Statics and Dynamics Quiz: Curvilinear Motion Polar Coordinates
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Curvilinear Motion Polar CoordinatesQuestion 1 of 8

A particle moves along a spiral path described by r=2θr = 2\theta (in meters, with θ\theta in radians). At the instant when θ=π/2\theta = \pi/2 rad, the angular velocity is θ˙=3\dot{\theta} = 3 rad/s and the angular acceleration is θ¨=−1\ddot{\theta} = -1 rad/s².

What is the radial component of acceleration ara_r at this instant?

ar=2θ¨−2θ˙2=−20a_r = 2\ddot{\theta} - 2\dot{\theta}^2 = -20 m/s², found by substituting r¨=2θ¨\ddot{r} = 2\ddot{\theta} and r=2θr = 2\theta, then applying ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2
ar=2θ¨−rθ˙2=−2−π(9)=−2−9πa_r = 2\ddot{\theta} - r\dot{\theta}^2 = -2 - \pi(9) = -2 - 9\pi m/s², found by correctly computing r¨=2θ¨\ddot{r} = 2\ddot{\theta} and using r=2θ=πr = 2\theta = \pi at the given instant
ar=r¨+rθ˙2=−2+9πa_r = \ddot{r} + r\dot{\theta}^2 = -2 + 9\pi m/s², found by using the incorrect sign convention in the radial acceleration formula and adding the centripetal-like term rather than subtracting it
ar=2θ˙2+2θ¨=18−2=16a_r = 2\dot{\theta}^2 + 2\ddot{\theta} = 18 - 2 = 16 m/s², found by differentiating r=2θr = 2\theta twice to get r¨=2θ¨=−2\ddot{r} = 2\ddot{\theta} = -2 but then incorrectly using r˙θ˙\dot{r}\dot{\theta} in place of rθ˙2r\dot{\theta}^2
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Curvilinear Motion Polar Coordinates

Practice Curvilinear Motion Polar Coordinates in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Curvilinear Motion Polar Coordinates, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Question 1

A particle moves along a spiral path described by r=2θr = 2\theta (in meters, with θ\theta in radians). At the instant when θ=π/2\theta = \pi/2 rad, the angular velocity is θ˙=3\dot{\theta} = 3 rad/s and the angular acceleration is θ¨=−1\ddot{\theta} = -1 rad/s².

What is the radial component of acceleration ara_r at this instant?

  1. ar=2θ¨−2θ˙2=−20a_r = 2\ddot{\theta} - 2\dot{\theta}^2 = -20 m/s², found by substituting r¨=2θ¨\ddot{r} = 2\ddot{\theta} and r=2θr = 2\theta, then applying ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2
  2. ar=2θ¨−rθ˙2=−2−π(9)=−2−9πa_r = 2\ddot{\theta} - r\dot{\theta}^2 = -2 - \pi(9) = -2 - 9\pi m/s², found by correctly computing r¨=2θ¨\ddot{r} = 2\ddot{\theta} and using r=2θ=πr = 2\theta = \pi at the given instant (correct answer)
  3. ar=r¨+rθ˙2=−2+9πa_r = \ddot{r} + r\dot{\theta}^2 = -2 + 9\pi m/s², found by using the incorrect sign convention in the radial acceleration formula and adding the centripetal-like term rather than subtracting it
  4. ar=2θ˙2+2θ¨=18−2=16a_r = 2\dot{\theta}^2 + 2\ddot{\theta} = 18 - 2 = 16 m/s², found by differentiating r=2θr = 2\theta twice to get r¨=2θ¨=−2\ddot{r} = 2\ddot{\theta} = -2 but then incorrectly using r˙θ˙\dot{r}\dot{\theta} in place of rθ˙2r\dot{\theta}^2
Explanation: When analyzing motion in polar coordinates, your go-to formulas for acceleration components are ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2 and aθ=rθ¨+2r˙θ˙a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}. The key challenge is correctly computing r¨\ddot{r} by differentiating the path equation, then carefully substituting the actual values at the given instant. Since r=2θr = 2\theta, differentiating once gives r˙=2θ˙\dot{r} = 2\dot{\theta}, and differentiating again gives r¨=2θ¨\ddot{r} = 2\ddot{\theta}. At θ=π/2\theta = \pi/2: r=2(π/2)=πr = 2(\pi/2) = \pi m, r¨=2(−1)=−2\ddot{r} = 2(-1) = -2 m/s². Plugging into the radial acceleration formula: ar=r¨−rθ˙2=−2−π(3)2=−2−9π≈−30.3a_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=πr = \pi at the given instant rather than leaving it as a generic expression. A is tempting because r¨=2θ¨\ddot{r} = 2\ddot{\theta} is correct, but it substitutes r=2θr = 2\theta symbolically and treats θ\theta as a number (likely π/2≈1\pi/2 \approx 1) without properly evaluating r=πr = \pi, yielding the wrong numerical result of −20-20 m/s². C uses the wrong sign in the formula — adding rθ˙2r\dot{\theta}^2 instead of subtracting it. The centripetal term always subtracts in ara_r; mixing up this sign is one of the most common errors in polar kinematics. D correctly finds r¨=−2\ddot{r} = -2, but then substitutes r˙θ˙\dot{r}\dot{\theta} instead of rθ˙2r\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 rr, r˙\dot{r}, and r¨\ddot{r}.

