Lesson 8 · 35 min
Connecting x–y and n–t Components
Rectangular and path components are two descriptions of the same vectors. Real problems often hand you one and ask about the other: a projectile (natural in \(x\)–\(y\)) whose path curvature you need, or a car (natural in \(n\)–\(t\)) whose acceleration you want on a map. This lesson is the bridge.
Learning objectives
- Write \(\et\) and \(\en\) in terms of \(\ihat\) and \(\jhat\) from the velocity.
- Convert \(x\)–\(y\) acceleration components to \(a_t\) and \(a_n\) using the dot and cross products, and find \(\rho\).
- Decide from \(\avec\cdot\vvec\) whether a particle is speeding up or slowing down.
- Convert \(a_t\) and \(a_n\) back to \(a_x\) and \(a_y\) for a known direction of travel.
From \(x\)–\(y\) to \(n\)–\(t\)
Suppose you know \(\vvec = v_x\ihat + v_y\jhat\) and \(\avec = a_x\ihat + a_y\jhat\) at some instant. The tangent direction is the direction of the velocity, so
\[ \et = \frac{\vvec}{v} = \frac{v_x\,\ihat + v_y\,\jhat}{v}, \qquad v = \sqrt{v_x^2 + v_y^2} \]The tangential acceleration is the part of \(\avec\) along \(\et\), a dot product. The normal acceleration is the part perpendicular to \(\et\); in the plane it is found with the "cross product" \(v_x a_y - v_y a_x\) (the size of \(\vvec \times \avec\)):
Rectangular to path components
\[ a_t = \avec\cdot\et = \frac{v_x a_x + v_y a_y}{v}, \qquad a_n = \frac{|v_x a_y - v_y a_x|}{v} = \sqrt{|\avec|^2 - a_t^2} \] \[ \rho = \frac{v^2}{a_n} \]\(\en\) is \(\et\) turned \(90^\circ\): \(\en = (-v_y\,\ihat + v_x\,\jhat)/v\) if \(v_x a_y - v_y a_x \gt 0\) (the path turns counter-clockwise), and the opposite vector if it is negative.
The sign of \(a_t\), and so of \(\avec\cdot\vvec\), tells you what the speed is doing:
- \(\avec\cdot\vvec \gt 0\): \(\avec\) has a forward part, and the particle is speeding up.
- \(\avec\cdot\vvec \lt 0\): \(\avec\) has a backward part, and it is slowing down.
- \(\avec\cdot\vvec = 0\): \(\avec \perp \vvec\), and the speed is momentarily steady.
Example 8.1 — A projectile described in path coordinates
A ball is launched at \(25\ \text{m/s}\), \(40^\circ\) above the horizontal. One second later, find its tangential and normal accelerations and the radius of curvature of its path.
Show solution
Rectangular components at \(t = 1\ \text{s}\).
\[ \begin{aligned} v_x &= 25\cos 40^\circ = 19.15\ \text{m/s}, & a_x &= 0 \\ v_y &= 25\sin 40^\circ - 9.81(1) = 6.260\ \text{m/s}, & a_y &= -9.81\ \text{m/s}^2 \end{aligned} \] \[ v = \sqrt{19.15^2 + 6.260^2} = 20.15\ \text{m/s} \]Tangential.
\[ a_t = \frac{v_x a_x + v_y a_y}{v} = \frac{0 + 6.260(-9.81)}{20.15} = -3.048\ \text{m/s}^2 \]Negative: the ball is still climbing, so gravity is slowing it down.
Normal.
\[ a_n = \frac{|v_x a_y - v_y a_x|}{v} = \frac{|19.15(-9.81) - 0|}{20.15} = 9.325\ \text{m/s}^2 \]Check: \(\sqrt{3.048^2 + 9.325^2} = 9.81\ \text{m/s}^2 = g\). ✓
Radius of curvature. \(\rho = v^2/a_n = 20.15^2/9.325 = 43.54\ \text{m}\). (At the top of the flight it would be \(19.15^2/9.81 = 37.39\ \text{m}\): the parabola is tightest at its peak.)
Example 8.2 — The robot gripper again
In Example 3.1, the gripper had \(\vvec = 6\ihat + 10\jhat\ \text{m/s}\) and \(\avec = 6\ihat + 6\jhat\ \text{m/s}^2\) at \(t = 2\ \text{s}\). Find \(a_t\), \(a_n\) and \(\rho\).
Show solution
\(v_x a_y - v_y a_x = -24 \lt 0\): the path is turning clockwise, to the right of the direction of travel, and the gripper is speeding up strongly (\(a_t \gt 0\)).
From \(n\)–\(t\) back to \(x\)–\(y\)
Now suppose you know \(a_t\) and \(a_n\) and the direction of travel \(\psi\) (measured counter-clockwise from \(+x\)). Write both unit vectors in rectangular form and add:
\[ \et = \cos\psi\,\ihat + \sin\psi\,\jhat, \qquad \en = \pm\left(-\sin\psi\,\ihat + \cos\psi\,\jhat\right) \]with \(+\) for a left (counter-clockwise) turn and \(-\) for a right turn. Then \(\avec = a_t\,\et + a_n\,\en\). For a left turn:
Path to rectangular components (turning left, counter-clockwise)
\[ a_x = a_t\cos\psi - a_n\sin\psi, \qquad a_y = a_t\sin\psi + a_n\cos\psi \]For a right turn, change the sign of every \(a_n\) term.
Example 8.3 — A car on a map
A car heading \(30^\circ\) north of east is turning left, with \(a_t = 2\ \text{m/s}^2\) and \(a_n = 3\ \text{m/s}^2\). With \(x\) east and \(y\) north, find \(a_x\) and \(a_y\).
Show solution
\(\psi = 30^\circ\), left turn:
\[ \begin{aligned} a_x &= 2\cos 30^\circ - 3\sin 30^\circ = 1.732 - 1.5 = 0.232\ \text{m/s}^2 \\ a_y &= 2\sin 30^\circ + 3\cos 30^\circ = 1 + 2.598 = 3.598\ \text{m/s}^2 \end{aligned} \]Check: \(\sqrt{0.232^2 + 3.598^2} = 3.606 = \sqrt{2^2 + 3^2}\ \text{m/s}^2\). ✓ The size of a vector does not depend on the components you use.
Check your understanding
Key takeaways
- \(\et = \vvec/v\); \(\en\) is \(\et\) turned \(90^\circ\) toward the side the path bends.
- \(a_t = (v_x a_x + v_y a_y)/v\) and \(a_n = |v_x a_y - v_y a_x|/v\); then \(\rho = v^2/a_n\).
- \(\avec\cdot\vvec \gt 0\) means speeding up, \(\lt 0\) slowing down, \(= 0\) steady speed.
- Back to \(x\)–\(y\) (left turn): \(a_x = a_t\cos\psi - a_n\sin\psi\), \(a_y = a_t\sin\psi + a_n\cos\psi\).
- Next: Lesson 9 puts it all together: choosing a system and solving engineering problems.