So we differentiate each centre-of-mass position in time — the same
"position → velocity" move we used for the pendulum bob (Topic 11), now on the arm.
Link 1 — the easy one
Link 1's centre of mass is at \(x_{c1}=\ell_1\cos\theta_1,\ y_{c1}=\ell_1\sin\theta_1\) — only
\(\theta_1\) appears. Differentiate and square:
Exactly the pendulum-bob result: a point at radius \(\ell_1\) turning at
\(\dot\theta_1\) moves at \(v_{c1}=\ell_1\dot\theta_1\).
The meaty one
Link 2 — two angles change at once
Link 2's CoM depends on both angles, so both \(\dot\theta_1\) and \(\dot\theta_2\)
appear. Start from the FK position and differentiate (chain rule, with
\(\frac{d}{dt}(\theta_1+\theta_2)=\dot\theta_1+\dot\theta_2\)):
The speed \(v_{c2}\) of link 2's centre of mass — what
\(T=\tfrac12 m_2 v_{c2}^2\) needs.
📄 Full line-by-line derivation
(printable PDF) — both centres of mass, the square-and-add, and the
\(\cos\theta_2\) simplification. A pen-and-paper companion; this page already has the steps.
Square and add: the result
Squaring \(\dot x_{c2}\) and \(\dot y_{c2}\) and adding, the cross term simplifies with
\(\cos\theta_1\cos(\theta_1+\theta_2)+\sin\theta_1\sin(\theta_1+\theta_2)=\cos\theta_2\):
🔭 Looking ahead: this \(v_{c2}^2\) is exactly the kinetic-energy term
next theme drops into \(T=\tfrac12 m_2 v_{c2}^2\) — and the \(2L_1\ell_2\cos\theta_2\) term
becomes the off-diagonal coupling of the mass matrix \(M(q)\).
The pattern: velocity = (matrix) × (joint speeds)
Every velocity came out as the joint speeds \(\dot\theta_1,\dot\theta_2\) multiplied by
geometry. In matrix form:
The matrix \(J(q)\) is the Jacobian — it maps joint speeds to
point speeds, and it depends on the pose \(q\). You do not need its full theory now; just notice
the clean structure: point velocity is linear in the joint speeds.
How this plugs into the energy method
$$q \;\to\; \text{CoM positions} \;\to\; \text{CoM velocities} \;\to\; T \text{ and } V$$
Theme 4 hands Theme 5 exactly what it needs: where each mass is, and
how fast it moves. Then Lagrange's equation finishes the job — and the robot equation
\(M(q)\ddot q + C(q,\dot q)\dot q + G(q)=\tau\) falls out.
📐 Worked example
Evaluate \(v_{c2}^2\) at a pose
For \(L_1=1\ \text{m}\), \(\ell_2=0.5\ \text{m}\), at the instant \(\theta_2=60^\circ\),
\(\dot\theta_1=2\ \text{rad/s}\), \(\dot\theta_2=1\ \text{rad/s}\), find \(v_{c2}^2\)
(\(\cos 60^\circ=0.5\)).
1Evaluate the three terms (\(\dot\theta_1+\dot\theta_2=3\)):
The cross term is non-zero because \(\theta_2\ne 90^\circ\) — the two joints'
motions reinforce each other. That coupling is the heart of the robot's mass matrix.
✏️ Try it yourself
For \(L_1=2\ \text{m}\), \(\ell_2=1\ \text{m}\), at \(\theta_2=90^\circ\),
\(\dot\theta_1=1\ \text{rad/s}\), \(\dot\theta_2=1\ \text{rad/s}\), find \(v_{c2}^2\)
(\(\cos 90^\circ=0\)).
Terms.
\(L_1^2\dot\theta_1^2=4(1)=4\); \(\ell_2^2(\dot\theta_1+\dot\theta_2)^2=1(4)=4\)Cross term.
\(2(2)(1)\cos 90^\circ(\dots)=0\) — it vanishes because \(\cos 90^\circ=0\)Total.
\(v_{c2}^2=4+4+0=8\ \text{m}^2/\text{s}^2\) (\(v_{c2}\approx 2.83\ \text{m/s}\)). At
\(\theta_2=90^\circ\) the coupling term disappears.
Common mistakes to avoid
Mistake
Fix
Differentiating only \(\theta_1\)
Both \(\theta_1\) and \(\theta_2\) change in
time — link 2's CoM needs both.
Dropping the \((\dot\theta_1+\dot\theta_2)\) factor
Differentiating
\(\sin(\theta_1+\theta_2)\) brings down \(\dot\theta_1+\dot\theta_2\) by the chain rule.
Forgetting the cross term in \(v_{c2}^2\)
Squaring a sum gives a middle term:
\(2L_1\ell_2\cos\theta_2\,\dot\theta_1(\dot\theta_1+\dot\theta_2)\).
Using \(L_2\) instead of \(\ell_2\) for the CoM
The mass sits at \(\ell_2\)
along link 2, not at its tip \(L_2\).
\(v_{c1}^2=\ell_1^2\dot\theta_1^2\) — like the pendulum bob
Link 2
\(v_{c2}^2\) with the \(\cos\theta_2\) coupling term
The Jacobian
\(\dot{\mathbf p}=J(q)\dot q\) — speeds are linear in joint speeds
Next topic
DH parameters & 3-D (enrichment)
For 2–3 link arms, direct geometry is fastest. For 6-joint arms there is a systematic
bookkeeping recipe — Denavit–Hartenberg parameters — and the same ideas extend to 3-D. A short
"know it exists" tour closes the theme.