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Tangential and normal acceleration

Acceleration written in the curve's own frame as $a_T\mathbf{T} + a_N\mathbf{N}$, with $a_T$ the rate the speed changes, $a_N = \kappa v^{2}$ the rate the path turns, and no component at all along the binormal.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to split an acceleration vector into its tangential and normal components, compute each from the velocity and acceleration with a dot and a cross product, use Pythagoras to recover a missing component, and read the normal component as the curvature times the square of the speed. You will also be able to say why there is never a component along the binormal.

2. What you already have

Velocity and acceleration from lesson 9, and the moving frame with its curvature from lesson 11. This lesson does one thing with them: it writes the acceleration vector in the frame the curve carries, instead of in the axes somebody chose, and the two numbers that come out turn out to be the two questions a driver cares about.

3. The words this lesson will use

The tangential component $a_T$ is the part of the acceleration along $\mathbf{T}$; the normal component $a_N$ is the part along $\mathbf{N}$. Together they are the decomposition of the acceleration in the moving frame. A centripetal acceleration is a normal component with no tangential part, as on a steady circular bend. $a_N$ is sometimes called the centripetal term for that reason, and $v^{2}/r$ is the same quantity written for a circle of radius $r$.

4. One arrow, two questions

Write the velocity as speed times direction, $\mathbf{v} = v\,\mathbf{T}$ with $v = |\mathbf{v}|$. Differentiating with the product rule,

$$\mathbf{a} = \frac{dv}{dt}\,\mathbf{T} + v\,\frac{d\mathbf{T}}{dt} = \frac{dv}{dt}\,\mathbf{T} + \kappa v^{2}\,\mathbf{N},$$

using $d\mathbf{T}/dt = (d\mathbf{T}/ds)(ds/dt) = \kappa v \mathbf{N}$. So

$$\boxed{\;\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, \qquad a_T = \frac{dv}{dt}, \qquad a_N = \kappa v^{2}.\;}$$

What each half means. $a_T$ answers am I speeding up or slowing down — it is the accelerator and the brake. $a_N$ answers am I turning, and how hard — it is the steering wheel. They are independent: a car can do either, both or neither.

There is no third component. Nothing points along $\mathbf{B}$. The derivation produced only two terms, so the acceleration always lies in the osculating plane — the plane the curve is momentarily lying in. A curve cannot accelerate out of its own plane of bending; if the motion leaves that plane, it is because the plane itself is turning.

Computing the two numbers. Since $\mathbf{T}$ and $\mathbf{N}$ are perpendicular unit vectors,

$$a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|}, \qquad a_N = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|}, \qquad |\mathbf{a}|^{2} = a_T^{2} + a_N^{2}.$$

The last of these is Pythagoras, and it is how a missing component is recovered from the other two.

5. The speed squared, and why bends are dangerous

$a_N = \kappa v^{2}$, and for a circular bend of radius $r$ the curvature is $1/r$, so

$$a_N = \frac{v^{2}}{r}.$$

The square is the whole story of road design. Taking a bend at twice the speed asks for four times the sideways acceleration, and the sideways force available from the tyres has not changed. Halving the radius of a bend does the same thing as multiplying the speed by $\sqrt{2}$.

It is also why motorway curves look almost straight. Designing for a high speed with a bounded $a_N$ forces $r = v^{2}/a_N$ to be enormous — hundreds of metres — and a bend of that radius, seen from inside it, reads as a road that is barely turning at all.

The same formula explains why a curve is banked. Tilting the road lets some of the normal force from the surface supply the sideways acceleration, instead of asking friction for all of it. What the bank angle is chosen to match is exactly $v^{2}/r$ at the design speed.

6. Where this goes wrong

Reading $a_T$ as the magnitude of the acceleration. It is one component of it. $|\mathbf{a}|$ is the hypotenuse of the two, and equals $|a_T|$ only on a straight path.

Combining the components by adding or subtracting them. They are perpendicular, so they combine by Pythagoras. $|\mathbf{a}| - a_T$ is not $a_N$.

Expecting $a_N$ to be negative on a bend to the right. $\mathbf{N}$ already points into the bend, whichever way that is, so $a_N = \kappa v^{2}$ is never negative. The signed information lives in $\mathbf{N}$, not in $a_N$.

Looking for a binormal component. There is not one. If your arithmetic produces a component along $\mathbf{B}$, the arithmetic is wrong.

