Curvilinear Motion: Polar Coordinates - Statics and Dynamics

1. Which of the following correctly describes the acceleration vector of the particle at this instant?
   Answer: $\vec{a} = -20\hat{e}_r + 20\hat{e}_\theta$ m/s², where $a_r = \ddot{r} - r\dot{\theta}^2 = 0 - 5(2)^2 = -20$ m/s² and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 5(4) + 2(0)(2) = 20$ m/s²; both $\ddot{r} = 0$ and the Coriolis term vanish explicitly because $r = \text{const}$ implies $\dot{r} = 0$ When a particle moves in polar coordinates, its acceleration has two components derived by differentiating the position vector twice: $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$. The key insight for circular motion problems is that a constant radius means $\dot{r} = 0$ and $\ddot{r} = 0$, which simplifies these formulas — but you must still write them out completely and justify each simplification explicitly. Answer D does exactly this. Since $r = 5$ m = const, $\dot{r} = 0$ and $\ddot{r} = 0$. Substituting: $a_r = 0 - 5(2)^2 = -20$ m/s² and $a_\theta = 5(4) + 2(0)(2) = 20$ m/s², giving $\vec{a} = -20\hat{e}_r + 20\hat{e}_\theta$ m/s². Every term is accounted for and every zero is justified — that's rigorous physics. Answer A arrives at the same numerical result but drops $\ddot{r}$ and $2\dot{r}\dot{\theta}$ without explanation. On an exam, silently omitting terms suggests you don't know why they vanish — D is preferred precisely because it demonstrates full understanding. Answer B corrupts the $a_\theta$ formula by adding a spurious $r\dot{\theta}$ term, as if an extra velocity-dependent contribution exists in the transverse direction. No such term appears in the correct derivation. Answer C incorrectly flips the sign of the centripetal term, treating $-r\dot{\theta}^2$ as positive. The centripetal acceleration always points inward (toward the center), meaning $a_r$ must be negative. Study tip: Always write both full polar acceleration formulas first, then substitute $\dot{r} = 0$ and $\ddot{r} = 0$ with an explicit reason. Showing your zeros is what separates a complete solution from a lucky answer.
2. What is the radial component of acceleration $a_r$ at this instant?
   Answer: $a_r = 2\ddot{\theta} - r\dot{\theta}^2 = -2 - \pi(9) = -2 - 9\pi$ m/s², found by correctly computing $\ddot{r} = 2\ddot{\theta}$ and using $r = 2\theta = \pi$ at the given instant When analyzing motion in polar coordinates, your go-to formulas for acceleration components are $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$. The key challenge is correctly computing $\ddot{r}$ by differentiating the path equation, then carefully substituting the actual values at the given instant. Since $r = 2\theta$, differentiating once gives $\dot{r} = 2\dot{\theta}$, and differentiating again gives $\ddot{r} = 2\ddot{\theta}$. At $\theta = \pi/2$: $r = 2(\pi/2) = \pi$ m, $\ddot{r} = 2(-1) = -2$ m/s². Plugging into the radial acceleration formula: $a_r = \ddot{r} - r\dot{\theta}^2 = -2 - \pi(3)^2 = -2 - 9\pi \approx -30.3$ m/s². That's exactly what B computes — it correctly evaluates $r = \pi$ at the given instant rather than leaving it as a generic expression. A is tempting because $\ddot{r} = 2\ddot{\theta}$ is correct, but it substitutes $r = 2\theta$ symbolically and treats $\theta$ as a number (likely $\pi/2 \approx 1$) without properly evaluating $r = \pi$, yielding the wrong numerical result of $-20$ m/s². C uses the wrong sign in the formula — adding $r\dot{\theta}^2$ instead of subtracting it. The centripetal term always subtracts in $a_r$; mixing up this sign is one of the most common errors in polar kinematics. D correctly finds $\ddot{r} = -2$, but then substitutes $\dot{r}\dot{\theta}$ instead of $r\dot{\theta}^2$, confusing two entirely different terms from the acceleration formula. Your study tip: always write the polar acceleration formulas from scratch before substituting — it forces you to catch sign errors and prevents mixing up $r$, $\dot{r}$, and $\ddot{r}$.
