Classical-Mechanics · Unit 20 · Video 1 · Interactive Practice

Rigid Bodies and Fixed-Axis Rotation: Angular Velocity and Angular Acceleration

IKey Formulas

Formula	Name	What it says
$\|\vec{r}_P - \vec{r}_Q\| = \text{constant}$	Rigid body	Every pairwise distance is fixed in time
$r_i = \|\vec{r}_i\|$	Perpendicular distance from the axis	Radius of the circle element $i$ traces
$\vec{\omega} = \dfrac{d\theta}{dt}\,\hat{\mathbf{k}} = \omega_z\,\hat{\mathbf{k}}$	Angular velocity	One vector for the whole body, along the axis
$\vec{\alpha} = \dfrac{d^{2}\theta}{dt^{2}}\,\hat{\mathbf{k}} = \alpha_z\,\hat{\mathbf{k}}$	Angular acceleration	$\alpha_z$ is the second derivative of that same $\theta$

Key Insight: Rigidity freezes every pairwise distance, so no element can gain angle on another: one $\omega_z$ and one $\alpha_z$ describe the entire body, and only the radius $r_i$ distinguishes one mass element from the next.

IIEvery Element on Its Own Circle

An element's circle has radius equal to its perpendicular distance from the axis, not its distance from the center.

Element

x

(m) 3.00 Element

z

(m) 1.50

IIIRigidity Forces One Angular Velocity

If two elements turned at different rates, the distance between them would change — which rigidity forbids.

Time

t

(s) 0.0 Outer element

💡 The same argument fixes one $\vec{\alpha}$ for the body: angular accelerations that differed would make the angular velocities differ an instant later.

IVSpinning Up a Compact Disc

The motor's $\alpha_z$ is the slope of $\omega_z(t)$ , and the angle $\theta$ accumulates as the disc turns.

Time

t

(s) 0.0

VQuiz Questions

Problem 1 · The Radius of an Element's Circle

Given: A rigid body rotates about the fixed $z$ -axis. One mass element sits at $(x, y, z) = (3.0,\ 4.0,\ 6.0)\ \mathrm{m}$ — find the radius $r_i$ of the circle it traces.

✅ Correct!

r_i = \sqrt{x^2 + y^2} = \sqrt{9 + 16} = 5.0\ \mathrm{m}

— the perpendicular distance from the axis.

❌ Not quite.

z

measures displacement along the axis, not away from it; sliding the element up or down the axis leaves its circle exactly the same size.

❌ Close, but that is a different length.

\sqrt{3^2 + 4^2 + 6^2} = 7.8\ \mathrm{m}

is the distance from the origin;

r_i

is measured perpendicular to the axis.

❌ Not quite. Only the two coordinates perpendicular to the axis enter:

r_i = \sqrt{x^2 + y^2}

Show solution

The element moves on a circle about the $z$ -axis, so its radius is the perpendicular distance from that axis:

r_i = |\vec{r}_i| = \sqrt{x^2 + y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{25} = 5.0\ \mathrm{m}

The $z$ -coordinate is missing on purpose. Every point of the body at the same $x$ and $y$ , whatever its height, rides the same circle of radius $5.0\ \mathrm{m}$ .

For contrast, the distance from the origin is $\sqrt{9 + 16 + 36} = \sqrt{61} = 7.8\ \mathrm{m}$ — a length that plays no role in fixed-axis rotation.

Problem 2 · One Body, One Angular Velocity

Given: A rigid body turns about a fixed axis. Element $i$ , at $r_i = 0.10\ \mathrm{m}$ , has $\omega_z = 12\ \mathrm{rad/s}$ at this instant; element $j$ sits at $r_j = 0.30\ \mathrm{m}$ — find $\omega_z$ for element $j$ at that same instant.

✅ Correct! Rigidity gives the whole body one angle

\theta(t)

up to a constant offset, so

\omega_z = d\theta/dt

is shared:

12\ \mathrm{rad/s}

for every element.

❌ Not quite. Scaling

\omega_z

with the radius would let one element gain angle on the other, so the distance between them would change — exactly what rigidity forbids. (The quantity that does grow with

r

is the element's speed along its circle, the subject of the next video.)

❌ Not quite. Compare the two elements' angular separation: rigidity requires it to stay constant.

Show solution

Suppose the two angular velocities differed. In every instant the faster element would gain angle on the slower one, their angular separation would change, and with it the straight-line distance between them.

That contradicts the definition of a rigid body — the distance between any two of its points is constant in time. Hence

\omega_{z,j} = \omega_{z,i} = 12\ \mathrm{rad/s}

and the same argument gives one $\alpha_z$ for the whole body. The radii $r_i$ and $r_j$ never enter; they distinguish the elements' circles, not their angular rates.

