Curvilinear Motion: n-t Coordinates - Statics and Dynamics

1. What is the magnitude of the total acceleration of the particle at this instant? (Recall that the radius of curvature for $y = f(x)$ is $\rho = \frac{(1 + (y')^2)^{3/2}}{|y''|}$.)
   Answer: $a \approx 4.31 \text{ m/s}^2$, found by computing $\rho = (1+(2)^2)^{3/2}/|2| \approx 5.59$ m, then $a_n = v^2/\rho = 9/5.59 \approx 1.61$ m/s², and $a = \sqrt{4^2 + 1.61^2} \approx 4.31$ m/s² When a particle moves along a curved path, its total acceleration has two perpendicular components: the tangential acceleration $a_t$ (rate of speed change) and the normal acceleration $a_n = v^2/\rho$ (due to the path's curvature). The total magnitude is $a = \sqrt{a_t^2 + a_n^2}$. You need both components — never assume one is negligible without checking. For $y = x^2$, compute the derivatives: $y' = 2x$ and $y'' = 2$. At $x = 1$: $y' = 2$, $y'' = 2$. Plugging into the radius of curvature formula: $\rho = \frac{(1 + (2)^2)^{3/2}}{|2|} = \frac{(5)^{3/2}}{2} = \frac{11.18}{2} \approx 5.59 \text{ m}$. Then the normal acceleration is $a_n = \frac{v^2}{\rho} = \frac{(3)^2}{5.59} \approx 1.61 \text{ m/s}^2$. Finally, $a = \sqrt{(4)^2 + (1.61)^2} \approx \sqrt{16 + 2.59} \approx 4.31 \text{ m/s}^2$, confirming C. Choice A ignores the normal acceleration entirely — a critical error, since the path is curved and $a_n$ is not negligible here. Choice B uses $1 + 2(1)$ inside the formula instead of $1 + (y')^2 = 1 + (2)^2 = 5$, incorrectly substituting $y'$ rather than $(y')^2$. This is a common algebra mistake. Choice D correctly computes $\rho$ but then uses $v/\rho$ instead of $v^2/\rho$ — forgetting to square the speed. Study tip: Memorize $a_n = v^2/\rho$, not $v/\rho$, and always square the derivative $y'$ when computing $\rho$. These two spots are where most errors occur.
2. At $t = 1$ s, what is the magnitude of the total acceleration of the particle?
   Answer: $a = \sqrt{(6)^2 + (0.8)^2} \approx 6.05 \text{ m/s}^2$, where $a_t = \dot{v} = 8t - 2 = 6$ m/s² and $a_n = v^2/\rho = (2)^2/5 = 0.8$ m/s² When a particle moves along a curved path, its acceleration has two perpendicular components: the tangential acceleration $a_t$, which describes how fast the speed is changing, and the normal acceleration $a_n$, which describes how fast the direction is changing. The total magnitude combines them as $a = \sqrt{a_t^2 + a_n^2}$. The tangential acceleration is always the time derivative of speed: $a_t = \dot{v} = \frac{d}{dt}(4t^2 - 2t) = 8t - 2$. At $t = 1$ s, this gives $a_t = 6$ m/s². The speed at that moment is $v = 4(1)^2 - 2(1) = 2$ m/s, so the normal acceleration is $a_n = \frac{v^2}{\rho} = \frac{(2)^2}{5} = 0.8$ m/s². The total acceleration is therefore $a = \sqrt{(6)^2 + (0.8)^2} \approx 6.05$ m/s², confirming B is correct. A makes a subtle but critical algebraic error: it adds $a_n$ without squaring it inside the radical — writing $\sqrt{36 + 0.8}$ instead of $\sqrt{36 + 0.64}$. Both components must be squared before summing. C confuses the speed $v = 2$ m/s with the tangential acceleration, plugging the speed value directly in place of $a_t$. Speed and its rate of change are fundamentally different quantities. D incorrectly uses $a_n = 1/\rho$, ignoring that the normal acceleration formula requires $v^2$ in the numerator — curvature alone is not an acceleration. A reliable habit: always differentiate $v(t)$ to find $a_t$, never substitute $v$ itself, and always square both components before taking the square root.