Question 2

A particle's position in polar coordinates satisfies r=e0.5θr = e^{0.5\theta} (meters) with θ\theta in radians. At a certain instant, θ=0\theta = 0, θ˙=4\dot{\theta} = 4 rad/s, and θ¨=0\ddot{\theta} = 0.

At this instant, which statement correctly describes the relationship between the radial and transverse velocity components?

  1. vr=vθ=2v_r = v_\theta = 2 m/s; the speed components are equal because r˙=rθ˙\dot{r} = r\dot{\theta} holds whenever r=ekθr = e^{k\theta} with k=0.5k = 0.5, making the velocity vector point at 45° to the radial direction
  2. vr=8v_r = 8 m/s and vθ=4v_\theta = 4 m/s; the radial component is double the transverse component because r˙=2θ˙\dot{r} = 2\dot{\theta} follows from differentiating r=e0.5θr = e^{0.5\theta} and confusing the chain-rule coefficient with an additive factor
  3. vr=2v_r = 2 m/s and vθ=4v_\theta = 4 m/s; the radial component is half the transverse component because r˙=0.5rθ˙\dot{r} = 0.5r\dot{\theta} while vθ=rθ˙=1⋅4=4v_\theta = r\dot{\theta} = 1 \cdot 4 = 4 m/s, so vr=0.5(1)(4)=2v_r = 0.5(1)(4) = 2 m/s (correct answer)
  4. vr=4v_r = 4 m/s and vθ=2v_\theta = 2 m/s; the transverse component is half the radial component because vθ=rθ˙/2v_\theta = r\dot{\theta}/2 due to the curvature correction term that must be applied when rr depends exponentially on θ\theta
Explanation: When analyzing particle motion in polar coordinates, your first move should always be to identify the velocity components using the standard formulas: vr=r˙v_r = \dot{r} and vθ=rθ˙v_\theta = r\dot{\theta}. The key is correctly applying the chain rule to find r˙\dot{r}. Since r=e0.5θr = e^{0.5\theta}, differentiating with respect to time gives r˙=0.5e0.5θθ˙\dot{r} = 0.5e^{0.5\theta}\dot{\theta}. At θ=0\theta = 0, we get r=e0=1r = e^0 = 1 m, so r˙=0.5(1)(4)=2\dot{r} = 0.5(1)(4) = 2 m/s. The transverse component is simply vθ=rθ˙=(1)(4)=4v_\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 vrv_r but incorrectly claims vθ=2v_\theta = 2 m/s as well. The transverse velocity is rθ˙=4r\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 r˙=2θ˙\dot{r} = 2\dot{\theta} as if the chain-rule coefficient multiplies θ˙\dot{\theta} alone rather than the full product rθ˙r\dot{\theta}, yielding inflated and dimensionally inconsistent numbers. Choice D invents a fictional "curvature correction" to vθv_\theta. No such correction exists — vθ=rθ˙v_\theta = r\dot{\theta} is exact in polar coordinates regardless of how rr depends on θ\theta. Your study tip: when you see r=f(θ)r = f(\theta), always compute r˙\dot{r} via the chain rule as drdθθ˙\frac{dr}{d\theta}\dot{\theta}, then plug into the standard polar velocity formulas without modification.