3. What is the angle $\phi$ that the net force vector makes with the radial direction (measured from $-\hat{e}_r$ toward $\hat{e}_\theta$)?
   Answer: $\phi = \arctan\!\left(\dfrac{a_\theta}{|a_r|}\right) = \arctan\!\left(\dfrac{4}{16}\right) = \arctan(0.25) \approx 14.0°$, using $a_r = -r\dot{\theta}^2 = -4(4) = -16$ m/s² and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 4(1) + 0 = 4$ m/s² When working with polar coordinate dynamics, your first job is always to write down the correct acceleration formulas before plugging in any numbers. The two components are: $a_r = \ddot{r} - r\dot{\theta}^2 \qquad a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$ With $\ddot{r} = 0$, $r = 4$ m, $\dot{r} = 0$, $\dot{\theta} = 2$ rad/s, and $\ddot{\theta} = 1$ rad/s², these give: $a_r = 0 - 4(2)^2 = -16 \text{ m/s}^2, \qquad a_\theta = 4(1) + 2(0)(2) = 4 \text{ m/s}^2$ The negative $a_r$ means the radial force points inward ($-\hat{e}_r$). The angle $\phi$ measured from $-\hat{e}_r$ toward $\hat{e}_\theta$ is therefore $\phi = \arctan\!\left(\dfrac{a_\theta}{|a_r|}\right) = \arctan\!\left(\dfrac{4}{16}\right) \approx 14.0°$, confirming C. Choice A corrupts the radial formula by subtracting a spurious $r\ddot{\theta}$ term, leaving $|a_r| = 12$ instead of 16 — the radial acceleration has no $\ddot{\theta}$ dependence whatsoever. Choice B gets both acceleration magnitudes right but flips the arctangent ratio, effectively measuring the angle from $\hat{e}_\theta$ rather than from $-\hat{e}_r$, which swaps numerator and denominator. Choice D correctly finds $a_r = -16$ m/s² but inflates the transverse acceleration to 12 m/s² by tacking on an extra $r\dot{\theta}$ term that has no place in the $a_\theta$ formula. Study tip: Memorize both polar acceleration formulas as a matched pair and check each term's origin — every symbol has a physical meaning, so phantom terms (like $r\dot{\theta}$ in $a_\theta$ or $r\ddot{\theta}$ in $a_r$) should raise an immediate red flag.
4. What is $|\dot{r}|$ at this instant?
   Answer: $|\dot{r}| = \sqrt{v^2 - (r\dot{\theta})^2} = \sqrt{100 - 36} = 8$ m/s, using the correct decomposition $v^2 = \dot{r}^2 + (r\dot{\theta})^2$ with $r\dot{\theta} = 2(3) = 6$ m/s When a particle moves in polar coordinates, its velocity has two orthogonal components: a radial component $v_r = \dot{r}$ (along the $\hat{e}_r$ direction) and a transverse component $v_\theta = r\dot{\theta}$ (along the $\hat{e}_\theta$ direction). Because these components are perpendicular, speed is found using the Pythagorean theorem: $v^2 = \dot{r}^2 + (r\dot{\theta})^2$. This is the foundational formula to recall whenever you're given polar-coordinate kinematics. Applying it here, the transverse speed is $r\dot{\theta} = (2)(3) = 6$ m/s. Solving for $|\dot{r}|$: $|\dot{r}| = \sqrt{v^2 - (r\dot{\theta})^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ m/s. That's answer A, the correct choice. B is wrong because it subtracts the components algebraically — $10 - 6 = 4$ m/s — as if velocity were a scalar sum. Orthogonal vectors don't add or subtract that way; you must use the Pythagorean relationship. C makes the mistake of dropping the radius $r$, plugging in $\dot{\theta} = 3$ rad/s directly instead of the transverse speed $r\dot{\theta} = 6$ m/s. Remember: $\dot{\theta}$ has units of rad/s, not m/s — multiplying by $r$ converts it to a speed. D adds $(r\dot{\theta})^2$ instead of subtracting it, which would make $\dot{r}$ larger than $v$ — physically impossible, since $\dot{r}$ is just one component of the total velocity. Study tip: Always write $v^2 = \dot{r}^2 + (r\dot{\theta})^2$ first, then identify which quantity is unknown. Never forget the $r$ in $r\dot{\theta}$ — it's one of the most common errors on polar-coordinate problems.