Problem 3 · From the Angle to Its Two Derivatives

Given: A disc starts from rest and, while the motor drives it, its angle follows $\theta(t) = 2.0\,t^{2}$ (in $\mathrm{rad}$ , with $t$ in $\mathrm{s}$ ) — find $\omega_z$ at $t = 3.0\ \mathrm{s}$ and $\alpha_z$ during this phase.

What is $\omega_z$ at $t = 3.0\ \mathrm{s}$ ?

What is $\alpha_z$ during this phase?

✅ Correct!

\omega_z = d\theta/dt = 4.0\,t

gives

12\ \mathrm{rad/s}

t = 3.0\ \mathrm{s}

, and

\alpha_z = d^2\theta/dt^2 = 4.0\ \mathrm{rad/s^2}

, constant while the motor drives.

❌ Check the derivative.

\theta(3.0) = 18\ \mathrm{rad}

is the angle turned, not the rate; differentiate first, then substitute

t = 3.0\ \mathrm{s}

❌ Check the second derivative. Differentiate

\omega_z = 4.0\,t

once more — the result is a number, and it is not the coefficient

2.0

you started from.

Show solution

Step 1 — the angular velocity is the first derivative of the same $\theta$ :

\omega_z = \frac{d\theta}{dt} = \frac{d}{dt}\left(2.0\,t^{2}\right) = 4.0\,t \quad\Longrightarrow\quad \omega_z(3.0) = 4.0(3.0) = 12\ \mathrm{rad/s}

Step 2 — the angular acceleration is the second derivative:

\alpha_z = \frac{d^{2}\theta}{dt^{2}} = \frac{d}{dt}\left(4.0\,t\right) = 4.0\ \mathrm{rad/s^2}

So $\vec{\omega} = 12\,\hat{\mathbf{k}}\ \mathrm{rad/s}$ and $\vec{\alpha} = 4.0\,\hat{\mathbf{k}}\ \mathrm{rad/s^2}$ at that instant, and both describe every element of the disc at once.

Check: $\theta(3.0) = 2.0(9.0) = 18\ \mathrm{rad}$ — an angle, about $2.9$ turns, not a rate.

Problem 4 · Two Vectors Along the Axis

Given: A flywheel turns about the fixed $z$ -axis with $\theta(t) = 8.0\,t - 0.50\,t^{2}$ (in $\mathrm{rad}$ ), where $\theta$ is measured in the right-handed sense about $\hat{\mathbf{k}}$ — find both vectors at $t = 2.0\ \mathrm{s}$ .

What is $\vec{\omega}$ at $t = 2.0\ \mathrm{s}$ ?

What is $\vec{\alpha}$ at $t = 2.0\ \mathrm{s}$ ?

✅ Excellent!

\omega_z = 8.0 - t

+6.0\ \mathrm{rad/s}

t = 2.0\ \mathrm{s}

, so

\vec{\omega}

points along

+\hat{\mathbf{k}}

, while

\alpha_z = -1.0\ \mathrm{rad/s^2}

at every instant sends

\vec{\alpha}

along

-\hat{\mathbf{k}}

— antiparallel to

\vec{\omega}

❌ Check the first derivative.

\theta(2.0) = 14\ \mathrm{rad}

is the angle and

8.0\ \mathrm{rad/s}

is the rate at

t = 0

; you need

\omega_z = d\theta/dt

evaluated at

t = 2.0\ \mathrm{s}

❌ Check the second derivative. Differentiating

-0.50\,t^{2}

twice brings down a factor of

2

, and the sign of the result decides which way along the axis

\vec{\alpha}

points.

Show solution

Step 1 — differentiate once:

\omega_z = \frac{d\theta}{dt} = 8.0 - 1.0\,t \quad\Longrightarrow\quad \omega_z(2.0) = 8.0 - 2.0 = +6.0\ \mathrm{rad/s}

Step 2 — differentiate again:

\alpha_z = \frac{d^{2}\theta}{dt^{2}} = -1.0\ \mathrm{rad/s^2} \quad (\text{the same at every } t)

Step 3 — attach the direction. Both vectors lie along the axis, and the sign of the $z$ -component picks the sense:

\vec{\omega} = \omega_z\,\hat{\mathbf{k}} = 6.0\,\hat{\mathbf{k}}\ \mathrm{rad/s}, \qquad \vec{\alpha} = \alpha_z\,\hat{\mathbf{k}} = -1.0\,\hat{\mathbf{k}}\ \mathrm{rad/s^2}

With $\theta$ measured in the right-handed sense, $\omega_z > 0$ puts $\vec{\omega}$ along $\hat{\mathbf{k}}$ — curl the fingers of your right hand with increasing $\theta$ and the thumb points that way. Here $\alpha_z < 0$ , so $\vec{\alpha}$ points along $-\hat{\mathbf{k}}$ , opposite to $\vec{\omega}$ .

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