Question 3

In polar coordinates, a particle's position satisfies r=5r = 5 m (constant) and θ\theta varies with time. At a certain instant, θ˙=2\dot{\theta} = 2 rad/s and θ¨=4\ddot{\theta} = 4 rad/s².

Which of the following correctly describes the acceleration vector of the particle at this instant?

  1. a⃗=−20e^r+20e^θ\vec{a} = -20\hat{e}_r + 20\hat{e}_\theta m/s², because ar=−rθ˙2=−5(4)=−20a_r = -r\dot{\theta}^2 = -5(4) = -20 m/s² and aθ=rθ¨=5(4)=20a_\theta = r\ddot{\theta} = 5(4) = 20 m/s²; the r¨\ddot{r} and r˙\dot{r} terms are omitted without justification since rr is stated to be constant
  2. a⃗=−20e^r+30e^θ\vec{a} = -20\hat{e}_r + 30\hat{e}_\theta m/s², because ar=−rθ˙2=−20a_r = -r\dot{\theta}^2 = -20 m/s² and aθ=rθ¨+rθ˙=5(4)+5(2)=30a_\theta = r\ddot{\theta} + r\dot{\theta} = 5(4) + 5(2) = 30 m/s², with the extra rθ˙r\dot{\theta} term incorrectly added as though differentiating e^θ\hat{e}_\theta contributes an additional velocity-dependent term to aθa_\theta
  3. a⃗=+20e^r+20e^θ\vec{a} = +20\hat{e}_r + 20\hat{e}_\theta m/s², because ar=rθ˙2=5(4)=+20a_r = r\dot{\theta}^2 = 5(4) = +20 m/s² (taken as positive, directed outward) and aθ=rθ¨=5(4)=20a_\theta = r\ddot{\theta} = 5(4) = 20 m/s², treating the centripetal term as a positive outward quantity
  4. a⃗=−20e^r+20e^θ\vec{a} = -20\hat{e}_r + 20\hat{e}_\theta m/s², where ar=r¨−rθ˙2=0−5(2)2=−20a_r = \ddot{r} - r\dot{\theta}^2 = 0 - 5(2)^2 = -20 m/s² and aθ=rθ¨+2r˙θ˙=5(4)+2(0)(2)=20a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 5(4) + 2(0)(2) = 20 m/s²; both r¨=0\ddot{r} = 0 and the Coriolis term vanish explicitly because r=constr = \text{const} implies r˙=0\dot{r} = 0 (correct answer)
Explanation: When a particle moves in polar coordinates, its acceleration has two components derived by differentiating the position vector twice: ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2 and aθ=rθ¨+2r˙θ˙a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}. The key insight for circular motion problems is that a constant radius means r˙=0\dot{r} = 0 and r¨=0\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=5r = 5 m = const, r˙=0\dot{r} = 0 and r¨=0\ddot{r} = 0. Substituting: ar=0−5(2)2=−20a_r = 0 - 5(2)^2 = -20 m/s² and aθ=5(4)+2(0)(2)=20a_\theta = 5(4) + 2(0)(2) = 20 m/s², giving a⃗=−20e^r+20e^θ\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 r¨\ddot{r} and 2r˙θ˙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 aθa_\theta formula by adding a spurious rθ˙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 −rθ˙2-r\dot{\theta}^2 as positive. The centripetal acceleration always points inward (toward the center), meaning ara_r must be negative. Study tip: Always write both full polar acceleration formulas first, then substitute r˙=0\dot{r} = 0 and r¨=0\ddot{r} = 0 with an explicit reason. Showing your zeros is what separates a complete solution from a lucky answer.

Question 4

A particle of mass m=2m = 2 kg moves under a single radial force FrF_r (no transverse force applied). Its motion in polar coordinates is described by r=3t2r = 3t^2 m and θ=t\theta = t rad, where tt is in seconds.

