Lesson 5 · 35 min

Acceleration

When a robot arm swings round and extends at the same time, its gripper accelerates even though every joint runs at a steady rate. Differentiating the velocity once more shows why, one term at a time.

Learning objectives

Derive \(\avec = (\ddot r - r\dot\theta^2)\,\er + (r\ddot\theta + 2\dot r\dot\theta)\,\et + \ddot z\,\ez\).
Name the centripetal, Coriolis and angular-acceleration terms, and predict when each one is zero.
Connect circular motion in polar coordinates with the path (\(t\), \(n\)) components.
Find the velocity and acceleration of a motion given as \(r(t)\), \(\theta(t)\), \(z(t)\) at a given instant.

Differentiating the velocity

Start from \(\vvec = \dot r\,\er + r\dot\theta\,\et + \dot z\,\ez\) and differentiate each term with the product rule. Every time a unit vector is differentiated, replace \(d\er/dt\) by \(\dot\theta\,\et\) and \(d\et/dt\) by \(-\dot\theta\,\er\):

\[ \begin{aligned} \frac{d}{dt}\left(\dot r\,\er\right) &= \ddot r\,\er + \dot r\dot\theta\,\et \\ \frac{d}{dt}\left(r\dot\theta\,\et\right) &= \dot r\dot\theta\,\et + r\ddot\theta\,\et - r\dot\theta^2\,\er \\ \frac{d}{dt}\left(\dot z\,\ez\right) &= \ddot z\,\ez \end{aligned} \]

The middle line has three terms because \(r\dot\theta\,\et\) is a product of three changing factors: \(r\), \(\dot\theta\) and \(\et\). Add the three lines and collect the \(\er\), \(\et\) and \(\ez\) parts:

Acceleration in cylindrical coordinates

\[ \avec = \left(\ddot r - r\dot\theta^2\right)\colR{\er} + \left(r\ddot\theta + 2\dot r\dot\theta\right)\colT{\et} + \ddot z\,\colZ{\ez} \] \[ a_r = \ddot r - r\dot\theta^2, \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta, \qquad a_z = \ddot z, \qquad a = \sqrt{a_r^2 + a_\theta^2 + a_z^2} \]

Each piece has a name and a physical meaning:

\[ \begin{aligned} a_r &= \ddot r \underbrace{{}- r\dot\theta^2}_{\text{centripetal}} \\ a_\theta &= \underbrace{r\ddot\theta}_{\text{angular accel.}} + \underbrace{2\dot r\dot\theta}_{\text{Coriolis}} \end{aligned} \]

The five acceleration terms
Term	Direction	Name	Nonzero when…
\(\ddot r\)	\(\er\)	Radial (sliding) acceleration	the rate of moving in or out changes
\(-r\dot\theta^2\)	\(-\er\), toward the axis	Centripetal acceleration	the particle turns (\(\dot\theta \ne 0\)) away from the axis
\(r\ddot\theta\)	\(\et\)	Angular-acceleration term	the turning rate changes (\(\ddot\theta \ne 0\))
\(2\dot r\dot\theta\)	\(\et\)	Coriolis acceleration	the particle moves in or out while turning
\(\ddot z\)	\(\ez\)	Axial acceleration	the climbing rate changes

Why the 2 in the Coriolis term? One \(\dot r\dot\theta\) comes from \(\er\) turning while the particle moves outward, so the radial velocity \(\dot r\,\er\) keeps changing direction. The other comes from \(v_\theta = r\dot\theta\) growing because \(r\) grows. Both point along \(\et\), so they add.

Circular motion as a special case

On a circle of radius \(r\) about \(O\), \(\dot r = \ddot r = 0\). What is left is

\[ a_r = -r\dot\theta^2 = -\frac{v^2}{r}, \qquad a_\theta = r\ddot\theta = \dot v \qquad (v = r\dot\theta) \]

These are the path components you know: the normal acceleration \(a_n = v^2/\rho\) points to the center (here \(-\er\), with \(\rho = r\)), and the tangential acceleration \(a_t = \dot v\) points along the path (here \(\et\) for counter-clockwise travel). For a circle centered on \(O\), polar and path components are the same thing. For any other path they differ, and the polar form is usually easier when the motion is given through \(r\) and \(\theta\).

See the terms in motion

The simulator moves a particle with \(r(t)\), \(\theta(t)\) and \(z(t)\) that you choose, and draws \(\vvec\) and \(\avec\) at the particle together with their components along \(\er\), \(\et\) and \(\ez\). The live numbers list every term of the acceleration as it plays.

Figure 5.1 Choose a motion, press Play, or drag the time slider. The green arrow is the velocity \(\vvec\), tangent to the path; the red arrow is the acceleration \(\avec\). The thinner arrows are their components along \(\er\) (orange), \(\et\) (violet) and \(\ez\) (blue), drawn tip to tail so that each chain adds up to its vector. In Robot arm, \(\avec\) points toward the column but tilts forward, in the direction of rotation: that sideways part is the Coriolis term \(2\dot r\dot\theta\). In Custom, set \(\dot r = 0\) and the Coriolis term disappears.

Worked examples

Example 5.1 — A robot arm that extends while it slews

The robot of Example 4.1 rotates at a constant \(\dot\theta = 2\ \text{rad/s}\), extends at a constant \(\dot r = 0.5\ \text{m/s}\) and rises at a constant \(\dot z = 0.3\ \text{m/s}\). Find the acceleration of the gripper at the instant when \(r = 1\ \text{m}\).