5. At $t = 1$ s, what is the magnitude of the net force acting on the particle?
   Answer: $F = m\sqrt{a_r^2 + a_\theta^2} = 2\sqrt{(3)^2 + (12)^2} = 2\sqrt{153} \approx 24.7$ N, found using $a_r = \ddot{r} - r\dot{\theta}^2 = 6 - 3 = 3$ m/s² and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 + 2(6)(1) = 12$ m/s² When a particle moves in polar coordinates, you need two acceleration components — radial and transverse — even when no external transverse force is applied. The key formulas are $a_r = \ddot{r} - r\dot{\theta}^2$ and $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$. These components arise purely from the geometry of curvilinear motion, not just from applied forces. For this problem, derive the kinematics from $r = 3t^2$ and $\theta = t$: you get $\dot{r} = 6t$, $\ddot{r} = 6$, $\dot{\theta} = 1$, and $\ddot{\theta} = 0$. At $t = 1$ s: $a_r = 6 - (3)(1)^2 = 3$ m/s² and $a_\theta = (3)(0) + 2(6)(1) = 12$ m/s². Since these are orthogonal components, the net acceleration magnitude is $\sqrt{3^2 + 12^2} = \sqrt{153}$, giving $F = 2\sqrt{153} \approx 24.7$ N. This confirms B is correct. A makes a conceptually seductive mistake: because no transverse force is externally applied, it assumes $a_\theta = 0$. But $a_\theta$ is a kinematic quantity — it equals zero only if the motion itself demands it, not simply because you didn't push sideways. Here, the Coriolis term $2\dot{r}\dot{\theta}$ is nonzero, so $a_\theta = 12$ m/s² and a transverse force must exist. C correctly identifies both components but drops the factor of 2 from the Coriolis term, halving $a_\theta$ to 6 m/s² — a purely algebraic error. D adds $a_r$ and $a_\theta$ as scalars rather than combining them as perpendicular vectors using the Pythagorean theorem. Remember: in polar coordinates, "no applied transverse force" is a physics claim you must verify kinematically — never assume it.
6. What is $\ddot{r}$ at this instant?
   Answer: $\ddot{r} = \frac{4\ddot{\theta} - 2\dot{r}^2}{2r} = \frac{4(3) - 2(4)}{2(2)} = \frac{4}{4} = 1$ m/s², found by differentiating the constraint $r^2 = 4\theta$ twice to obtain $2\dot{r}^2 + 2r\ddot{r} = 4\ddot{\theta}$, then solving for $\ddot{r}$ with $r = 2$ m and $\dot{r} = 2$ m/s When a particle's path is given as a constraint relating $r$ and $\theta$, your job is to extract velocity and acceleration information by differentiating that constraint with respect to time — not by guessing at kinematic formulas. Start with $r^2 = 4\theta$. Differentiating once: $2r\dot{r} = 4\dot{\theta}$. At $\theta = 1$ rad, $r = \sqrt{4(1)} = 2$ m. Substituting gives $\dot{r} = \frac{4\dot{\theta}}{2r} = \frac{4(2)}{2(2)} = 2$ m/s. Differentiating the first derivative again using the product rule on $2r\dot{r}$: $2\dot{r}^2 + 2r\ddot{r} = 4\ddot{\theta}$. Solving for $\ddot{r}$: $\ddot{r} = \frac{4\ddot{\theta} - 2\dot{r}^2}{2r} = \frac{4(3) - 2(4)}{2(2)} = \frac{4}{4} = 1$ m/s². That's answer A, and it's correct. Answer B introduces a spurious $2\dot{\theta}^2$ term, as if $\dot{\theta}$ appeared inside the differentiation of $r\dot{r}$ — it doesn't. The constraint only involves $\theta$, not $\dot{\theta}$ in a way that generates that extra term. Answer C drops the factor of 2 in front of $\dot{r}^2$ when rearranging, a careless algebra error that changes the result entirely. Answer D sets up the correct formula but substitutes $\dot{r}^2 = 1$ instead of $\dot{r}^2 = (2)^2 = 4$ — a classic numerical slip where the student forgets to square the velocity value. Study tip: Always compute intermediate quantities — especially $r$, $\dot{r}$ — explicitly before substituting into the second-derivative equation. Writing each step prevents the kind of substitution error that traps answer D.