At t=1t = 1 s, what is the magnitude of the net force acting on the particle?

  1. F=m∣ar∣=2∣6−3(1)2∣=6F = m|a_r| = 2|6 - 3(1)^2| = 6 N, obtained by correctly computing ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2 but incorrectly assuming aθ=0a_\theta = 0 because no transverse force was stated to be applied externally
  2. F=mar2+aθ2=2(3)2+(12)2=2153≈24.7F = m\sqrt{a_r^2 + a_\theta^2} = 2\sqrt{(3)^2 + (12)^2} = 2\sqrt{153} \approx 24.7 N, found using ar=r¨−rθ˙2=6−3=3a_r = \ddot{r} - r\dot{\theta}^2 = 6 - 3 = 3 m/s² and aθ=rθ¨+2r˙θ˙=0+2(6)(1)=12a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 + 2(6)(1) = 12 m/s² (correct answer)
  3. F=mar2+aθ2=2(3)2+(6)2≈13.4F = m\sqrt{a_r^2 + a_\theta^2} = 2\sqrt{(3)^2 + (6)^2} \approx 13.4 N, obtained by correctly computing ar=3a_r = 3 m/s² but halving the Coriolis term to get aθ=6a_\theta = 6 m/s² by omitting the factor of 2 in 2r˙θ˙2\dot{r}\dot{\theta}
  4. F=m(ar+aθ)=2(3+12)=30F = m(a_r + a_\theta) = 2(3 + 12) = 30 N, computed by adding the scalar values of ara_r and aθa_\theta algebraically instead of combining them as orthogonal vector components
Explanation: 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 ar=r¨−rθ˙2a_r = \ddot{r} - r\dot{\theta}^2 and aθ=rθ¨+2r˙θ˙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=3t2r = 3t^2 and θ=t\theta = t: you get r˙=6t\dot{r} = 6t, r¨=6\ddot{r} = 6, θ˙=1\dot{\theta} = 1, and θ¨=0\ddot{\theta} = 0. At t=1t = 1 s: ar=6−(3)(1)2=3a_r = 6 - (3)(1)^2 = 3 m/s² and aθ=(3)(0)+2(6)(1)=12a_\theta = (3)(0) + 2(6)(1) = 12 m/s². Since these are orthogonal components, the net acceleration magnitude is 32+122=153\sqrt{3^2 + 12^2} = \sqrt{153}, giving F=2153≈24.7F = 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 aθ=0a_\theta = 0. But aθa_\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 2r˙θ˙2\dot{r}\dot{\theta} is nonzero, so aθ=12a_\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 aθa_\theta to 6 m/s² — a purely algebraic error. D adds ara_r and aθa_\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.

Question 5

A bead slides outward along a straight frictionless rod that rotates in a horizontal plane. At a given instant, r=0.5r = 0.5 m, r˙=1.2\dot{r} = 1.2 m/s, r¨=0\ddot{r} = 0, θ˙=3\dot{\theta} = 3 rad/s, and θ¨=0\ddot{\theta} = 0.

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?

  1. The claim is correct. On a frictionless rod, no transverse force exists; therefore aθ=0a_\theta = 0 by Newton's second law, and the Coriolis term 2r˙θ˙2\dot{r}\dot{\theta} must also vanish, implying the rod cannot simultaneously have r˙≠0\dot{r} \neq 0 and θ˙≠0\dot{\theta} \neq 0.
  2. The claim is correct in an inertial frame. The transverse acceleration aθ=rθ¨+2r˙θ˙=7.2a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 7.2 m/s² is a kinematic result, but since the rod is straight and frictionless the net transverse force averages to zero over a full revolution, validating the student's statement in the mean.
  3. The claim is incorrect. aθ=rθ¨+2r˙θ˙=0.5(0)+2(1.2)(3)=7.2a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0.5(0) + 2(1.2)(3) = 7.2 m/s², but this transverse acceleration is fictitious — it appears only because polar coordinates are non-inertial, so no real transverse force is needed to produce it.
  4. The claim is incorrect. aθ=rθ¨+2r˙θ˙=0+2(1.2)(3)=7.2a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 + 2(1.2)(3) = 7.2 m/s² ≠0\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. (correct answer)
Explanation: 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 aθ=rθ¨+2r˙θ˙a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}. Plugging in the given values: aθ=(0.5)(0)+2(1.2)(3)=7.2 m/s2a_\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 aθ=0a_\theta = 0, and then doubles down by claiming r˙\dot{r} and θ˙\dot{\theta} cannot coexist — they clearly can, and the Coriolis term 2r˙θ˙2\dot{r}\dot{\theta} captures exactly this situation. B is wrong because averaging over a full revolution doesn't make aθa_\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 aθa_\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.