Show solution

Every rate is constant, so \(\ddot r = \ddot\theta = \ddot z = 0\).

\[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 0 - (1)(2)^2 = -4\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 0 + 2(0.5)(2) = 2\ \text{m/s}^2 \\ a_z &= \ddot z = 0 \end{aligned} \] \[ \avec = (-4\,\er + 2\,\et)\ \text{m/s}^2, \qquad a = \sqrt{(-4)^2 + 2^2} = 2\sqrt{5} \approx 4.472\ \text{m/s}^2 \]

Interpret. Although every joint runs at a constant rate, the gripper accelerates at \(4.472\ \text{m/s}^2\). The \(-4\,\er\) part is centripetal (toward the column); the \(2\,\et\) part is entirely Coriolis, because \(\ddot\theta = 0\). Together they tilt \(\avec\) by \(\tan^{-1}(2/4) \approx 26.57^\circ\) from \(-\er\) toward the direction of rotation. In Figure 5.1, choose Robot arm and press Jump to Example 5.1.

Example 5.2 — Riding a spiral slide

The chute of a spiral slide is a helix of radius \(r = 2\ \text{m}\) around a vertical pole. A child goes round the pole at a constant \(\dot\theta = 0.8\ \text{rad/s}\) and descends at a constant \(\dot z = -0.5\ \text{m/s}\). Find the child's velocity, speed and acceleration.

Show solution

The radius is fixed, so \(\dot r = \ddot r = 0\); the rates \(\dot\theta\) and \(\dot z\) are constant, so \(\ddot\theta = \ddot z = 0\).

\[ \vvec = r\dot\theta\,\et + \dot z\,\ez = (1.6\,\et - 0.5\,\ez)\ \text{m/s}, \qquad v = \sqrt{1.6^2 + 0.5^2} \approx 1.676\ \text{m/s} \] \[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 0 - (2)(0.8)^2 = -1.28\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 0, \qquad a_z = \ddot z = 0 \end{aligned} \]

Interpret. The speed is constant, yet the child accelerates, because the direction of \(\vvec\) keeps changing. The acceleration is purely centripetal: \(1.28\ \text{m/s}^2\) (about \(0.13g\)), horizontal and pointing straight at the pole. The chute slopes down at \(\tan^{-1}(0.5/1.6) \approx 17.35^\circ\), and one turn takes \(2\pi/0.8 \approx 7.854\ \text{s}\), during which the child drops \(3.927\ \text{m}\). Lesson 7 finds the force the chute exerts.

Motion given as functions of time

Many problems give the motion as \(r(t)\) and \(\theta(t)\) (and \(z(t)\)). The procedure is always the same:

Differentiate each coordinate twice: \(r, \dot r, \ddot r\) and \(\theta, \dot\theta, \ddot\theta\) (and \(\dot z, \ddot z\)).
Evaluate all of them at the instant asked for.
Substitute into \(v_r, v_\theta\) and \(a_r, a_\theta\), and combine for the magnitudes.

Example 5.3 — A collar on an accelerating rod

A rod turns in a horizontal plane with \(\theta = 0.5t^2\ \text{rad}\), and a collar slides along it with \(r = (1 + 0.2t^2)\ \text{m}\), where \(t\) is in seconds. Find the collar's velocity and acceleration at \(t = 2\ \text{s}\).

Show solution

Derivatives at \(t = 2\ \text{s}\):

\[ \begin{aligned} r &= 1 + 0.2t^2 = 1.8\ \text{m}, & \dot r &= 0.4t = 0.8\ \text{m/s}, & \ddot r &= 0.4\ \text{m/s}^2 \\ \theta &= 0.5t^2 = 2\ \text{rad}, & \dot\theta &= t = 2\ \text{rad/s}, & \ddot\theta &= 1\ \text{rad/s}^2 \end{aligned} \]

Velocity:

\[ v_r = \dot r = 0.8\ \text{m/s}, \quad v_\theta = r\dot\theta = (1.8)(2) = 3.6\ \text{m/s}, \quad v = \sqrt{0.8^2 + 3.6^2} \approx 3.688\ \text{m/s} \]

Acceleration:

\[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 0.4 - (1.8)(2)^2 = -6.8\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = (1.8)(1) + 2(0.8)(2) = 1.8 + 3.2 = 5.0\ \text{m/s}^2 \\ a &= \sqrt{(-6.8)^2 + 5.0^2} \approx 8.440\ \text{m/s}^2 \end{aligned} \]

The Coriolis term (\(3.2\ \text{m/s}^2\)) is larger than the angular-acceleration term (\(1.8\ \text{m/s}^2\)). Figure 5.2 shows this motion.

Figure 5.2 Example 5.3 seen from above: \(r = (1 + 0.2t^2)\ \text{m}\) and \(\theta = 0.5t^2\ \text{rad}\), for \(0 \le t \le 2.5\ \text{s}\). The rod (gray) turns faster and faster while the collar slides out. The red acceleration arrow is split into \(a_r\,\er\) (orange) and \(a_\theta\,\et\) (violet); the live numbers split \(a_r\) and \(a_\theta\) into their terms. Press Jump to t = 2 s to check the example.

Check your understanding

Key takeaways

\(\avec = (\ddot r - r\dot\theta^2)\,\er + (r\ddot\theta + 2\dot r\dot\theta)\,\et + \ddot z\,\ez\).
\(-r\dot\theta^2\) is centripetal (toward the axis), \(2\dot r\dot\theta\) is Coriolis (moving in or out while turning) and \(r\ddot\theta\) comes from a changing turning rate.
Constant joint rates do not mean zero acceleration: a slewing, extending arm still accelerates.
On a circle about \(O\): \(a_r = -v^2/r\) and \(a_\theta = \dot v\), the normal and tangential accelerations.
For motion given as functions of time: differentiate twice, evaluate at the instant, substitute.
Next, Lesson 6 handles particles guided along a path \(r = f(\theta)\).