Curvilinear motion: rectangular, normal-tangential and polar coordinates

Most motion is not along a straight line. A parcel leaving a chute follows a curved path, a tote on a spiral conveyor winds round a column, and the gripper of a robot arm sweeps through an arc while it extends. Describing these motions needs vectors, and it needs a coordinate system suited to the problem. This lesson develops the three systems used throughout dynamics: rectangular components, which suit projectiles; normal and tangential components, which suit motion along a known path; and polar components, which suit anything that rotates and extends at the same time. The physics is the same in all three. The choice of coordinates only decides how hard the arithmetic is.

Position, velocity and acceleration as vectors

Let the position of a particle relative to a fixed origin be the vector \(\mathbf{r}(t)\). As the particle moves, the tip of \(\mathbf{r}\) traces out the path. The velocity is the time derivative of position, \(\mathbf{v} = d\mathbf{r}/dt\), and it is always tangent to the path, pointing in the direction of motion. Its magnitude is the speed, \(v = ds/dt\), where \(s\) is the distance measured along the path.

The acceleration is \(\mathbf{a} = d\mathbf{v}/dt\). Unlike the velocity, the acceleration is generally not tangent to the path. A car going round a bend at constant speed has a velocity that is always tangent to the road, but its velocity vector is turning, and that change in direction is an acceleration pointing toward the inside of the bend. This is the single most important idea of curvilinear motion: a particle can accelerate without changing speed, simply by changing direction. Anything that follows a curve needs a force pushing it toward the inside of the curve, and the machine has to supply that force.

Rectangular components

In a fixed \(x\)-\(y\)-\(z\) frame with unit vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\), the position is \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\). Because the unit vectors are fixed in direction, differentiating is easy:

A dot over a symbol means a derivative with respect to time. Each component behaves like an independent rectilinear motion, which is why the rectangular system is so useful when the accelerations along fixed directions are known. The classic case is projectile motion.

Projectile motion

A body moving freely near the Earth's surface, with air resistance neglected, has an acceleration of \(g\) straight down and nothing else. Taking \(x\) horizontal and \(y\) vertically up, \(a_x = 0\) and \(a_y = -g\). The horizontal motion is at constant velocity and the vertical motion is at constant acceleration:

The two motions are linked only through time. A typical problem gives enough information to find the time of flight from one direction and uses it in the other. Eliminating \(t\) shows that the path is a parabola. Parcels leaving the end of a belt, gravel thrown from a slinger conveyor, water from a nozzle and a part ejected from a press all follow nearly parabolic paths, as long as they are dense enough that air resistance is small.

Light, bulky cartons are not dense enough to ignore air resistance completely. The program below compares the drag-free answer for a parcel leaving a sloping chute with a numerical solution that includes a drag force proportional to the square of the speed. It also computes the radius of curvature of the path at launch, which the next section explains.

# A parcel leaves the end of a chute at 3.2 m/s, angled 20 degrees below
# horizontal, 1.1 m above a collection belt. Where does it land?
# Compare the drag-free projectile answer with a numerical solution that
# includes quadratic air drag (light box, large face).
import math

g = 9.81
v0, ang, h = 3.2, math.radians(-20), 1.1
vx0, vy0 = v0 * math.cos(ang), v0 * math.sin(ang)

# Drag-free: y = h + vy0 t - g t^2/2 = 0
t_land = (vy0 + math.sqrt(vy0**2 + 2 * g * h)) / g
print(f"no drag:   flight {t_land:.3f} s, lands {vx0 * t_land:.3f} m out")

# With drag: a = -(c/m)|v| v - g j, c/m = 0.15 1/m for a light carton
cm = 0.15
dt, t, x, y, vx, vy = 1e-4, 0.0, 0.0, h, vx0, vy0
while y > 0:
    sp = math.hypot(vx, vy)
    ax, ay = -cm * sp * vx, -cm * sp * vy - g
    vx, vy = vx + ax * dt, vy + ay * dt
    x, y, t = x + vx * dt, y + vy * dt, t + dt
print(f"with drag: flight {t:.3f} s, lands {x:.3f} m out, speed {math.hypot(vx, vy):.2f} m/s")