3. At this instant, what is the ratio $a_n / a_t$, and what does this ratio imply about the direction of the total acceleration vector?
   Answer: $a_n/a_t = \frac{v^2/(\rho)}{a_t} = \frac{(7800)^2/(6.8\times10^6)}{0.5} \approx \frac{8.94}{0.5} \approx 17.9$; the total acceleration is nearly perpendicular to the path (almost entirely in the $\hat{e}_n$ direction), with $\theta \approx \arctan(1/17.9) \approx 3.2°$ from the normal When analyzing acceleration in curvilinear motion, always decompose into normal and tangential components. The normal acceleration $a_n = v^2/\rho$ points toward the center of curvature, while $a_t$ acts along the path. The angle the total acceleration makes with the normal is $\theta = \arctan(a_t/a_n)$ — not $\arctan(a_n/a_t)$. For this satellite, $a_n = v^2/\rho = (7800)^2/(6.8 \times 10^6) \approx 8.94$ m/s², giving a ratio $a_n/a_t = 8.94/0.5 \approx 17.9$. Since $a_n \gg a_t$, the total acceleration vector is pulled almost entirely toward the normal direction. The angle from the normal is $\theta = \arctan(1/17.9) \approx 3.2°$ — nearly perpendicular to the path. This is answer D, and it correctly interprets both the calculation and the geometry. Answer A gets the ratio right numerically but completely misreads the implication. When $a_n \gg a_t$, the vector does not bisect the axes at 45° — that would require $a_n = a_t$. The 45° claim reveals a fundamental misunderstanding of vector addition. Answer B correctly computes $a_n/a_t \approx 17.9$ using the right formula, but then draws the opposite conclusion — claiming the large ratio means the tangential component dominates. This is backwards: a large $a_n/a_t$ ratio means the normal component dominates. Answer C uses the wrong formula entirely, computing $v/\rho$ instead of $v^2/\rho$, dropping one factor of velocity and producing a physically meaningless result. Your study tip: memorize that $\theta$ is measured from the dominant component. A large ratio $a_n/a_t$ means the vector hugs the normal, not the tangent.
4. Comparing the two particles at their respective locations, which statement about their total acceleration vectors is correct?
   Answer: Both particles have the same total acceleration magnitude ($= 5$ m/s²) but their total acceleration vectors point in different directions in the n-t plane — the accelerating particle's vector is in the first quadrant and the decelerating particle's is in the fourth quadrant (taking $\hat{e}_n$ as positive and $\hat{e}_t$ as positive in the direction of motion) When analyzing acceleration in normal-tangential (n-t) coordinates, remember that the total acceleration has two independent components: the normal component $a_n = v^2/\rho$ (always pointing toward the center of curvature) and the tangential component $a_t$ (along the path, positive or negative depending on whether the particle speeds up or slows down). These components are perpendicular, so they combine like legs of a right triangle. For both particles here, $a_n = v^2/\rho = 36/9 = 4$ m/s² and $|a_t| = 3$ m/s². The total magnitude for each is $\sqrt{4^2 + 3^2} = \sqrt{25} = 5$ m/s² — identical, since magnitude depends only on the absolute values. However, the direction differs: the accelerating particle's vector lies in the first quadrant of the n-t plane (positive $\hat{e}_t$, positive $\hat{e}_n$), while the decelerating particle's vector lies in the fourth quadrant (negative $\hat{e}_t$, positive $\hat{e}_n$). This confirms D is correct. A is wrong because the sign of $a_t$ absolutely does affect the vector's direction — it determines which side of the $\hat{e}_n$ axis the total vector falls on. B is wrong because it treats $a_n$ and $a_t$ as if they act along the same axis and can "cancel" — they're perpendicular, so magnitudes always add via Pythagorean theorem, never algebraically. C is wrong because identical magnitudes of components don't produce identical vectors if one component has opposite sign; the directions differ. Your study tip: always sketch the n-t axes and plot both components as perpendicular arrows. The sign of $a_t$ rotates the resultant vector across the $\hat{e}_n$ axis — same magnitude, mirror-image direction.
