Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

WHY do we even need a bridge?

Setting up: arc length is the seed of everything

WHY radians matter: the relation $s = r\theta$ only works if $\theta$ is in radians. If you used degrees, you'd need ugly factors of $\pi/180$. Radians are designed so arc length = radius × angle, cleanly.

Deriving $v = r\omega$ from scratch

Derivation (HOW): Start from arc length $s = r\theta$. For a rigid body, $r$ is constant in time (the point doesn't drift closer to the axis). Differentiate with respect to time: $\frac{ds}{dt} = r\frac{d\theta}{dt}$ Now $\dfrac{ds}{dt} = v$ (speed = rate of change of distance along the path) and $\dfrac{d\theta}{dt} = \omega$ (angular velocity). Therefore: $\boxed{v = r\omega}$

• Why this step? We could differentiate because $r$ is constant — pull it out of the derivative. If $r$ changed, we'd get an extra term.
• WHAT it means: $v$ is the tangential speed — it points along the circle, perpendicular to the radius.

Deriving $a_t = r\alpha$ (tangential acceleration)

Derivation (HOW): Take $v = r\omega$ and differentiate again (rigid body $\Rightarrow r$ constant): $\frac{dv}{dt} = r\frac{d\omega}{dt} \quad\Rightarrow\quad a_t = r\alpha$ where $\alpha = d\omega/dt$ is the angular acceleration.

• Why this step? $a_t$ is the rate of change of speed. It exists only when the rotation is speeding up or slowing down. If $\omega$ is constant, $\alpha = 0$ and $a_t = 0$.
• WHAT it does: $a_t$ changes the magnitude of velocity, not its direction.

Deriving $a_c = r\omega^2$ (centripetal acceleration)

Derivation (HOW) — first principles: Even at constant speed, a point on a circle is accelerating because its velocity direction keeps changing. Write the position vector: $\vec{r} = r(\cos\theta,\ \sin\theta), \qquad \theta = \omega t$ Velocity (differentiate, $\omega$ constant): $\vec{v} = r\omega(-\sin\theta,\ \cos\theta)$ Acceleration (differentiate again): $\vec{a} = r\omega^2(-\cos\theta,\ -\sin\theta) = -\omega^2 \vec{r}$ The minus sign means $\vec{a}$ points toward the center (opposite to $\vec r$). Its magnitude: $\boxed{a_c = r\omega^2}$ Using $v = r\omega \Rightarrow \omega = v/r$, substitute: $a_c = r(v/r)^2 = v^2/r$. ✓

• Why this step? We differentiated the vector, not just the speed — that's how we captured the direction change, which is the whole source of centripetal acceleration.

Dual Coding: the geometry

The diagram shows a point on a rotating disk with $\vec v$ tangent, $\vec a_t$ tangent (along/against $v$), and $\vec a_c$ pointing to the center. Bigger $r$ → longer $\vec v$ arrow for the same $\omega$.

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Mnemonic

Feynman: explain to a 12-year-old

Flashcards

Connections

• Angular velocity and angular acceleration — the source quantities $\omega,\alpha$.
• Radian measure and arc length — why $s=r\theta$ underpins all of this.
• Centripetal force and circular motion — $F_c = m r\omega^2$ uses $a_c$.
• Uniform vs non-uniform circular motion — $a_t=0$ vs $a_t\ne 0$.
• Moment of inertia — kinetic energy $\frac12 m v^2$ summed using $v=r\omega$ gives $\frac12 I\omega^2$.
• Rolling motion — combines $v=r\omega$ at the contact point.

Concept Map

Hinglish (regional understanding)