Question 6

A particle moves in the rr–θ\theta plane. Its speed (magnitude of velocity) at a given instant is v=10v = 10 m/s. At that instant, r=2r = 2 m, θ˙=3\dot{\theta} = 3 rad/s, and r˙\dot{r} is unknown.

What is ∣r˙∣|\dot{r}| at this instant?

  1. ∣r˙∣=v2−(rθ˙)2=100−36=8|\dot{r}| = \sqrt{v^2 - (r\dot{\theta})^2} = \sqrt{100 - 36} = 8 m/s, using the correct decomposition v2=r˙2+(rθ˙)2v^2 = \dot{r}^2 + (r\dot{\theta})^2 with rθ˙=2(3)=6r\dot{\theta} = 2(3) = 6 m/s (correct answer)
  2. ∣r˙∣=v−rθ˙=10−6=4|\dot{r}| = v - r\dot{\theta} = 10 - 6 = 4 m/s, found by treating the speed as the algebraic sum of the radial and transverse speed components rather than combining them as orthogonal vectors
  3. ∣r˙∣=v2−θ˙2=100−9≈9.54|\dot{r}| = \sqrt{v^2 - \dot{\theta}^2} = \sqrt{100 - 9} \approx 9.54 m/s, obtained by omitting the factor rr in the transverse velocity component and using θ˙\dot{\theta} directly instead of rθ˙r\dot{\theta}
  4. ∣r˙∣=v2+(rθ˙)2=100+36≈11.66|\dot{r}| = \sqrt{v^2 + (r\dot{\theta})^2} = \sqrt{100 + 36} \approx 11.66 m/s, obtained by adding rather than subtracting the transverse component squared, as if r˙\dot{r} and rθ˙r\dot{\theta} were components of different vectors being combined
Explanation: When a particle moves in polar coordinates, its velocity has two orthogonal components: a radial component vr=r˙v_r = \dot{r} (along the e^r\hat{e}_r direction) and a transverse component vθ=rθ˙v_\theta = r\dot{\theta} (along the e^θ\hat{e}_\theta direction). Because these components are perpendicular, speed is found using the Pythagorean theorem: v2=r˙2+(rθ˙)2v^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 rθ˙=(2)(3)=6r\dot{\theta} = (2)(3) = 6 m/s. Solving for ∣r˙∣|\dot{r}|: ∣r˙∣=v2−(rθ˙)2=100−36=64=8|\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=410 - 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 rr, plugging in θ˙=3\dot{\theta} = 3 rad/s directly instead of the transverse speed rθ˙=6r\dot{\theta} = 6 m/s. Remember: θ˙\dot{\theta} has units of rad/s, not m/s — multiplying by rr converts it to a speed. D adds (rθ˙)2(r\dot{\theta})^2 instead of subtracting it, which would make r˙\dot{r} larger than vv — physically impossible, since r˙\dot{r} is just one component of the total velocity. Study tip: Always write v2=r˙2+(rθ˙)2v^2 = \dot{r}^2 + (r\dot{\theta})^2 first, then identify which quantity is unknown. Never forget the rr in rθ˙r\dot{\theta} — it's one of the most common errors on polar-coordinate problems.

Question 7

A particle's position in polar coordinates satisfies r2=4θr^2 = 4\theta (with rr in meters and θ\theta in radians). At the instant when θ=1\theta = 1 rad, θ˙=2\dot{\theta} = 2 rad/s and θ¨=3\ddot{\theta} = 3 rad/s².

