Angular and linear acceleration of a single link: 22-Mec-B12 Robotics, December 2019, Q5

Question 5: Angular and linear acceleration of a single link (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. One rigid link on one revolute joint, with its mass properties expressed about the mass centre and its joint angle prescribed as an explicit function of time.

Given data
Quantity	Symbol	Value
Total mass	$m$	$1$
Mass centre, in frame {1}	$^{1}P_{C}$	$[2,\,0,\,0]^{T}$
Inertia tensor about the mass centre	$^{C}I_{1}$	$\mathrm{diag}(1,\,2,\,2)$
Prescribed joint angle	$\theta_{1}(t)$	$bt+ct^{2}$ (radians)
Joint axis	$\hat Z_{1}$	coincident with $\hat Z_{0}$

Find. $^{1}\dot\omega_{1}(t)$ and $^{1}\dot v_{C}(t)$, both referred to frame {1}.

[Figure not reproduced: Figure 2 redrawn. Frame {1} is fixed in the link with $X_1$ along it and $Z_1$ along the joint axis, so the mass centre sits at a constant $[2,0,0]$ in {1} however the link moves. See the official exam paper.]

Approach. Differentiate the prescribed joint angle twice, propagate angular velocity and angular acceleration outward from the fixed base, and then use the rigid-body acceleration formula for a point fixed in a rotating frame.

Differentiate the joint trajectory. With $\theta_{1}(t)=bt+ct^{2}$, $$\dot\theta_{1}(t)=b+2ct,\qquad \ddot\theta_{1}(t)=2c\ \ \text{(constant)}.$$
Propagate the angular velocity. The base is fixed, $^{0}\omega_{0}=0$, and joint 1 is revolute about $\hat Z_{1}$, so $$^{1}\omega_{1}={}^{1}_{0}R\,{}^{0}\omega_{0}+\dot\theta_{1}\,\hat Z_{1}=\begin{bmatrix}0\\ 0\\ b+2ct\end{bmatrix}.$$
Propagate the angular acceleration. Differentiating, and noting that $\hat Z_{1}$ is constant in frame {1} so the $\omega\times\dot\theta\hat Z$ term vanishes for the first link, $$\boxed{^{1}\dot\omega_{1}=\ddot\theta_{1}\hat Z_{1}=\begin{bmatrix}0\\ 0\\ 2c\end{bmatrix}\ \text{rad/s}^{2}}$$ The angular acceleration is constant: the quadratic joint trajectory is a constant-acceleration profile.
Set up the linear acceleration. The origin of frame {1} sits on the joint axis and never translates, so $^{1}\dot v_{1}=0$ (gravity is excluded here; see the note below). For a point fixed in the rotating link, $$^{1}\dot v_{C}={}^{1}\dot\omega_{1}\times{}^{1}P_{C}+{}^{1}\omega_{1}\times\left({}^{1}\omega_{1}\times{}^{1}P_{C}\right).$$
Evaluate the tangential term. $$^{1}\dot\omega_{1}\times{}^{1}P_{C}=\begin{bmatrix}0\\ 0\\ 2c\end{bmatrix}\times\begin{bmatrix}2\\ 0\\ 0\end{bmatrix}=\begin{bmatrix}0\\ 4c\\ 0\end{bmatrix},$$ which is simply $r\ddot\theta$ with $r=2$, directed along $\hat Y_{1}$.
Evaluate the centripetal term. First $^{1}\omega_{1}\times{}^{1}P_{C}=[0,\,2\dot\theta_{1},\,0]^{T}$, and crossing again with $^{1}\omega_{1}$ gives $$^{1}\omega_{1}\times\left({}^{1}\omega_{1}\times{}^{1}P_{C}\right)=\begin{bmatrix}-2\dot\theta_{1}^{2}\\ 0\\ 0\end{bmatrix},$$ i.e. $r\dot\theta^{2}$ pointing back along $-\hat X_{1}$ toward the axis, as a centripetal term must.
Combine. Adding the two contributions and substituting $\dot\theta_{1}=b+2ct$, $$\boxed{^{1}\dot v_{C}(t)=\begin{bmatrix}-2\,(b+2ct)^{2}\\ 4c\\ 0\end{bmatrix}}$$ The tangential component is constant while the centripetal component grows as the square of time; the resultant magnitude is $\sqrt{4(b+2ct)^{4}+16c^{2}}$.
Numerical illustration and the wrench that follows. Taking $b=0.35\ \text{rad/s}$ and $c=0.60\ \text{rad/s}^{2}$, at $t=2\ \text{s}$ the joint rate is $2.75\ \text{rad/s}$ and $^{1}\dot v_{C}=[-15.125,\,2.400,\,0]^{T}$. The Newton–Euler wrench about the mass centre is then $F=m\,{}^{1}\dot v_{C}=[-15.125,\,2.400,\,0]^{T}$ and $N={}^{C}I_{1}{}^{1}\dot\omega_{1}+{}^{1}\omega_{1}\times{}^{C}I_{1}{}^{1}\omega_{1}=[0,\,0,\,2.400]^{T}$, so the joint torque is $\tau_{1}=N_{z}+\left({}^{1}P_{C}\times F\right)_{z}=2.4+4.8=7.2\ \text{N}\cdot\text{m}$. The $\omega\times I\omega$ term vanishes here because $\omega$ is an eigenvector of the diagonal inertia tensor.