7. Slowing down and turning are different accelerations

Everyday speech uses accelerating to mean going faster, and that single word is responsible for most of the trouble here. In this subject acceleration is a vector, and the decomposition says precisely how the everyday meaning fits inside the technical one: $a_T$ is the everyday sense, and it is one of two components.

The consequences run one way only. Zero acceleration does imply steady speed and a straight path, since both components must vanish. But steady speed implies nothing about the acceleration except that $a_T = 0$, and a straight path implies nothing except that $a_N = 0$. A learner who reads constant speed as no acceleration has thrown away the turning term, which on a bend is usually the larger of the two — and on a steady circular orbit is the only one there is.

8. Splitting an acceleration at an instant

1. Let $\mathbf{r}(t) = \langle t,\, t^{2},\, 0 \rangle$, so $\mathbf{v} = \langle 1,\, 2t,\, 0 \rangle$ and $\mathbf{a} = \langle 0,\, 2,\, 0 \rangle$. At $t = 1$: $\mathbf{v} = \langle 1,2,0 \rangle$, $v = \sqrt{5}$.

   Velocity, acceleration and speed at the instant.

2. $\mathbf{v} \cdot \mathbf{a} = 4$, so $a_T = 4/\sqrt{5}$; and $\mathbf{v} \times \mathbf{a} = \langle 0,0,2 \rangle$, so $a_N = 2/\sqrt{5}$.

   One dot product and one cross product.

3. Check: $16/5 + 4/5 = 4 = |\mathbf{a}|^{2}$. Both components are positive, so at this instant the particle is both speeding up and turning — and the acceleration vector $\langle 0,2,0 \rangle$ has been re-read as how much of that is the accelerator and how much the steering.

   Pythagoras checks the whole calculation.

9. A bend taken twice as fast

1. A car rounds a bend of radius $50$ metres at $10$ metres per second. Its speed is steady, so $a_T = 0$ and $a_N = 10^{2}/50 = 2$ metres per second squared.

   Steady speed leaves only the turning part.

2. The same bend at $20$ metres per second gives $a_N = 400/50 = 8$.

   Twice the speed, four times the acceleration.

3. The tyres must supply four times as much sideways force for a doubling of speed, and the bend has not changed at all. The square in $\kappa v^{2}$ is not a detail of the algebra; it is the reason speed limits fall on bends rather than on straights.

   The square is the whole of the design rule.

10. Your turn: how hard is a cyclist turning?

1. A cyclist rides at a steady $6$ metres per second round a bend of radius $12$ metres. Because the speed is steady, $a_T = 0$.

   The derivative of a constant speed is zero.

2. The normal component is $a_N = v^{2}/r = 36/12 = 3$ metres per second squared, and $|\mathbf{a}| = \sqrt{0 + 9} = 3$ as well.

3. Your turn: work this step out. Its working is at the end of the packet.

   So the whole acceleration is sideways and about a third of gravity, which is why the cyclist has to lean. Had the speed been dropping at $4$ metres per second squared as well, the magnitude would have been $\sqrt{16 + 9} = 5$ — the two effects add as legs of a right-angled triangle, never as numbers.

11. Guided practice

Match each quantity to what it equals, for a particle moving at speed $4$ along a curve.

12. Guided practice

A vehicle holds a steady speed of $2$ metres per second round a circular bend of radius $6$ metres. Fill in the two components of its acceleration and their resultant.

13. Practice

A particle's acceleration has magnitude $35$ and tangential component $21$. What is its normal component?

Answer:

14. Practice

A train takes a bend at a steady $25$ metres per second. Which component of its acceleration is zero?

• The tangential component
• The normal component
• Both components
• Neither component

15. Practice

Match each quantity to what it equals, for a particle moving at speed $8$ along a curve.

16. Somewhere new

An aircraft's instruments report an acceleration of magnitude $35$, of which $21$ is tangential, while the airspeed is $14$. What is the radius of curvature of its flight path at that instant?

Answer:

17. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

18. Test question

A vehicle holds a steady speed of $7$ metres per second round a circular bend of radius $4$ metres. Fill in the two components of its acceleration and their resultant.

19. What you can do now

You can split an acceleration into the part that changes the speed and the part that turns the path, and say which of them a steady speed removes. Next: leaving curves behind for functions of more than one variable, and the surfaces they describe.

Working for the steps left to you

10. Your turn: how hard is a cyclist turning?, step 3