled is more than $5$ rad because the wheel went out and came partway back.

Time at which $\omega = 0$

Set $\omega = 0$: $0 = (+6) + (-2)t$, so $t = 3$ s. At this instant, the wheel is momentarily not rotating. This is not the end of the problem. $\alpha$ is still $-2$ rad/s$^2$ at $t = 3$ s, so the wheel keeps responding to it and immediately starts rotating CW.

Read

Three different times, three different answers, all from the same setup. The wheel did not stop when $\omega$ hit zero, it reversed. The angular displacement at $t = 5$ s tells you net rotation, not total angle covered. The minus sign on $\omega$ at $t = 5$ s tells you direction, not how slow.

§3

Reading angular kinematic graphs.

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The three angular kinematic graphs ($\theta$-vs-$t$, $\omega$-vs-$t$, $\alpha$-vs-$t$) carry the same slope and area relationships as their linear counterparts. The same two rules cover everything:

Slope of a graph gives the next variable up. Slope of $\theta$-vs-$t$ is $\omega$. Slope of $\omega$-vs-$t$ is $\alpha$. If $\alpha$ is constant, the slope of $\alpha$-vs-$t$ is zero, and that's where this lesson stops.
Area under a graph gives the next variable down. Area under $\omega$-vs-$t$ between two times is $\Delta\theta$ over that interval. Area under $\alpha$-vs-$t$ between two times is $\Delta\omega$ over that interval.

The trap: reading the slope of $\omega$-vs-$t$ as $\omega$ itself, or reading the area under $\alpha$-vs-$t$ as $\Delta\theta$. The slope of $\omega$-vs-$t$ is $\alpha$, not $\omega$. The area under $\alpha$-vs-$t$ is $\Delta\omega$, not $\Delta\theta$. The graphs talk to each other in pairs, one level at a time.

For the worked example in §2 (with $\omega_0 = +6$ rad/s, $\alpha = -2$ rad/s$^2$):

$\omega$-vs-$t$ is a straight line starting at $+6$ rad/s with slope $-2$ rad/s$^2$, crossing zero at $t = 3$ s, reaching $-4$ rad/s at $t = 5$ s. The slope of that line is $\alpha$, a constant. The line itself is $\omega$.
$\theta$-vs-$t$ is a parabola opening downward, climbing from $0$ to a maximum at $t = 3$ s (where $\omega = 0$), then falling back. Maximum $\theta$ is $9$ rad; value at $t = 5$ s is $+5$ rad.
$\alpha$-vs-$t$ is a flat horizontal line at $-2$ rad/s$^2$. It does not change with time, not even at $t = 3$ s, where $\omega$ is zero. Zero $\omega$ does not mean zero $\alpha$.

One last graph rule, this one specific to rigid-body rotation: every point on the disk shares the same $\omega$-vs-$t$ graph and the same $\alpha$-vs-$t$ graph. The rim does not get a faster line. The hub does not get a slower one. One rigid body, one angular-kinematics-graph set.

Three rules to carry. (1) Slope of $\theta$-vs-$t$ is $\omega$; slope of $\omega$-vs-$t$ is $\alpha$. (2) Area under $\omega$-vs-$t$ is $\Delta\theta$; area under $\alpha$-vs-$t$ is $\Delta\omega$. (3) For a rigid body, every point shares one $\omega$ and one $\alpha$ at any instant. The rim and the hub take the same time to complete one revolution.

Interactive ↗

Graph Sign Detective Twenty-seven sign permutations for the three graphs of rotational motion. Read $\theta_0$ off the start of the $\theta$ graph, $\omega_0$ off the start of the $\omega$ graph, and $\alpha$ straight off its own constant graph. Decide positive, zero, or negative for each. The applet checks you and shows the rules whenever you want them.

§4

Where students go wrong.

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The three slips here all trace to reading a spinning body point by point instead of as one rigid object: one $\omega$ and one $\alpha$ shared by every point, tracked in a sign convention you fix once and keep.

Pitfall · 01

"The rim spins faster than the hub."

A point at the edge of a record covers more arc per second than a point near the center, so it feels faster. It is, in arc length. But angular velocity is the rate at which the radial spoke sweeps out angle, not the rate at which a point covers arc. For a rigid disk, the whole spoke rotates as one piece. Every point on it shares one $\omega$, every time.

Fix. Two different quantities to keep straight: linear speed (arc length per second, bigger for larger $r$) and angular speed ($\omega$, same for every point on the rigid body). Topic 5.2 will make the linkage $v = r\omega$ precise; in this lesson, just keep the two ideas separate. If the question asks for $\omega$, the answer does not depend on $r$.

Pitfall · 02

"$\omega$ is large, so $\alpha$ is large too."

$\omega$ and $\alpha$ are independent quantities. A wheel spinning steadily at $+10$ rad/s has $\alpha = 0$, even though $\omega$ is large. A wheel just released from rest has $\omega = 0$, even though $\alpha$ may be very large. The size of $\omega$ tells you nothing about the size of $\alpha$. On an $\omega$-vs-$t$ graph it's tempting to read the value as $\alpha$. The value is $\omega$. The slope is $\alpha$.

Fix. Treat them as two different ledgers. $\omega$ answers "how fast is the angle changing right now?" $\alpha$ answers "how fast is $\omega$ itself changing?" On a graph, the value gives $\omega$ and the slope gives $\alpha$. The two can be set to whatever you want, independently.

Pitfall · 03

"Mixing CW and CCW signs mid-problem."

A wheel decelerates by $\alpha = 2$ rad/s$^2$ for $3$ s starting from $\omega_0 = 6$ rad/s. What is the final $\omega$? A common answer: $6 + (2)(3) = 12$ rad/s, because "the wheel slowed down by $6$ rad/s." But "decelerates" usually means $\alpha$ is opposite to $\omega_0$. If CCW is positive and $\omega_0$ is $+6$, then a decelerating $\alpha$ is $-2$, not $+2$. The algebra has to know that. Pick a positive direction at the start and use it for every quantity in the problem: $\omega_0$, $\alpha$, $\omega$, $\Delta\theta$.

Fix. Declare CW or CCW positive in the first line of your work. Then write every quantity with its sign attached: $\omega_0 = +6$, $\alpha = -2$, and so on. Plug into $\omega = \omega_0 + \alpha t$ as written and let the algebra carry the signs. If you get confused, the fix is almost always going back and rewriting each quantity with its sign in your chosen convention.

§5

Skill Check.

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