Mistake Master
Mistake Master
Topic 5.1 said the loud part out loud: a rigid body has one $\omega$, shared by every point. Topic 5.2 says the quiet, consequential part: the linear quantities at each point are not shared. Same disk, same $\omega$, totally different $v$ at the hub and at the rim. The fix is a one-line conversion, applied three times, with the radius $r$ as the bridge: $s = r\theta$, $v = r\omega$, $a_t = r\alpha$.
Imagine a record on a turntable, spinning steadily. The whole record turns through the same angle each second, so every point on the record has the same $\omega$. That is the rigid-body promise from 5.1: one angular velocity, shared by the whole disk.
Now ask: how fast is the needle moving in m/s? How fast is the edge of the record moving in m/s? The needle traces a small circle each second; the edge traces a much larger one. The edge moves faster. The angular velocity is shared. The linear velocity is not.
The conversion is one line:
$$v = r\omega$$where $v$ is the tangential speed at a point, $r$ is that point's distance from the axis, and $\omega$ is the body's angular velocity. The same pattern works for arc length and tangential acceleration:
$$s = r\theta \quad \quad a_t = r\alpha$$Three quantities, one bridge. The radius $r$ converts a shared angular quantity into a point-specific linear one.
Three formulas, one structure. Each takes an angular quantity the whole body shares and multiplies by $r$ to get a linear quantity at a point.
Worked example. A grinding disk of radius $R = 0.30$ m spins at $\omega = +4.0$ rad/s with angular acceleration $\alpha = +2.0$ rad/s$^2$. Find the arc length traced, the tangential speed, and the tangential acceleration at the rim over the first $2.0$ seconds. Then repeat at the midpoint of the disk, $r = 0.15$ m.
The whole disk shares these: at $t = 2.0$ s, $\omega = \omega_0 + \alpha t = 4.0 + (2.0)(2.0) = 8.0$ rad/s. The angular displacement over the $2.0$ s is $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = (4.0)(2.0) + \tfrac{1}{2}(2.0)(2.0)^2 = 12$ rad.
Multiply each angular quantity by $r$. Arc length swept: $s = r\theta = (0.30)(12) = 3.6$ m. Tangential speed at $t = 2.0$ s: $v = r\omega = (0.30)(8.0) = 2.4$ m/s. Tangential acceleration: $a_t = r\alpha = (0.30)(2.0) = 0.60$ m/s$^2$.
Same $\theta$, $\omega$, $\alpha$ (the disk is rigid; angular quantities are shared). Apply the linkage at the new radius. $s = (0.15)(12) = 1.8$ m, $v = (0.15)(8.0) = 1.2$ m/s, $a_t = (0.15)(2.0) = 0.30$ m/s$^2$. Every linear quantity halved because $r$ halved.
One review-only note for completeness: a point on a rotating body also has a centripetal acceleration $a_c = r\omega^2$ pointing toward the axis. That belongs to Unit 2's circular-motion topic. Topic 5.2 focuses on the tangential piece $a_t = r\alpha$, which is what changes with $\alpha$.
Put two physical dots on a spinning disk: $P$ at the rim ($r = R$) and $Q$ at half the radius ($r = R/2$). Compare what they share and what they do not.
This is the single hardest move in 5.2: holding two things in your head at the same time. The body has one shared $\omega$. Each point has its own $v$. They are linked by $v = r\omega$, and the link is exactly what makes them different at different radii.
Two of the three traps here are the same mistake (dropping the factor of $r$) in different costumes. The third is the rigid-body promise from 5.1 misapplied to a linear quantity.
The units settle it: $\omega$ is in rad/s, $v$ is in m/s. Different units, different quantities. The instinct comes from informally saying "how fast is the disk going?" without distinguishing whether you mean angular or linear. For an extended body, those are different numbers.
Fix. Write $v = r\omega$ every time. The factor of $r$ carries the meters and turns rad/s into m/s.
The first half is right: the rim and the hub do share $\omega$. The second half misses what $v = r\omega$ says. With one $\omega$, the rim's $v$ scales with the rim's $r$; the hub's $v$ scales with the hub's tiny $r$. The two numbers are different. A child near the center of a merry-go-round moves slowly; one at the edge has to hold on.
Fix. Compare radii. "Rim is twice as far out, so rim's $v$ is twice as big." Angular quantities are shared because the body is rigid; linear quantities are not, because $r$ is in the conversion.
The radian is defined so arc length and angle are numerically equal at $r = 1$ m. That convenience hides the radius. At any other radius, $s = r\theta$, with $r$ in meters. Example: a turntable rotates through $\theta = 8$ rad. A point at $r = 0.10$ m sweeps not $8$ m but $s = (0.10)(8) = 0.80$ m.
Fix. Track the units. $\theta$ in radians, $r$ in meters, $s$ in meters. The product $r\theta$ has meters; $\theta$ alone does not.