# Radius of curvature at the start of the drag-free path: rho = v^2 / a_n
a_n = g * math.cos(ang)            # component of g normal to the velocity
print(f"radius of curvature at launch: {v0**2 / a_n:.3f} m")

no drag:   flight 0.375 s, lands 1.128 m out
with drag: flight 0.395 s, lands 1.069 m out, speed 4.91 m/s
radius of curvature at launch: 1.111 m

For this short throw, drag shortens the landing distance by about 6 cm, roughly 5 percent. That is small enough that the drag-free model is fine for a first layout, but large enough that the chute end and the receiving belt should be positioned with some adjustment. For longer throws the effect grows quickly, because drag acts for longer and the parcel is moving faster.

Normal and tangential components

When the path is known, as it is for a vehicle on a road, a tote on a curved conveyor or a cam follower in its groove, it is natural to describe the motion relative to the path itself. At each point on the path we place two unit vectors: \(\mathbf{u}_t\), tangent to the path in the direction of motion, and \(\mathbf{u}_n\), perpendicular to it and pointing toward the centre of curvature. The velocity is purely tangential, \(\mathbf{v} = v\,\mathbf{u}_t\). The acceleration has two parts:

The tangential component, \(a_t = \dot{v}\), is the rate of change of speed. It is positive when the particle speeds up and negative when it slows down. The normal component, \(a_n = v^2/\rho\), comes from the change in direction, where \(\rho\) is the radius of curvature of the path at that point. It always points toward the centre of curvature and is never negative. On a straight line \(\rho\) is infinite and \(a_n = 0\); on a circle \(\rho\) is the radius. The magnitude of the total acceleration is \(\sqrt{a_t^2 + a_n^2}\).

For a path given as \(y = f(x)\), the radius of curvature is

Highway and railway designers use this formula to set the transition curves that ease vehicles into bends, so that the normal acceleration builds up gradually rather than jumping from zero to its full value at the start of the curve. The same thinking applies to conveyor and cam design: a sudden change in curvature means a sudden change in the force needed to hold the load on its path.

The normal acceleration grows with the square of speed. Doubling the speed of a curved conveyor quadruples the inward force needed to keep the totes on it, which is why curved sections often run slower than straight ones, or are banked.

Polar coordinates

For motion described by a distance from a fixed point and an angle, such as a robot arm that rotates about its base while its gripper slides in or out, polar coordinates are the natural choice. The position is \(\mathbf{r} = r\,\mathbf{u}_r\), where \(\mathbf{u}_r\) points from the origin toward the particle and \(\mathbf{u}_\theta\) is perpendicular to it in the direction of increasing \(\theta\). These unit vectors rotate with the particle, so their derivatives are not zero: \(\dot{\mathbf{u}}_r = \dot\theta\,\mathbf{u}_\theta\) and \(\dot{\mathbf{u}}_\theta = -\dot\theta\,\mathbf{u}_r\). Differentiating carefully gives

Each term has a physical meaning. In the radial direction, \(\ddot r\) is the acceleration of the sliding motion and \(-r\dot\theta^2\) is the centripetal acceleration of a point going round in a circle. In the transverse direction, \(r\ddot\theta\) is the acceleration due to angular acceleration, and \(2\dot r\dot\theta\) is the Coriolis acceleration. The Coriolis term appears whenever something moves outward or inward along a rotating arm. It is why a person walking outward on a turning merry-go-round feels pushed sideways, and why a robot that extends its arm while slewing feels a sideways load on the gripper that disappears if it does one motion at a time.

A practical consequence: if the robot needs to move a delicate item, the programmer can reduce the sideways load by extending the arm and slewing it in separate phases, at the cost of a longer cycle. Many pick-and-place routines blend the motions anyway and simply limit the speeds so that the combined acceleration stays below what the gripper can hold.

Choosing a coordinate system

The three systems describe the same physical acceleration, so any one of them gives the right answer if used correctly. The test of a good choice is how many terms vanish. For the robot arm, polar coordinates make the problem a few lines long; in rectangular coordinates it would require differentiating sines and cosines of a changing angle. A minute spent choosing coordinates often saves ten minutes of algebra and avoids most sign errors.