Lesson 6 · 35 min
Paths Given as \(r = f(\theta)\)
A cam groove, a spiral guide or a circular slot fixes the shape of the path, \(r = f(\theta)\). A turning arm fixes \(\theta(t)\). The chain rule combines the two into \(\dot r\) and \(\ddot r\), and the shape alone fixes the direction of the velocity.
Learning objectives
- Find \(\dot r\) and \(\ddot r\) for a particle on a path \(r = f(\theta)\) from \(\dot\theta\) and \(\ddot\theta\), with the chain rule.
- Compute the velocity and acceleration of particles on spirals, cardioids and circles.
- Find the tangent angle \(\psi\) from \(\tan\psi = r/(dr/d\theta)\) and explain what it describes.
- Decide whether a problem gives the motion as \(r(t)\) or as a path \(r = f(\theta)\) plus \(\theta(t)\).
The chain rule on a path
In Lesson 5 the motion was given directly as \(r(t)\) and \(\theta(t)\). Often it is given in two pieces instead: a guide forces the particle onto a curve \(r = f(\theta)\), and an arm or a motor sets \(\theta\), \(\dot\theta\) and \(\ddot\theta\). Then \(r\) depends on time only through \(\theta\), and the chain rule gives its rates. Write \(f' = dr/d\theta\) and \(f'' = d^2r/d\theta^2\):
\[ \begin{aligned} \dot r &= \frac{dr}{d\theta}\,\frac{d\theta}{dt} = f'\,\dot\theta \\ \ddot r &= \frac{d}{dt}\left(f'\,\dot\theta\right) = \left(f''\,\dot\theta\right)\dot\theta + f'\,\ddot\theta = f''\,\dot\theta^2 + f'\,\ddot\theta \end{aligned} \]In the second line, \(f'\) itself depends on \(\theta\), so its time derivative is \(f''\,\dot\theta\) (the chain rule again), and the product rule gives the two terms.
Rates on a path \(r = f(\theta)\)
\[ \dot r = \frac{dr}{d\theta}\,\dot\theta, \qquad \ddot r = \frac{d^2 r}{d\theta^2}\,\dot\theta^2 + \frac{dr}{d\theta}\,\ddot\theta \]Then use \(v_r = \dot r\), \(v_\theta = r\dot\theta\), \(a_r = \ddot r - r\dot\theta^2\) and \(a_\theta = r\ddot\theta + 2\dot r\dot\theta\) as before.
Procedure. (1) Write \(f\), \(f'\) and \(f''\). (2) Evaluate them at the given \(\theta\), in radians. (3) Find \(\dot r\) and \(\ddot r\). (4) Substitute into the velocity and acceleration components.
| Path | \(r = f(\theta)\) | \(f' = dr/d\theta\) | \(f'' = d^2r/d\theta^2\) |
|---|---|---|---|
| Circle about \(O\) | \(r = a\) | \(0\) | \(0\) |
| Archimedean spiral | \(r = b\,\theta\) | \(b\) | \(0\) |
| Cardioid | \(r = a(1 + \cos\theta)\) | \(-a\sin\theta\) | \(-a\cos\theta\) |
| Circle through \(O\) | \(r = d\cos\theta\) | \(-d\sin\theta\) | \(-d\cos\theta\) |
| Logarithmic spiral | \(r = a\,e^{k\theta}\) | \(k\,r\) | \(k^2 r\) |
| Straight line \(y = d\) | \(r = d/\sin\theta\) | \(-d\cos\theta/\sin^2\theta\) | \(d\,(1 + \cos^2\theta)/\sin^3\theta\) |
The circle through \(O\), \(r = d\cos\theta\), has diameter \(d\) along the \(x\)-axis. It is the path of a pin in a fixed circular slot that passes through the pivot of a turning arm.
Explore a guided particle
The tangent angle \(\psi\)
The velocity is tangent to the path. Its direction relative to the radial line follows from its components:
\[ \tan\psi = \frac{v_\theta}{v_r} = \frac{r\dot\theta}{(dr/d\theta)\,\dot\theta} \]The \(\dot\theta\) cancels, so the angle depends only on the shape of the path:
Tangent angle
\[ \tan\psi = \frac{r}{dr/d\theta} \]\(\psi\) is measured from the extended radial line (\(\er\)) to the tangent, counter-clockwise (toward \(\et\)) when positive, clockwise when negative.
- On a circle about \(O\), \(dr/d\theta = 0\): \(\psi = 90^\circ\), and the path crosses every radial line at right angles.
- On the Archimedean spiral \(r = b\theta\), \(\tan\psi = \theta\): the spiral crosses the radial lines more and more squarely as it winds out.
- On the logarithmic spiral \(r = ae^{k\theta}\), \(\tan\psi = 1/k\) is the same everywhere. That is why it is called the equiangular spiral, and why it is used for cutters and cams that must meet their work at a constant angle.
In Lesson 8, \(\psi\) gives the direction of the force from a smooth guide: perpendicular to the tangent, at \(\psi\) from \(\et\).
Worked examples
Example 6.1 — A follower on a cardioid cam
A pin follows a groove shaped as the cardioid \(r = 0.5(1 + \cos\theta)\ \text{m}\), pushed by an arm turning at a constant \(\dot\theta = 2\ \text{rad/s}\). Find the pin's velocity and acceleration, and the tangent angle, when \(\theta = 60^\circ\).
