Path variables

At every point on a particle's path, attach a local pair of unit vectors: $\hat e_t$, the unit tangent, pointing in the instantaneous direction of motion, and $\hat e_n$, the unit normal, pointing from the particle toward the centre of curvature of the path at that instant. These two directions are always mutually perpendicular, and they rotate together with the particle as it moves along a curved path — unlike the fixed $x$-$y$ axes of rectangular coordinates. Because velocity is, by definition, tangent to the path, it has only one component in this frame:

where: $s$ is the arc length measured along the path from some reference point, so $v$ is simply the speed (never negative in magnitude, only in the sense of direction reversal along $\hat e_t$). Acceleration is where path variables earn their keep, because $\hat e_t$ itself is turning as the particle moves, and differentiating $\vec v = v\hat e_t$ with respect to time produces two distinct terms:

where: $a_t$ is the tangential acceleration, $a_n$ is the normal (centripetal) acceleration, and $\rho$ is the radius of curvature of the path at that instant — the radius of the circle that best matches the path's bend right there.

• $a_t$ tracks only the rate of change of speed. It is exactly zero whenever the particle moves at constant speed, no matter how sharply the path curves.
• $a_n$ tracks only the path's curving effect. It is exactly zero only where the path is momentarily straight ($\rho \to \infty$), no matter how fast the particle is going.
• These two causes are completely independent — one can be zero while the other is large, and both can act simultaneously, which is exactly what makes n-t coordinates more informative than a single "acceleration" number.

For a circular path of constant radius $R$, $\rho = R$ at every point, so uniform circular motion ($a_t = 0$) recovers the familiar centripetal result $a_n = v^2/R$ as a special case, not a separate formula to memorise.

In kinetics, Newton's second law resolved along $\hat e_t$ and $\hat e_n$ gives two independent scalar equations,

where: $\sum F_t$ and $\sum F_n$ are the net force components along the tangent and toward the centre of curvature respectively. The normal equation is often the most physically transparent route to a normal reaction, cable tension, or the speed at which contact is lost on a curved track, because the direction "toward the centre" is usually easy to identify from the geometry alone.

Extend the hill-crest example from the syllabus: a block of mass $m = 2\ \text{kg}$ crosses the crest of a circular hill of radius $\rho = 40\ \text{m}$. Instead of a fixed speed, suppose the block is also braking, decelerating at a constant rate so its tangential acceleration is $a_t = -1.5\ \text{m/s}^2$ at the exact instant its speed is $v = 15\ \text{m/s}$. Find the total acceleration magnitude and direction at that instant, and the normal force from the road, taking $g = 9.8\ \text{m/s}^2$.

Step 1. The normal acceleration depends only on speed and curvature, not on braking: $a_n = v^2/\rho = 15^2/40 = 225/40 = 5.625\ \text{m/s}^2$, directed toward the centre of curvature, which lies below the road at a hill crest.

Step 2. The tangential acceleration is given directly by the braking: $a_t = -1.5\ \text{m/s}^2$ (negative because it opposes the direction of travel). This is entirely independent of the curvature — braking harder would not change $a_n$, and a sharper hill would not change $a_t$.

Step 3. Combine the two components into the total acceleration magnitude: $a = \sqrt{a_t^2 + a_n^2} = \sqrt{1.5^2 + 5.625^2} = \sqrt{2.25 + 31.64} = \sqrt{33.89} = 5.82\ \text{m/s}^2$, at an angle $\phi = \tan^{-1}(|a_t|/a_n) = \tan^{-1}(1.5/5.625) = 14.9^\circ$ from the normal direction, tilted toward the rearward (braking) side.

Step 4. For the normal force, apply $\sum F_n = ma_n$ with the positive normal direction taken downward (toward the centre, below the crest): $mg - N = ma_n$, so $N = mg - ma_n = 2(9.8) - 2(5.625) = 19.6 - 11.25 = 8.35\ \text{N}$. The normal force is $N = 8.35\ \text{N}$, exactly the same value as in the constant-speed case, because $N$ depends only on $a_n$ — braking changes the block's total acceleration and its direction, but not the road's normal reaction at that instant.

• Assuming $a_t = 0$ just because the path is circular — $a_t$ depends only on whether speed is changing, never on path shape.
• Assuming $a_n = 0$ just because speed is constant — $a_n$ depends only on curvature, never on whether speed is changing.
• Using the wrong sign or direction for $\hat e_n$ — it always points toward the centre of curvature, which flips from "up" at a valley to "down" at a crest, and a sign slip here reverses the answer for $N$.
• Treating $\rho$ as automatically constant for a non-circular path — for a general curve $y = f(x)$, $\rho = [1+(dy/dx)^2]^{3/2}/|d^2y/dx^2|$ must be computed locally unless the path is truly circular.

Ask the tutor: Want to work through a vertical-circle loop-the-loop problem, or see the full derivation of $\rho = [1+(dy/dx)^2]^{3/2}/|d^2y/dx^2|$ for a general curve?