Lesson 6 · 40 min

Normal and Tangential Coordinates

A particle on a curved path accelerates toward the inside of the curve even at constant speed. Something has to push it there. Write the equation of motion along the path and across it, and the normal equation tells you how large that push must be: \(mv^2/\rho\). Whether the road, a cord or a rail can supply it decides whether the particle stays on its path.

Learning objectives

Write the equations of motion in normal and tangential components, \(\sum F_t = m\,dv/dt\) and \(\sum F_n = mv^2/\rho\).
Identify the real force, or forces, that supply the normal acceleration; there is no separate "centripetal" or "centrifugal" force.
Find the friction a car needs on a flat curve, and the largest speed at which it can take the curve.
Combine the tangential and normal components when a particle speeds up or slows down on a curve, and analyze a conical pendulum.

The equations of motion along and across the path

When the path is known, put the axes on it. The tangential axis \(t\) points along the motion, the normal axis \(n\) points toward the center of curvature, and the binormal \(b\) is perpendicular to both. The kinematics comes from the Curvilinear Motion module: \(a_t = \dot v\), \(a_n = v^2/\rho\), \(a_b = 0\).

Normal and tangential components

\[ \textstyle\sum F_t = m\colMA{a_t} = m\dfrac{dv}{dt} \qquad \sum F_n = m\colMA{a_n} = m\dfrac{v^2}{\rho} \qquad \sum F_b = 0 \]

\(\rho\) is the radius of curvature, and \(\sum F_n\) is the sum of the force components toward the center of curvature.

The tangential forces change the speed. The normal forces change the direction.
\(mv^2/\rho\) is not a force. It goes on the kinetic diagram, and the free-body diagram must show which real forces add up to it: friction, a normal force, a tension, gravity.
A passenger in a turning car feels pushed outward, but no outward force acts. Their body tends to go straight, and the seat and door push it inward to make it follow the curve.

A car on a flat curve

On a level curve the only horizontal force on a car is the friction of the road on its tires. Before you open Figure 6.1, make a prediction.

Figure 6.1 Top: a \(1200\ \text{kg}\) car drives along a straight road into a flat U-turn of radius \(\rho\) at constant speed. In the curve, friction \(\colF{F_f}\) on the tires points toward the center \(O\). Bottom: the free-body diagram of the car, seen from behind in the curve, and its kinetic diagram, \(\colMA{mv^2/\rho}\) toward the center. When the friction needed is more than the road can give, \(\mu_s N\), the two diagrams cannot be equal: the car skids, and friction, now opposing the slide, can no longer turn it.

Across the road (the \(b\) direction, vertical), \(N = mg\). Toward the center:

\[ \textstyle\sum F_n = m\dfrac{v^2}{\rho}{:}\quad \colF{F_f} = m\frac{v^2}{\rho} \]

The tires do not slip sideways, so this is static friction, and it can be at most \(\mu_s N = \mu_s mg\).

The fastest a car can take a flat curve

\[ m\frac{v^2}{\rho} \le \mu_s mg \quad\Rightarrow\quad v_{\max} = \sqrt{\mu_s g\rho} \]

The mass cancels: a truck and a bicycle on the same road have the same limit, if their tires grip alike. Double the speed and the friction needed quadruples.

Example 6.1 — Dry road, wet road

A flat curve has a radius of \(80\ \text{m}\). How fast can a car take it when the road is dry, \(\mu_s = 0.6\), and when it is wet, \(\mu_s = 0.3\)?

Show solution

\[ v_{\max} = \sqrt{0.6(9.81)(80)} = 21.7\ \text{m/s} = 78\ \text{km/h} \qquad v_{\max} = \sqrt{0.3(9.81)(80)} = 15.3\ \text{m/s} = 55\ \text{km/h} \]

Halving the friction cuts the safe speed by a factor of \(\sqrt 2\), not 2: \(v_{\max}\) goes as \(\sqrt{\mu_s}\).

Speeding up on a curve

When the speed changes on the curve, the acceleration has both components, and so does the force that causes it. For a car, both come from friction: \(F_t = m\,dv/dt\) along the road and \(F_n = mv^2/\rho\) across it. The tires grip only if the total friction is within the limit:

\[ F_f = m\sqrt{a_t^2 + a_n^2} \le \mu_s N \]

That is why braking hard or accelerating hard in a curve is what makes a car slide: it uses up grip that the turn needs.

