9. Mechanics — Dynamics

Radial and transverse velocity and acceleration

Radial and transverse velocity and acceleration resolve a particle's motion into components along and perpendicular to the radius vector in polar coordinates. This topic covers these kinematic formulas.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Kinematic Relations in Polar Coordinates

Let a particle move in a plane curve such that its position at time $t$ is given in polar coordinates by $(r, \theta)$. The radial velocity is $v_r = \dot{r}$, the transverse velocity is $v_t = r\dot{\theta}$, the radial acceleration is $f_r = \ddot{r} - r\dot{\theta}^2$, and the transverse acceleration is $f_t = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) = r\ddot{\theta} + 2\dot{r}\dot{\theta}$.

📐 Geometric Interpretation

The radial velocity represents the rate of change of the distance from the origin, while the transverse velocity represents the linear speed due to angular rotation perpendicular to the radius vector. Similarly, radial and transverse accelerations capture the net inward/outward and Coriolis-like rotational acceleration components.

⚙️ Algebraic & Structural Meaning

Deriving these expressions relies on differentiating unit vectors $\hat{r} = \cos\theta \hat{i} + \sin\theta \hat{j}$ and $\hat{\theta} = -\sin\theta \hat{i} + \cos\theta \hat{j}$ with respect to time $t$, utilizing the chain rule where $\dot{\theta} = \frac{d\theta}{dt}$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Differentiability of position coordinates	`N/A`	Strictly Mandatory	If $r(t)$ or $\theta(t)$ fails to be twice differentiable, velocities and accelerations cease to exist at those instants.
Non-zero radius vector	`N/A`	Essential for transverse acceleration components	Division by $r$ in transverse acceleration formulas becomes undefined at the pole $r = 0$, requiring Cartesian fallback or limiting forms.
Single-valued polar coordinates	`N/A`	Mandatory for uniqueness	Multivalued angles introduce discontinuities in angular velocity $\dot{\theta}$ unless properly unwrapped.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Transverse Velocity via Angular Momentum Form

$$v_t = r\dot{\theta} = \frac{h}{r}$$

Rule: undefined

Application: Directly applied in central force problems where $h$ is conserved in UP PGT mechanics.

Radial Acceleration for Circular Orbits

$$f_r = -r\dot{\theta}^2 = -\frac{v^2}{r}$$

Rule: For circular motion about the origin ($r = \text{constant} \implies \dot{r} = 0, \ddot{r} = 0$), radial acceleration collapses to $-r\dot{\theta}^2$.

Application: Frequent shortcut for uniform and non-uniform circular motion queries in TGT/PGT.

Transverse Acceleration Zero Condition

$$r^2\dot{\theta} = \text{constant} \implies f_t = 0$$

Rule: If a particle moves under a central force, transverse acceleration $f_t = 0$, which immediately implies $r^2\dot{\theta} = h$ (Kepler's Second Law equivalent).

Using Pedal Equation Substitution

$$p = r \sin \phi \implies \dot{r} = \dots$$

Rule: Substitute $r = f(\theta)$ and use $\frac{1}{p^2} = \frac{1}{r^2} + \frac{1}{r^4}\left(\frac{dr}{d\theta}\right)^2$ for pedal coordinate transformations.

Application: Saves massive algebra in complicated path velocity questions.

Direct Velocity Magnitude Formula

$$v = \sqrt{v_r^2 + v_t^2} = \sqrt{\dot{r}^2 + r^2\dot{\theta}^2}$$

Rule: Always compute radial and transverse velocities separately and combine using Pythagorean sum rather than transforming back to Cartesian unless necessary.

Rapid Elimination via Dimensions

$$[v_r] = [LT^{-1}], \quad [f_r] = [LT^{-2}]$

Rule: Eliminate options failing dimensional homogeneity for velocity and acceleration expressions instantly.

Application: Effective for knocking out 2 incorrect options in numerical UP PGT problems.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting the $r\dot{\theta}^2$ term in Radial Acceleration

Common Pitfall: Candidates often write radial acceleration simply as $\ddot{r}$, forgetting the centripetal component $-r\dot{\theta}^2$ arising from the rotation of the unit vector $\hat{r}$.