5. As the car travels deeper into the tightening curve at constant speed, which of the following correctly describes how the normal acceleration $a_n$, tangential acceleration $a_t$, and total acceleration $a$ change?
   Answer: $a_n$ increases (because $\rho$ decreases with $v$ constant), $a_t = 0$ remains zero (constant speed), and therefore $a$ increases — the total acceleration vector grows in magnitude and remains aligned with the $\hat{e}_n$ direction When analyzing motion along a curved path, the key framework is the normal-tangential (n-t) coordinate system, where acceleration splits into two independent components: $a_n = \frac{v^2}{\rho}$ (pointing toward the center of curvature) and $a_t = \dot{v}$ (along the direction of motion). Since the car maintains constant speed, $\dot{v} = 0$, which means $a_t = 0$ throughout — speed isn't changing, so there's no tangential acceleration. However, the normal acceleration depends on both speed and radius of curvature. As the curve tightens, $\rho$ decreases, so $a_n = \frac{v^2}{\rho}$ grows larger even though $v$ stays fixed. The total acceleration is therefore $a = \sqrt{a_n^2 + a_t^2} = a_n$, which increases and remains pointed in the $\hat{e}_n$ direction. That confirms C is correct. A confuses $a_n$ with $a_t$. The formula $a_n = \dot{v}$ is simply wrong — that's the definition of tangential acceleration. Normal acceleration depends on curvature, not the rate of speed change. B correctly identifies that $a_n$ increases, but falsely claims $a_t$ also increases. A tighter curve does not require tangential effort — tangential acceleration is purely about changing speed, which isn't happening here. D inverts the relationship entirely. A smaller $\rho$ means a sharper curve, which increases $a_n$, not decreases it. A reliable study tip: always treat $a_n$ and $a_t$ as independent quantities governed by separate formulas. Constant speed kills $a_t$ immediately — then focus entirely on how $\rho$ affects $a_n$.
6. At the top of the loop, what is the magnitude of the net force per unit mass (i.e., the net acceleration magnitude) acting on the pilot, and in which direction does the normal acceleration component point?
   Answer: Net acceleration magnitude $= \sqrt{(20)^2 + (37.5)^2} \approx 42.5 \text{ m/s}^2$; the normal acceleration points downward (toward the center of the loop, which is below the aircraft at the top) When a particle moves along a curved path, its acceleration has two independent components: a tangential component $a_t$ (along the path, changing speed) and a normal component $a_n = v^2/\rho$ (perpendicular to the path, changing direction). These components are kinematic — they describe the actual acceleration of the particle in the inertial frame, determined purely by the geometry of motion. Gravity and normal forces are the causes of this acceleration; they are not added on top of it. At the top of the loop, the center of the circle lies directly below the aircraft, so the normal (centripetal) acceleration always points downward toward that center. Computing it: $a_n = \frac{v^2}{\rho} = \frac{(150)^2}{600} = 37.5 \text{ m/s}^2$. The net acceleration magnitude is then $\sqrt{a_t^2 + a_n^2} = \sqrt{(20)^2 + (37.5)^2} \approx 42.5 \text{ m/s}^2$, pointing downward in the normal direction. This makes A correct. B is wrong because it claims the normal acceleration points upward. "Centrifugal" tendency is a fictitious force in a rotating frame — in the inertial frame used here, centripetal acceleration always points toward the center, which is downward at the loop's top. C is wrong because it replaces $v^2/\rho$ with $g$, confusing the condition where gravity alone provides centripetal force (a special case) with the general kinematic formula. D is wrong because it adds $g$ directly to $v^2/\rho$ to get $a_n$. Gravity is already accounted for within the net force equation that produces $v^2/\rho$ — you never tack it onto the kinematic result. Study tip: Keep kinematics and kinetics separate. Always compute $a_n = v^2/\rho$ first from geometry, then use Newton's second law to find which forces (gravity, normal force, thrust) combine to produce that acceleration.
