Motion in a Plane (Resolved Components / Polar)

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2020)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

UPSC appeared once (2020) asking a derivation of the polar-component form of velocity and acceleration. The topic tests whether you can pass from Cartesian differentiation of $x = r\cos\theta$, $y = r\sin\theta$ to the clean radial-transverse decomposition used throughout dynamics. A clean, well-labelled derivation earns all marks in one go.

Minimum Theory

Polar coordinates in the plane

Let a particle’s position be $(r, \theta)$ in polar coordinates, with unit vectors $\hat{e}_r$ (radially outward) and $\hat{e}_\theta$ (transverse, $90°$ anticlockwise from $\hat{e}_r$).

$\hat{e}_r = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}, \qquad \hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}$

Time derivatives of the unit vectors:

$\dot{\hat{e}}_r = \dot{\theta}\,\hat{e}_\theta, \qquad \dot{\hat{e}}_\theta = -\dot{\theta}\,\hat{e}_r$

Position, velocity, acceleration

$\mathbf{r} = r\,\hat{e}_r$

Velocity:

$\mathbf{v} = \dot{r}\,\hat{e}_r + r\dot{\theta}\,\hat{e}_\theta$

• Radial component: $v_r = \dot{r}$
• Transverse component: $v_\theta = r\dot{\theta}$

Acceleration (differentiate $\mathbf{v}$):

$\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\,\hat{e}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\,\hat{e}_\theta$

• Radial component: $a_r = \ddot{r} - r\dot{\theta}^2$
• Transverse component: $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$

Cartesian derivation check

With $x = r\cos\theta$, $y = r\sin\theta$:

$\dot{x} = \dot{r}\cos\theta - r\dot{\theta}\sin\theta, \qquad \dot{y} = \dot{r}\sin\theta + r\dot{\theta}\cos\theta$

$\ddot{x} = (\ddot{r} - r\dot{\theta}^2)\cos\theta - (r\ddot{\theta} + 2\dot{r}\dot{\theta})\sin\theta$

$\ddot{y} = (\ddot{r} - r\dot{\theta}^2)\sin\theta + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\cos\theta$

confirming $\mathbf{a} = (\ddot{r}-r\dot{\theta}^2)\hat{e}_r + (r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{e}_\theta$.

Special cases

Circular motion ($r =$ const, $\dot{r} = 0$, $\ddot{r} = 0$): $a_r = -r\dot{\theta}^2 = -\frac{v^2}{r}, \qquad a_\theta = r\ddot{\theta}$

Radial motion ($\theta =$ const): $\mathbf{v} = \dot{r}\,\hat{e}_r, \qquad \mathbf{a} = \ddot{r}\,\hat{e}_r$

Question Archetypes

polar-derivation (1 question; 2020)

Recognition Cues

• The question asks to “derive” or “obtain” expressions for velocity or acceleration components.
• Polar or plane-polar coordinates $(r, \theta)$ are mentioned explicitly.
• May ask you to start from Cartesian and arrive at the polar form.
• Sometimes phrased as “find the radial and transverse components of acceleration.”

Solution Template

1. Write the position vector $\mathbf{r} = r\,\hat{e}_r$ and define $\hat{e}_r$, $\hat{e}_\theta$ in Cartesian terms.
2. Compute $\dot{\hat{e}}_r$ and $\dot{\hat{e}}_\theta$ by differentiating with respect to $t$ (chain rule through $\theta$).
3. Differentiate $\mathbf{r}$ once for velocity; collect $\hat{e}_r$ and $\hat{e}_\theta$ terms.
4. Differentiate velocity for acceleration; substitute $\dot{\hat{e}}_r$, $\dot{\hat{e}}_\theta$; collect terms.
5. Identify $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$; simplify $a_\theta = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$ if useful.

Worked Example

2020 Paper 1, 2020-P1-Q3a (10 marks)

Derive expressions for the radial and transverse components of the velocity and acceleration of a particle moving in a plane, using polar coordinates $(r, \theta)$.

Step 1. Set up unit vectors.

Define: $\hat{e}_r = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}, \qquad \hat{e}_\theta = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}.$

Note $\hat{e}_r \cdot \hat{e}_r = 1$, $\hat{e}_\theta \cdot \hat{e}_\theta = 1$, $\hat{e}_r \cdot \hat{e}_\theta = 0$.

Step 2. Time derivatives of unit vectors.

$\dot{\hat{e}}_r = \frac{d}{dt}(\cos\theta\,\hat{i} + \sin\theta\,\hat{j}) = (-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})\dot{\theta} = \dot{\theta}\,\hat{e}_\theta$

$\dot{\hat{e}}_\theta = \frac{d}{dt}(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j}) = (-\cos\theta\,\hat{i} - \sin\theta\,\hat{j})\dot{\theta} = -\dot{\theta}\,\hat{e}_r$

Step 3. Velocity.

$\mathbf{r} = r\,\hat{e}_r$

$\mathbf{v} = \dot{\mathbf{r}} = \dot{r}\,\hat{e}_r + r\dot{\hat{e}}_r = \dot{r}\,\hat{e}_r + r\dot{\theta}\,\hat{e}_\theta$

Radial component: $v_r = \dot{r}$; transverse component: $v_\theta = r\dot{\theta}$.

Step 4. Acceleration.

$\mathbf{a} = \dot{\mathbf{v}} = \frac{d}{dt}(\dot{r}\,\hat{e}_r) + \frac{d}{dt}(r\dot{\theta}\,\hat{e}_\theta)$

$= \ddot{r}\,\hat{e}_r + \dot{r}\dot{\hat{e}}_r + (\dot{r}\dot{\theta} + r\ddot{\theta})\hat{e}_\theta + r\dot{\theta}\dot{\hat{e}}_\theta$

$= \ddot{r}\,\hat{e}_r + \dot{r}\dot{\theta}\,\hat{e}_\theta + (\dot{r}\dot{\theta} + r\ddot{\theta})\hat{e}_\theta + r\dot{\theta}(-\dot{\theta}\,\hat{e}_r)$

$= (\ddot{r} - r\dot{\theta}^2)\hat{e}_r + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{e}_\theta$

Step 5. Final result.

$\boxed{a_r = \ddot{r} - r\dot{\theta}^2, \qquad a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})}$

Common Traps

• Forgetting the $-r\dot{\theta}^2$ centripetal term in $a_r$ (it comes from $r\dot{\theta}\dot{\hat{e}}_\theta = -r\dot{\theta}^2\hat{e}_r$).
• Writing $\dot{\hat{e}}_r = \dot{\theta}\hat{e}_r$ (wrong sign/direction); it must be $\dot{\theta}\hat{e}_\theta$.
• Omitting the cross-term $\dot{r}\dot{\theta}$ in $a_\theta$, giving $r\ddot{\theta}$ instead of $r\ddot{\theta} + 2\dot{r}\dot{\theta}$.
• Confusing $\dot{r}$ (rate of change of distance) with $|\mathbf{v}|$ (speed); they differ when $\dot{\theta}\neq 0$.

Marks-Aware Writing

This is a 10-mark derivation. Examiners look for:

1. Defined notation — state $\hat{e}_r$, $\hat{e}_\theta$ explicitly (2 marks).
2. Unit-vector derivatives — both $\dot{\hat{e}}_r$ and $\dot{\hat{e}}_\theta$ shown with working (3 marks).
3. Velocity derivation — one clean differentiation step (2 marks).
4. Acceleration derivation — product rule applied correctly, all four sub-terms collected (3 marks).

Write every line of algebra; do not skip the collection step. Underline or box the final components.

Practice Set

Only one historical question on this atom (shown above).