Example 6.2 — Accelerating out of a curve

A \(1500\ \text{kg}\) car on a flat curve of radius \(100\ \text{m}\) is traveling at \(20\ \text{m/s}\) and speeding up at \(3\ \text{m/s}^2\). What friction force acts on it, and what is the least coefficient of static friction that prevents sliding?

Show solution

\[ a_n = \frac{v^2}{\rho} = \frac{20^2}{100} = 4\ \text{m/s}^2 \qquad a_t = 3\ \text{m/s}^2 \qquad a = \sqrt{3^2 + 4^2} = 5\ \text{m/s}^2 \] \[ F_f = ma = 1500(5) = 7500\ \text{N} \qquad \mu_s \ge \frac{a}{g} = \frac{5}{9.81} = 0.51 \]

At a steady \(20\ \text{m/s}\), \(\mu_s = 0.41\) would be enough.

The conical pendulum

A ball on a cord of length \(L\) swings in a horizontal circle, the cord sweeping out a cone at a steady angle \(\theta\) from the vertical. The ball's path has radius \(\rho = L\sin\theta\) and its acceleration is \(v^2/\rho\), horizontal and toward the axis. Only two forces act: the weight and the cord's tension.

Figure 6.2 A conical pendulum in motion: seen from the side, a little from above (top left), and from directly above (top right). Press Play. The velocity \(\colV{\vvec}\) is always tangent to the circle; the acceleration \(\colMA{\avec}\), of constant size \(v^2/\rho\), always points horizontally at the axis and turns with the ball, so the speed never changes but the direction does. Seen from the side, both arrows shorten as they turn toward or away from you; in the top view, your side is at the bottom. Below, the free-body and kinetic diagrams of the ball, drawn in the vertical plane through the axis and the ball, which turns with it: the vertical part of the tension holds up the weight, \(T\cos\theta = \colW{mg}\), and the horizontal part is the whole of \(\colMA{mv^2/\rho}\).

\[ \textstyle\sum F_b = 0{:}\quad T\cos\theta - mg = 0 \qquad\qquad \sum F_n = m\dfrac{v^2}{\rho}{:}\quad T\sin\theta = m\frac{v^2}{L\sin\theta} \]

Conical pendulum

\[ T = \frac{mg}{\cos\theta} \qquad v = \sqrt{gL\sin\theta\tan\theta} \qquad \tau = 2\pi\sqrt{\frac{L\cos\theta}{g}} \]

\(\tau\) is the time for one revolution. The tension always exceeds the weight, and grows without limit as the cone flattens: no speed makes a cord swing exactly horizontally.

Example 6.3 — A ball on a cord

A \(0.5\ \text{kg}\) ball on a \(1.2\ \text{m}\) cord swings as a conical pendulum with the cord at \(30^\circ\) to the vertical. Find the tension, the ball's speed, and the time for one revolution.

Show solution

\[ T = \frac{0.5(9.81)}{\cos 30^\circ} = 5.66\ \text{N} \qquad \rho = 1.2\sin 30^\circ = 0.6\ \text{m} \] \[ v = \sqrt{9.81(1.2)\sin 30^\circ\tan 30^\circ} = 1.84\ \text{m/s} \qquad \tau = \frac{2\pi\rho}{v} = \frac{2\pi(0.6)}{1.843} = 2.05\ \text{s} \]

Check your understanding

Key takeaways

\(\sum F_t = m\,dv/dt\) changes the speed; \(\sum F_n = mv^2/\rho\) changes the direction.
\(mv^2/\rho\) belongs on the kinetic diagram. Real forces, such as friction, normal forces and tensions, must add up to it.
On a flat curve, friction supplies it: \(v_{\max} = \sqrt{\mu_s g\rho}\), whatever the mass.
Speeding up or braking on a curve adds a tangential component, so the total friction needed is \(m\sqrt{a_t^2 + a_n^2}\).
Conical pendulum: \(T\cos\theta = mg\), \(T\sin\theta = mv^2/\rho\).
Next, Lesson 7 banks the curve, and sends particles over crests and around loops.