What is r¨\ddot{r} at this instant?

  1. r¨=4θ¨−2r˙22r=4(3)−2(4)2(2)=44=1\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 r2=4θr^2 = 4\theta twice to obtain 2r˙2+2rr¨=4θ¨2\dot{r}^2 + 2r\ddot{r} = 4\ddot{\theta}, then solving for r¨\ddot{r} with r=2r = 2 m and r˙=2\dot{r} = 2 m/s (correct answer)
  2. r¨=2θ˙2+2rθ¨−2r˙22r=2(4)+2(2)(3)−2(4)4=8+12−84=3\ddot{r} = \frac{2\dot{\theta}^2 + 2r\ddot{\theta} - 2\dot{r}^2}{2r} = \frac{2(4) + 2(2)(3) - 2(4)}{4} = \frac{8+12-8}{4} = 3 m/s², obtained by incorrectly expanding the second derivative of r2=4θr^2 = 4\theta with an extra 2θ˙22\dot{\theta}^2 term not present in the correct derivation
  3. r¨=2θ¨−r˙2/r2=6−22=2\ddot{r} = \frac{2\ddot{\theta} - \dot{r}^2/r}{2} = \frac{6 - 2}{2} = 2 m/s², found by improperly simplifying the second differentiation of 2rr˙=4θ˙2r\dot{r} = 4\dot{\theta}, dropping the factor of 2 in the r˙2\dot{r}^2 term and mishandling the denominator
  4. r¨=4θ¨−2r˙22r=4(3)−2(1)2(2)=104=2.5\ddot{r} = \frac{4\ddot{\theta} - 2\dot{r}^2}{2r} = \frac{4(3) - 2(1)}{2(2)} = \frac{10}{4} = 2.5 m/s², found by setting up the correct formula but substituting r˙2=1\dot{r}^2 = 1 instead of the correct r˙2=(2)2=4\dot{r}^2 = (2)^2 = 4, a numerical substitution error
Explanation: When a particle's path is given as a constraint relating rr and θ\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 r2=4θr^2 = 4\theta. Differentiating once: 2rr˙=4θ˙2r\dot{r} = 4\dot{\theta}. At θ=1\theta = 1 rad, r=4(1)=2r = \sqrt{4(1)} = 2 m. Substituting gives r˙=4θ˙2r=4(2)2(2)=2\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 2rr˙2r\dot{r}: 2r˙2+2rr¨=4θ¨2\dot{r}^2 + 2r\ddot{r} = 4\ddot{\theta}. Solving for r¨\ddot{r}: r¨=4θ¨−2r˙22r=4(3)−2(4)2(2)=44=1\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 2θ˙22\dot{\theta}^2 term, as if θ˙\dot{\theta} appeared inside the differentiation of rr˙r\dot{r} — it doesn't. The constraint only involves θ\theta, not θ˙\dot{\theta} in a way that generates that extra term. Answer C drops the factor of 2 in front of r˙2\dot{r}^2 when rearranging, a careless algebra error that changes the result entirely. Answer D sets up the correct formula but substitutes r˙2=1\dot{r}^2 = 1 instead of r˙2=(2)2=4\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 rr, r˙\dot{r} — explicitly before substituting into the second-derivative equation. Writing each step prevents the kind of substitution error that traps answer D.

Question 8

A particle moves such that in polar coordinates r¨=0\ddot{r} = 0, r=4r = 4 m, r˙=0\dot{r} = 0, θ˙=2\dot{\theta} = 2 rad/s, and θ¨=1\ddot{\theta} = 1 rad/s². The particle has mass m=3m = 3 kg.

What is the angle ϕ\phi that the net force vector makes with the radial direction (measured from −e^r-\hat{e}_r toward e^θ\hat{e}_\theta)?