7. Which of the following correctly identifies the magnitudes of the normal and tangential accelerations at the instant when the speed is 25 m/s?
   Answer: $a_t = 2 \text{ m/s}^2$ and $a_n = 7.81 \text{ m/s}^2$, because the tangential acceleration equals the rate of speed change and the normal component is $v^2/\rho$ evaluated at 25 m/s When a particle moves along a curved path with changing speed, its acceleration has two independent components: a tangential component that captures how fast speed is changing, and a normal component that captures how sharply the path curves. These are defined as $a_t = \frac{dv}{dt}$ and $a_n = \frac{v^2}{\rho}$, and you must evaluate each at the specific instant of interest. Here, the speed increases uniformly from 20 m/s to 30 m/s in 5 s, so the tangential acceleration is constant: $a_t = \frac{30 - 20}{5} = 2 \text{ m/s}^2$. At the instant when $v = 25 \text{ m/s}$, the normal acceleration is $a_n = \frac{(25)^2}{80} = \frac{625}{80} \approx 7.81 \text{ m/s}^2$. This confirms A is correct. B is wrong because it uses $v/\rho$ instead of $v^2/\rho$ for the normal acceleration — a fundamental formula error. The centripetal term always involves velocity squared. C is wrong on two counts: it incorrectly computes $a_t$ as total speed change divided by total distance (mixing up $\Delta v / \Delta t$ with $\Delta v / \Delta x$), and while its $a_n$ happens to be correct, the flawed reasoning disqualifies it. D applies the correct $v^2/\rho$ formula but then divides by 2 for no physical reason, halving the correct answer and inventing a nonexistent rule. A reliable strategy: always write out $a_t = dv/dt$ and $a_n = v^2/\rho$ before plugging in numbers, and be careful to evaluate $a_n$ at the specified instant's speed, not an average or modified value.
8. At $t = 2$ s, what is the magnitude of the particle's normal acceleration?
   Answer: $a_n = R\dot{\theta}^2 = 2(9)^2 = 162 \text{ m/s}^2$, where $\dot{\theta} = 3t^2 - 3 = 9$ rad/s at $t = 2$ s and $v = R\dot{\theta} = 18$ m/s, giving $a_n = v^2/R = 324/2 = 162$ m/s² When a particle moves along a circular path, you must distinguish between two fundamentally different acceleration components: the tangential acceleration (along the path) and the normal acceleration (toward the center). The normal acceleration depends on how fast the particle is moving around the curve — specifically, it equals $a_n = \frac{v^2}{R} = R\dot{\theta}^2$. Notice that angular velocity is squared — this is the centripetal relationship, and getting it right is the entire challenge of this problem. At $t = 2$ s, the angular velocity is $\dot{\theta} = 3t^2 - 3 = 3(4) - 3 = 9$ rad/s. Plugging into the formula: $a_n = R\dot{\theta}^2 = 2(9)^2 = 2(81) = 162 \text{ m/s}^2$. You can verify this equivalently using $v = R\dot{\theta} = 18$ m/s, so $a_n = v^2/R = 324/2 = 162 \text{ m/s}^2$. Both routes confirm B is correct. A uses $a_n = R\ddot{\theta}$, which is actually the formula for tangential acceleration, not normal acceleration. Plugging in the angular acceleration $\ddot{\theta} = 6t = 12$ rad/s² gives the wrong component entirely. C uses $a_n = R\dot{\theta}$ without squaring — this is simply the linear speed $v$, not an acceleration at all. D combines $\dot{\theta}$ and $\ddot{\theta}$ in quadrature, which has no physical basis; normal and tangential accelerations come from different kinematic sources and are never mixed this way. A reliable memory aid: $a_n$ points inward and involves $\dot{\theta}^2$; $a_t$ points along the path and involves $\ddot{\theta}$. Keep these two straight and circular motion problems become straightforward.

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