Velocity from position

Kinetic energy needs a speed. But we describe a machine by its coordinate. This is the one skill that turns "where it is" into "how fast it goes."

Source: course notes, Week 2 (the energy method).

Before you start

What you need first

Kinetic energy $T=\tfrac12 m v^2$ — so we need $v$.
Coordinate & velocity $q,\dot q$ — and that $\dot q$ means $dq/dt$.
The chain rule from calculus.

What you'll be able to do

Turn a part's position into its velocity.
Use $v^2 = \dot x^2 + \dot y^2$ to get a speed for $T$.
Find the speed of a swinging pendulum bob.

The gap to bridge

Last topic, kinetic energy came out as $T=\tfrac12 m v^2$ — it needs the speed $v$ of each part. But we describe a machine by its coordinate $q$ (an angle, a displacement), not by speeds.

So we need a bridge: given where a part is (its position as a function of $q$), find how fast it moves (its speed $v$). That bridge is one chain-rule step.

The recipe

Position → velocity → speed

1Write the position of the mass as a function of the coordinate $q$: its $x$ and $y$.

2Differentiate in time to get the velocity components $\dot x,\dot y$ (chain rule — see below).

3Square and add for the speed squared (exactly what $T$ needs):

$$v^{2} = \dot x^{2} + \dot y^{2}$$

The chain rule. Position depends on $q$, and $q$ depends on time, so $\dot x = \dfrac{dx}{dq}\,\dot q$. For example, if $x = L\sin\theta$ then $\dot x = L\cos\theta\,\dot\theta$ — the $\dot\theta$ must appear.

📐 Worked example

Speed of a pendulum bob

A bob hangs on a light rod of length $L$, at angle $\theta$ from straight-down. Find its speed $v$ (and hence $v^2$ for the kinetic energy).

Bob position: $x=L\sin\theta$ across, $y=L\cos\theta$ down. The velocity $v$ is tangent to the swing.

1Position (pivot at origin, $y$ measured down):

$$x = L\sin\theta, \qquad y = L\cos\theta$$

2Differentiate in time (chain rule, $L$ constant):

$$\dot x = L\cos\theta\,\dot\theta, \quad \dot y = -L\sin\theta\,\dot\theta$$

3Square and add — $\sin^2+\cos^2=1$ tidies it up:

$$v^{2} = \dot x^{2}+\dot y^{2} = L^{2}\dot\theta^{2}$$

So the bob's speed is $v = L\dot\theta$, and its kinetic energy is $T=\tfrac12 m v^2 = \tfrac12 m L^2\dot\theta^2$ — ready to drop into the energy method.

✏️ Try it yourself

A point sits on the rim of a wheel of radius $R$, at angle $\phi$: its position is $x = R\cos\phi$, $y = R\sin\phi$. Find $v^2$.

Step 1. $\dot x = -R\sin\phi\,\dot\phi, \qquad \dot y = R\cos\phi\,\dot\phi$ Step 2. $v^2 = \dot x^2 + \dot y^2 = R^2\dot\phi^2(\sin^2\phi+\cos^2\phi) = R^2\dot\phi^2$ Read it: a point on the rim moves at $v = R\dot\phi$ — same shape as the pendulum bob ($L\dot\theta$), because both go round a circle.

Common mistakes to avoid

Mistake	Fix
Dropping the $\dot\theta$ after differentiating	Chain rule: $\frac{d}{dt}\sin\theta = \cos\theta\,\dot\theta$. The $\dot\theta$ must be there.
Using $v$ when you need $v^2$	Kinetic energy needs $v^2=\dot x^2+\dot y^2$.
Forgetting a coordinate's contribution	If a part depends on two coordinates, differentiate with respect to both (each gets a $\dot q$).
Treating $L$ (a fixed length) as changing	Constants come straight through the time-derivative; only $\theta$ changes with time here.

Recap — the whole topic on one screen

$$\text{position}(q) \;\xrightarrow{\ d/dt\ }\; \dot x,\dot y \;\xrightarrow{\ \text{square+add}\ }\; v^{2}=\dot x^{2}+\dot y^{2}$$

Idea	What you own now
The bridge	Position in terms of $q$ → speed in terms of $\dot q$
The tool	Chain rule: $\dot x = \frac{dx}{dq}\dot q$
The result	$v^2=\dot x^2+\dot y^2$, ready for $T=\tfrac12 mv^2$
The pendulum	$v^2 = L^2\dot\theta^2$ ⟹ $T=\tfrac12 mL^2\dot\theta^2$

Next topic

The Lagrangian & Lagrange's equation

Now we have both halves — $T$ (with the speeds we just found) and $V$. Next we combine them into the Lagrangian $\mathcal{L}=T-V$ and turn a crank — Lagrange's equation — that drops out the equation of motion.

→ The Lagrangian & Lagrange's equation

Mistake	Fix
Dropping the \(\dot\theta\) after differentiating	Chain rule: \(\frac{d}{dt}\sin\theta = \cos\theta\,\dot\theta\). The \(\dot\theta\) must be there.
Using \(v\) when you need \(v^2\)	Kinetic energy needs \(v^2=\dot x^2+\dot y^2\).
Forgetting a coordinate's contribution	If a part depends on two coordinates, differentiate with respect to both (each gets a \(\dot q\)).
Treating \(L\) (a fixed length) as changing	Constants come straight through the time-derivative; only \(\theta\) changes with time here.

Idea	What you own now
The bridge	Position in terms of \(q\) → speed in terms of \(\dot q\)
The tool	Chain rule: \(\dot x = \frac{dx}{dq}\dot q\)
The result	\(v^2=\dot x^2+\dot y^2\), ready for \(T=\tfrac12 mv^2\)
The pendulum	\(v^2 = L^2\dot\theta^2\) ⟹ \(T=\tfrac12 mL^2\dot\theta^2\)