Show solution
Path and its derivatives at \(\theta = 60^\circ\):
\[ \begin{aligned} r &= 0.5(1 + \cos 60^\circ) = 0.75\ \text{m} \\ f' &= -0.5\sin 60^\circ \approx -0.4330\ \text{m/rad}, \qquad f'' = -0.5\cos 60^\circ = -0.25\ \text{m/rad}^2 \end{aligned} \]Rates (\(\dot\theta = 2\), \(\ddot\theta = 0\)):
\[ \dot r = f'\dot\theta = (-0.4330)(2) \approx -0.8660\ \text{m/s}, \qquad \ddot r = f''\dot\theta^2 + f'\ddot\theta = (-0.25)(4) + 0 = -1.0\ \text{m/s}^2 \]Velocity:
\[ v_r = -0.8660\ \text{m/s}, \qquad v_\theta = r\dot\theta = (0.75)(2) = 1.5\ \text{m/s}, \qquad v = \sqrt{0.75 + 2.25} = \sqrt{3} \approx 1.732\ \text{m/s} \]Acceleration:
\[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = -1.0 - (0.75)(4) = -4.0\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 0 + 2(-0.8660)(2) \approx -3.464\ \text{m/s}^2 \\ a &= \sqrt{16 + 12} = \sqrt{28} \approx 5.292\ \text{m/s}^2 \end{aligned} \]Tangent angle: \(\tan\psi = r/f' = 0.75/(-0.4330) \approx -1.732\), so \(\psi = -60^\circ\): the tangent is turned \(60^\circ\) clockwise from the extended radial line. The pin is moving inward (\(v_r \lt 0\)), and the Coriolis term \(2\dot r\dot\theta\) is negative.
Example 6.2 — A spiral feeder
A part rides in a spiral groove \(r = 0.05\theta\ \text{m}\) (\(\theta\) in radians), pushed by an arm turning at a constant \(5\ \text{rad/s}\). Find its speed, its acceleration and the tangent angle when \(\theta = 2\pi\).
Show solution
At \(\theta = 2\pi\): \(r = 0.05(2\pi) \approx 0.3142\ \text{m}\), \(f' = 0.05\ \text{m/rad}\), \(f'' = 0\). So \(\dot r = (0.05)(5) = 0.25\ \text{m/s}\) and \(\ddot r = 0\).
\[ \begin{aligned} v_r &= 0.25\ \text{m/s}, \qquad v_\theta = r\dot\theta = (0.3142)(5) \approx 1.571\ \text{m/s}, \qquad v \approx 1.591\ \text{m/s} \\ a_r &= 0 - (0.3142)(5)^2 \approx -7.854\ \text{m/s}^2 \\ a_\theta &= 0 + 2(0.25)(5) = 2.5\ \text{m/s}^2, \qquad a \approx 8.242\ \text{m/s}^2 \end{aligned} \]\(\tan\psi = r/f' = \theta = 2\pi\), so \(\psi \approx 80.96^\circ\): after one turn the groove is already nearly at right angles to the arm. Lesson 8 finds the forces the arm and the groove exert on this part.
Example 6.3 — An arm that speeds up, on a circular slot
A pin moves in a fixed circular slot \(r = 0.6\cos\theta\ \text{m}\), which passes through the arm's pivot \(O\). At the instant when \(\theta = 30^\circ\), the arm turns at \(\dot\theta = 2\ \text{rad/s}\) and is speeding up at \(\ddot\theta = 4\ \text{rad/s}^2\). Find the pin's speed and acceleration.
Show solution
At \(\theta = 30^\circ\): \(r = 0.6\cos 30^\circ \approx 0.5196\ \text{m}\), \(f' = -0.6\sin 30^\circ = -0.3\ \text{m/rad}\), \(f'' = -0.6\cos 30^\circ \approx -0.5196\ \text{m/rad}^2\).
\[ \begin{aligned} \dot r &= f'\dot\theta = (-0.3)(2) = -0.6\ \text{m/s} \\ \ddot r &= f''\dot\theta^2 + f'\ddot\theta = (-0.5196)(4) + (-0.3)(4) \approx -3.278\ \text{m/s}^2 \end{aligned} \]Velocity: \(v_r = -0.6\ \text{m/s}\), \(v_\theta = (0.5196)(2) \approx 1.039\ \text{m/s}\), so \(v = \sqrt{0.36 + 1.08} = 1.2\ \text{m/s}\).
\[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = -3.278 - (0.5196)(4) \approx -5.357\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = (0.5196)(4) + 2(-0.6)(2) \approx 2.078 - 2.4 = -0.3215\ \text{m/s}^2 \\ a &= \sqrt{5.357^2 + 0.3215^2} \approx 5.367\ \text{m/s}^2 \end{aligned} \]Check with path coordinates. The slot is a circle of radius \(0.3\ \text{m}\) about \((0.3, 0)\), and a pin on it turns about that center at twice the arm's rate (an inscribed angle is half the central angle): \(4\ \text{rad/s}\), speeding up at \(8\ \text{rad/s}^2\). So \(v = (0.3)(4) = 1.2\ \text{m/s}\), \(a_n = (0.3)(4)^2 = 4.8\ \text{m/s}^2\), \(a_t = (0.3)(8) = 2.4\ \text{m/s}^2\) and \(a = \sqrt{4.8^2 + 2.4^2} \approx 5.367\ \text{m/s}^2\). Both methods agree.
Check your understanding
Key takeaways
- On a path \(r = f(\theta)\): \(\dot r = f'\dot\theta\) and \(\ddot r = f''\dot\theta^2 + f'\ddot\theta\). Evaluate \(f\), \(f'\), \(f''\) at the given \(\theta\) in radians.
- A constant \(\dot\theta\) does not make \(\ddot r\) zero: \(f''\dot\theta^2\) remains.
- \(\tan\psi = r/(dr/d\theta)\) gives the angle from the extended radial line to the tangent. It depends only on the path.
- Next, Lesson 7 turns these accelerations into forces with Newton's second law.