7. A student claims: 'Since the rod is straight and frictionless, the transverse acceleration of the bead must be zero.' Which of the following best evaluates this claim?
   Answer: The claim is incorrect. $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 + 2(1.2)(3) = 7.2$ m/s² $\neq 0$. The rod exerts a normal (transverse) force on the bead to produce this acceleration; calling the rod 'frictionless' only means no force acts along the rod, not that the transverse force is zero. When analyzing bead-on-rotating-rod problems, you need to carefully distinguish between two different things the word "frictionless" tells you — and one thing it absolutely does not tell you. The transverse acceleration in polar coordinates is $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$. Plugging in the given values: $a_\theta = (0.5)(0) + 2(1.2)(3) = 7.2 \text{ m/s}^2$. This is real, nonzero acceleration — meaning a real, nonzero transverse force must exist. That force is the normal force the rod walls exert on the bead as it slides outward. D is correct because it recognizes this: "frictionless" only eliminates force along the rod (tangential contact), not perpendicular to the rod (normal contact). The rod must push the bead sideways to keep it on the rotating path. A is wrong on two counts: it incorrectly concludes $a_\theta = 0$, and then doubles down by claiming $\dot{r}$ and $\dot{\theta}$ cannot coexist — they clearly can, and the Coriolis term $2\dot{r}\dot{\theta}$ captures exactly this situation. B is wrong because averaging over a full revolution doesn't make $a_\theta$ zero at any given instant. The problem asks about conditions at a specific moment, not a time-averaged result. C contains a subtle but serious error: it calls $a_\theta$ "fictitious" because polar coordinates are supposedly non-inertial. Polar coordinates are a coordinate system used within an inertial frame — they are not a rotating reference frame. The acceleration and the force are both completely real. Study tip: On rotating-constraint problems, always ask yourself: "What does 'frictionless' actually eliminate?" It kills forces along the constraint surface, never the normal force perpendicular to it.
8. At this instant, which statement correctly describes the relationship between the radial and transverse velocity components?
   Answer: $v_r = 2$ m/s and $v_\theta = 4$ m/s; the radial component is half the transverse component because $\dot{r} = 0.5r\dot{\theta}$ while $v_\theta = r\dot{\theta} = 1 \cdot 4 = 4$ m/s, so $v_r = 0.5(1)(4) = 2$ m/s When analyzing particle motion in polar coordinates, your first move should always be to identify the velocity components using the standard formulas: $v_r = \dot{r}$ and $v_\theta = r\dot{\theta}$. The key is correctly applying the chain rule to find $\dot{r}$. Since $r = e^{0.5\theta}$, differentiating with respect to time gives $\dot{r} = 0.5e^{0.5\theta}\dot{\theta}$. At $\theta = 0$, we get $r = e^0 = 1$ m, so $\dot{r} = 0.5(1)(4) = 2$ m/s. The transverse component is simply $v_\theta = r\dot{\theta} = (1)(4) = 4$ m/s. This makes C correct — the radial component is exactly half the transverse component, a direct consequence of the 0.5 coefficient in the exponent. Choice A reaches the right value for $v_r$ but incorrectly claims $v_\theta = 2$ m/s as well. The transverse velocity is $r\dot{\theta} = 4$ m/s, not 2. The 45° angle conclusion falls apart because the components are not equal. Choice B reverses the logic entirely — claiming $\dot{r} = 2\dot{\theta}$ as if the chain-rule coefficient multiplies $\dot{\theta}$ alone rather than the full product $r\dot{\theta}$, yielding inflated and dimensionally inconsistent numbers. Choice D invents a fictional "curvature correction" to $v_\theta$. No such correction exists — $v_\theta = r\dot{\theta}$ is exact in polar coordinates regardless of how $r$ depends on $\theta$. Your study tip: when you see $r = f(\theta)$, always compute $\dot{r}$ via the chain rule as $\frac{dr}{d\theta}\dot{\theta}$, then plug into the standard polar velocity formulas without modification.

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