  1. ϕ=arctan⁡ ⁣(aθ∣ar∣)=arctan⁡ ⁣(412)≈18.4°\phi = \arctan\!\left(\dfrac{a_\theta}{|a_r|}\right) = \arctan\!\left(\dfrac{4}{12}\right) \approx 18.4°, obtained by correctly computing aθ=4a_\theta = 4 m/s² but evaluating ar=−(rθ˙2−r¨)=−(16−6)=−10a_r = -(r\dot{\theta}^2 - \ddot{r}) = -(16 - 6) = -10 m/s² by erroneously subtracting a nonexistent r¨=rθ¨=6\ddot{r} = r\ddot{\theta} = 6 term from the radial formula, yielding ∣ar∣=10|a_r| = 10 — wait, the answer states 12; obtained by using ∣ar∣=rθ˙2−rθ¨=16−4=12|a_r| = r\dot{\theta}^2 - r\ddot{\theta} = 16 - 4 = 12 due to incorrectly subtracting the angular acceleration contribution
  2. ϕ=arctan⁡ ⁣(∣ar∣aθ)=arctan⁡ ⁣(164)=arctan⁡(4)≈76.0°\phi = \arctan\!\left(\dfrac{|a_r|}{a_\theta}\right) = \arctan\!\left(\dfrac{16}{4}\right) = \arctan(4) \approx 76.0°, obtained by inverting the numerator and denominator when forming the arctangent ratio, measuring the angle from e^θ\hat{e}_\theta toward −e^r-\hat{e}_r instead of from −e^r-\hat{e}_r toward e^θ\hat{e}_\theta
  3. ϕ=arctan⁡ ⁣(aθ∣ar∣)=arctan⁡ ⁣(416)=arctan⁡(0.25)≈14.0°\phi = \arctan\!\left(\dfrac{a_\theta}{|a_r|}\right) = \arctan\!\left(\dfrac{4}{16}\right) = \arctan(0.25) \approx 14.0°, using ar=−rθ˙2=−4(4)=−16a_r = -r\dot{\theta}^2 = -4(4) = -16 m/s² and aθ=rθ¨+2r˙θ˙=4(1)+0=4a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} = 4(1) + 0 = 4 m/s² (correct answer)
  4. ϕ=arctan⁡ ⁣(aθ∣ar∣)=arctan⁡ ⁣(1216)≈36.9°\phi = \arctan\!\left(\dfrac{a_\theta}{|a_r|}\right) = \arctan\!\left(\dfrac{12}{16}\right) \approx 36.9°, obtained by computing ar=−16a_r = -16 m/s² correctly but computing the transverse acceleration as aθ=rθ¨+2r˙θ˙+rθ˙=4+0+8=12a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} + r\dot{\theta} = 4 + 0 + 8 = 12 m/s², incorrectly adding an extra rθ˙r\dot{\theta} term to the transverse formula
Explanation: 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: ar=r¨−rθ˙2aθ=rθ¨+2r˙θ˙a_r = \ddot{r} - r\dot{\theta}^2 \qquad a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} With r¨=0\ddot{r} = 0, r=4r = 4 m, r˙=0\dot{r} = 0, θ˙=2\dot{\theta} = 2 rad/s, and θ¨=1\ddot{\theta} = 1 rad/s², these give: ar=0−4(2)2=−16 m/s2,aθ=4(1)+2(0)(2)=4 m/s2a_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 ara_r means the radial force points inward (−e^r-\hat{e}_r). The angle ϕ\phi measured from −e^r-\hat{e}_r toward e^θ\hat{e}_\theta is therefore ϕ=arctan⁡ ⁣(aθ∣ar∣)=arctan⁡ ⁣(416)≈14.0°\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 rθ¨r\ddot{\theta} term, leaving ∣ar∣=12|a_r| = 12 instead of 16 — the radial acceleration has no θ¨\ddot{\theta} dependence whatsoever. Choice B gets both acceleration magnitudes right but flips the arctangent ratio, effectively measuring the angle from e^θ\hat{e}_\theta rather than from −e^r-\hat{e}_r, which swaps numerator and denominator. Choice D correctly finds ar=−16a_r = -16 m/s² but inflates the transverse acceleration to 12 m/s² by tacking on an extra rθ˙r\dot{\theta} term that has no place in the aθa_\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 rθ˙r\dot{\theta} in aθa_\theta or rθ¨r\ddot{\theta} in ara_r) should raise an immediate red flag.