Counterexample: $r = a, \theta = \omega t \implies f_r = -a\omega^2$

Why it fails: Ignoring $-r\dot{\theta}^2$ leads to missing the centripetal acceleration in circular motion.

🛡️ Exam Defense: Memorize the exact formula: $f_r = \ddot{r} - r\dot{\theta}^2$.

⚠️ COMMON ERROR #02

Misinterpreting Transverse Acceleration Formula

Common Pitfall: Confusing transverse acceleration $r\ddot{\theta} + 2\dot{r}\dot{\theta}$ with $\frac{1}{r}\frac{d}{dt}(r\dot{\theta})$.

Counterexample: $f_t = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) \neq \frac{d}{dt}(r\dot{\theta})$

Why it fails: Leads to incorrect expansion and sign errors during exams.

🛡️ Exam Defense: Always expand $f_t$ as $\frac{1}{r} \frac{d}{dt} (r^2 \dot{\theta}) = r\ddot{\theta} + 2\dot{r}\dot{\theta}$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Radial Velocity ($v_r$)	Smooth position vector $r(t)$	$v_r = \dot{r} = \frac{dr}{dt}$	Speed along the radius vector away from origin	Keywords: 'along the radius', 'rate of change of distance'
Transverse Velocity ($v_t$)	Smooth angular motion $\theta(t)$	$v_t = r\dot{\theta}$	Speed perpendicular to radius vector	Keywords: 'perpendicular to radius', 'angular speed component'
Radial Acceleration ($f_r$)	Twice differentiable polar motion	$f_r = \ddot{r} - r\dot{\theta}^2$	Net inward/outward acceleration along radius	Keywords: 'radial acceleration', 'centripetal term'
Transverse Acceleration ($f_t$)	Twice differentiable polar motion	$f_t = r\ddot{\theta} + 2\dot{r}\dot{\theta} = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$	Acceleration perpendicular to radius vector	Keywords: 'transverse acceleration', 'central force check ($f_t = 0$)'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If a particle moves along the curve $r = a e^{\theta}$ with constant angular velocity $\dot{\theta} = \omega$, find its transverse velocity at any time $t$.

(A) $a \omega e^{\theta}$(B) $a \omega^2 e^{\theta}$(C) $2a \omega e^{\theta}$(D) $a \omega \theta$

📚 Method A: Formal Textbook Derivation

Step 1: Identify given quantities $r = a e^{\theta}$ and $\dot{\theta} = \omega$. Step 2: Transverse velocity formula is $v_t = r\dot{\theta}$. Step 3: Substitute $r$ and $\dot{\theta}$ to get $v_t = (a e^{\theta})\omega = a\omega e^{\theta}$.

⚡ Method B: 25-Second Exam Shortcut

Direct substitution: $v_t = r\omega = a\omega e^{\theta}$.

💡 Tip: Do not confuse transverse velocity with transverse acceleration.

CANONICAL PROBLEM 02 Dual-Method Breakdown

A particle moves in a plane such that its radial acceleration is zero. What condition must satisfy the angular momentum per unit mass $h = r^2\dot{\theta}$?

(A) $h$ is zero(B) $h$ is constant(C) $\ddot{r} = r\dot{\theta}^2$(D) $\dot{r} = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Radial acceleration is $f_r = \ddot{r} - r\dot{\theta}^2$. Step 2: Set $f_r = 0$. Step 3: This yields $\ddot{r} = r\dot{\theta}^2$.

⚡ Method B: 25-Second Exam Shortcut

Direct translation of equation $f_r = 0 \implies \ddot{r} = r\dot{\theta}^2$.

💡 Tip: Read carefully whether radial or transverse acceleration is set to zero.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The radial component of velocity for a particle moving in polar coordinates is given by $v_r = $

A $r\dot{\theta}$

B $\dot{r}$

C $\ddot{r}$

D $r\ddot{\theta}$

Derivation & Analysis:

Radial velocity is defined as the rate of change of radius vector with respect to time, i.e., $v_r = \dot{r}$.

