Lesson 6 · 35 min
Acceleration in Path Coordinates
Press the accelerator on a straight road and you are pushed back into your seat. Hold a steady speed round a bend and you are pushed sideways. Path coordinates separate these two effects exactly: one term for changing speed, one for changing direction.
Learning objectives
- Derive \(\avec = \dot v\,\et + (v^2/\rho)\,\en\) from \(\vvec = v\,\et\), using \(d\et/dt = (v/\rho)\,\en\).
- Explain what the tangential and normal accelerations each do, and find the magnitude of the total acceleration.
- Apply the results to circular motion using \(v = \omega r\), \(a_t = \alpha r\) and \(a_n = \omega^2 r\).
- Use \(a_t = dv/dt\) or \(a_t = v\,dv/ds\) when the tangential acceleration depends on time or on position.
Differentiating \(\vvec = v\,\et\)
Both factors of \(\vvec = v\,\et\) can change: the speed \(v\), and the direction \(\et\). The product rule keeps both:
\[ \avec = \frac{d\vvec}{dt} = \dot v\,\et + v\,\frac{d\et}{dt} \]The first term is familiar. The second is new, and it is not zero because \(\et\) turns as the particle moves round a bend. How fast does it turn?
How fast does \(\et\) turn?
Let \(\psi\) be the direction of \(\et\). When the particle moves a short distance \(ds\) along a path of radius of curvature \(\rho\), the tangent turns through the angle \(d\psi = ds/\rho\) (arc length = radius × angle, on the osculating circle). A unit vector turned through a small angle \(d\psi\) changes by \(d\psi\) in the perpendicular direction, toward the side it turns to, which is \(\en\):
\[ d\et = d\psi\,\en = \frac{ds}{\rho}\,\en \quad\Rightarrow\quad \frac{d\et}{dt} = \frac{1}{\rho}\frac{ds}{dt}\,\en = \frac{v}{\rho}\,\en \]Substituting into the product rule gives the central result of this module:
Acceleration in path coordinates
\[ \avec = a_t\,\colT{\et} + a_n\,\colN{\en}, \qquad a_t = \dot v = \frac{dv}{dt} = v\frac{dv}{ds}, \qquad a_n = \frac{v^2}{\rho} \] \[ |\avec| = \sqrt{a_t^2 + a_n^2} \]- Tangential acceleration \(a_t\) changes the speed. It is positive when speeding up and negative when slowing down, exactly like rectilinear acceleration along the path. The form \(v\,dv/ds\) is the curved-path version of \(a\,ds = v\,dv\).
- Normal acceleration \(a_n\) changes the direction. It is never negative and always points toward the center of curvature. It is also called centripetal acceleration.
Special cases
| Motion | \(a_t\) | \(a_n\) | Direction of \(\avec\) |
|---|---|---|---|
| Straight line, changing speed | \(\dot v\) | \(0\) (\(\rho = \infty\)) | along the path |
| Curve, constant speed | \(0\) | \(v^2/\rho\) | toward the center of curvature |
| Curve, changing speed | \(\dot v\) | \(v^2/\rho\) | inside the curve, tilted forward (speeding up) or back (slowing) |
| Circle of radius \(r\), spin \(\omega\), \(\alpha\) | \(\alpha r\) | \(\omega^2 r = v^2/r\) | \(\en\) toward the center of the circle |
For a point on a rotating body at distance \(r\) from the axis, \(v = \omega r\), where \(\omega\) is the angular velocity in rad/s; its rate of change is the angular acceleration \(\alpha = \dot\omega\). Then \(a_t = \dot v = \alpha r\) and \(a_n = v^2/r = \omega^2 r = v\omega\).
Worked examples
Example 6.1 — Braking on a curve
A car travels at \(20\ \text{m/s}\) round a curve of radius \(150\ \text{m}\) and brakes, slowing at \(1.5\ \text{m/s}^2\). Find the magnitude of its acceleration.
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Tangential. The car slows, so \(a_t = -1.5\ \text{m/s}^2\) (opposite to \(\et\)).
Normal. \(a_n = v^2/\rho = 20^2/150 = 2.667\ \text{m/s}^2\), toward the center of the curve.
Total.
\[ |\avec| = \sqrt{(-1.5)^2 + 2.667^2} = 3.060\ \text{m/s}^2 \]Interpret. The tyres must provide both parts through friction. Braking on a curve uses up grip that would otherwise be available for turning, which is why drivers are taught to brake before a bend, not in it.
Example 6.2 — Speeding up with \(a_t\) given as a function of position
A car starts from rest and moves along a curved road, gaining speed with \(a_t = 0.05s\ \text{m/s}^2\), where \(s\) is the distance travelled in metres. When \(s = 60\ \text{m}\) it is on a part of the road where \(\rho = 50\ \text{m}\). Find its speed and the magnitude of its acceleration there.
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Speed. \(a_t\) depends on \(s\), so use \(a_t\,ds = v\,dv\) and integrate from rest at \(s = 0\):
\[ \int_0^v v\,dv = \int_0^s 0.05s\,ds \quad\Rightarrow\quad \frac{v^2}{2} = 0.025s^2 \quad\Rightarrow\quad v = \sqrt{0.05}\,s \]At \(s = 60\ \text{m}\): \(v = 0.2236 \times 60 = 13.42\ \text{m/s}\).
Acceleration. \(a_t = 0.05(60) = 3\ \text{m/s}^2\) and \(a_n = v^2/\rho = 180/50 = 3.6\ \text{m/s}^2\), so
\[ |\avec| = \sqrt{3^2 + 3.6^2} = 4.686\ \text{m/s}^2 \]Note. You cannot use \(v = v_0 + a_t t\) here: \(a_t\) is not constant.
Example 6.3 — A point on a spinning rotor
A point on a rotor blade is \(0.8\ \text{m}\) from the axis. At an instant the rotor spins at \(\omega = 10\ \text{rad/s}\) and is speeding up at \(\alpha = 3\ \text{rad/s}^2\). Find the point's speed and acceleration.
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Interpret. Spinning parts are dominated by the normal (centripetal) term. The blade root must carry the force needed to give every bit of the blade this acceleration, which is why rotors are designed around their top speed.
Check your understanding
Key takeaways
- \(\avec = \dot v\,\et + (v^2/\rho)\,\en\). The normal term comes from \(\et\) turning: \(d\et/dt = (v/\rho)\,\en\).
- \(a_t = dv/dt = v\,dv/ds\) changes the speed; \(a_n = v^2/\rho \ge 0\) changes the direction and points toward the center of curvature.
- \(|\avec| = \sqrt{a_t^2 + a_n^2}\). Constant speed on a curve still means \(a_n \ne 0\).
- Circular motion: \(v = \omega r\), \(a_t = \alpha r\), \(a_n = \omega^2 r\).
- Next: Lesson 7 shows how to find \(\rho\) when the path is given as an equation.