MCQ 2: The transverse component of velocity is expressed as $v_t = $

A $r\dot{\theta}$

B $\dot{r}$

C $\frac{1}{r}\dot{\theta}$

D $r^2\dot{\theta}$

Derivation & Analysis:

Transverse velocity is the product of radius and angular velocity, $v_t = r\dot{\theta}$.

MCQ 3: Radial acceleration $f_r$ in polar coordinates is given by

A $\ddot{r} + r\dot{\theta}^2$

B $\ddot{r} - r\dot{\theta}^2$

C $r\ddot{\theta} + 2\dot{r}\dot{\theta}$

D $r\ddot{\theta} - r\dot{\theta}^2$

Derivation & Analysis:

By differentiating position vector twice in polar coordinates, radial acceleration is $f_r = \ddot{r} - r\dot{\theta}^2$.

MCQ 4: Transverse acceleration $f_t$ can be written in alternative form as

A $\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$

B $r\frac{d}{dt}(\dot{\theta})$

C $\frac{d^2r}{dt^2}$

D $2\dot{r}\dot{\theta}$

Derivation & Analysis:

Transverse acceleration is identically equal to $\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) = r\ddot{\theta} + 2\dot{r}\dot{\theta}$.

MCQ 5: If a particle moves with a constant radius $r = a$, its radial velocity is

A a

B 0

C $a\omega$

D undefined

Derivation & Analysis:

If $r = a$ is constant, then $\dot{r} = 0$, making radial velocity zero.

MCQ 6: Under a central force, which of the following is identically zero?

A Radial velocity $v_r$

B Radial acceleration $f_r$

C Transverse acceleration $f_t$

D Total acceleration $f$

Derivation & Analysis:

A central force acts entirely along the radius vector, so the transverse component of acceleration $f_t = 0$.

MCQ 7: If $r^2\dot{\theta} = h$ (constant), then transverse acceleration $f_t$ is

A h

B 0

C $2h$

D $h/r^2$

Derivation & Analysis:

Since $f_t = \frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$ and $r^2\dot{\theta}$ is constant, its derivative is zero, so $f_t = 0$.

MCQ 8: Assertion (A): Radial acceleration contains a term $-r\dot{\theta}^2$. Reason (R): This term represents the centripetal acceleration due to rotation about the origin.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

The term $-r\dot{\theta}^2$ in radial acceleration is indeed the centripetal acceleration component arising from the unit vector's rotation.

MCQ 9: If a particle describes a circle of radius $a$ with constant angular speed $\omega$, its total acceleration magnitude is

A $a\omega^2$

B $2a\omega^2$

C 0

D $a\omega$

Derivation & Analysis:

For circular motion with $r=a$ and constant $\omega$, $\ddot{r}=0, \dot{r}=0, \ddot{\theta}=0$. Thus $f_r = -a\omega^2$ and $f_t = 0$, giving magnitude $a\omega^2$.

Q10

MCQ 10: If $r = a(1 + \cos\theta)$ and $\dot{\theta} = \omega$, the radial velocity $\dot{r}$ is equal to

A $-a \omega \sin\theta$

B $a \omega \sin\theta$

C $-a \omega \cos\theta$

D 0

Derivation & Analysis:

Using chain rule, $\dot{r} = \frac{dr}{d\theta}\dot{\theta} = (-a\sin\theta)\omega = -a\omega\sin\theta$.

Common questions

Unlike fixed Cartesian unit vectors $\hat{i}$ and $\hat{j}$, the polar unit vectors $\hat{r}$ and $\hat{\theta}$ change their spatial direction as the angle $\theta$ changes with time. Specifically, $\frac{d\hat{r}}{dt} = \dot{\theta}\hat{\theta}$ and $\frac{d\hat{\theta}}{dt} = -\dot{\theta}\hat{r}$.

Confusing partial differentiation with total time derivatives or missing the factor of $2$ in the Coriolis-like term $2\dot{r}\dot{\theta}$ of transverse acceleration.

Memorize standard velocity/acceleration formulas in polar, pedal, and intrinsic coordinates, and leverage shortcuts like $r^2\dot{\theta} = h$ for central forces to skip lengthy intermediate integrations.

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