Check: the phrase “from rest at $t=0$” is not consistent with a non-zero $b$. The prescribed trajectory gives $\dot\theta_{1}(0)=b$, so the link is at rest at $t=0$ only if $b=0$. The general answer above is kept in terms of both constants, which is what “as a function of $t$” asks for and what the marking scheme of this recurring Craig exercise expects. If the “from rest” clause is taken literally, set $b=0$ and the results reduce to $^{1}\dot\omega_{1}=[0,0,2c]^{T}$ and $^{1}\dot v_{C}=[-8c^{2}t^{2},\,4c,\,0]^{T}$. Both readings are stated here in line with the exam’s instruction to record any assumption made.

Check: gravity is excluded. The question asks for the acceleration of the mass centre, so the kinematic result above is the answer. If the standard Newton–Euler trick of loading the base with $^{0}\dot v_{0}=g\hat Y_{0}$ is used to fold gravity into the same recursion, add $^{1}_{0}R\,g\hat Y_{0}$ to $^{1}\dot v_{C}$; that changes the joint torque but not the angular acceleration.

Final results
Quantity	Symbol	Value
Joint rate	$\dot\theta_{1}(t)$	$b+2ct$
Joint acceleration	$\ddot\theta_{1}(t)$	$2c$ (constant)
Angular velocity of the link	$^{1}\omega_{1}$	$[0,\,0,\,b+2ct]^{T}$
Angular acceleration of the link	$^{1}\dot\omega_{1}$	$[0,\,0,\,2c]^{T}$
Linear acceleration of the mass centre	$^{1}\dot v_{C}$	$[-2(b+2ct)^{2},\;4c,\;0]^{T}$
Tangential component	—	$4c$ along $\hat Y_{1}$, constant
Centripetal component	—	$2(b+2ct)^{2}$ toward the joint axis
Illustrative value at $t=2\ \text{s}$ ($b=0.35$, $c=0.60$)	$^{1}\dot v_{C}$	$[-15.125,\;2.400,\;0]^{T}$

Concept & Theory

Topic. Rigid-body kinematics of a single rotating link; the outward Newton–Euler recursion.

Key relations. $^{i+1}\omega_{i+1}={}^{i+1}_{\;\;i}R\,{}^{i}\omega_{i}+\dot\theta_{i+1}\hat Z_{i+1}$; $^{i+1}\dot\omega_{i+1}={}^{i+1}_{\;\;i}R\,{}^{i}\dot\omega_{i}+{}^{i+1}_{\;\;i}R\,{}^{i}\omega_{i}\times\dot\theta_{i+1}\hat Z_{i+1}+\ddot\theta_{i+1}\hat Z_{i+1}$; $\dot v_{C}=\dot\omega\times P_{C}+\omega\times(\omega\times P_{C})$; $F=m\dot v_{C}$ and $N=I\dot\omega+\omega\times I\omega$.

Why this works. Because frame {1} is rigidly attached to the link, the mass centre has fixed coordinates in it and only the frame’s rotation produces acceleration. The two surviving terms then have direct physical readings: $\dot\omega\times P_{C}$ is the tangential acceleration $r\ddot\theta$, and $\omega\times(\omega\times P_{C})$ is the centripetal acceleration $r\dot\theta^{2}$ pointing at the axis. Writing the answer in frame {1} rather than frame {0} is what keeps it compact — in {0} the same vector would carry sines and cosines of $\theta_{1}(t)$.

Common pitfall. Dropping the centripetal term because the joint acceleration is constant — it depends on $\dot\theta^{2}$, not on $\ddot\theta$, and it dominates at speed. Also, the inertia tensor is given about the mass centre; using it about the joint axis without the parallel-axis shift over-states the torque.

Source. Craig, Introduction to Robotics, 4th ed., §6.4–6.5 (iterative Newton–Euler dynamic formulation, Example 6.5); Spong et al., 2nd ed., ch. 7.

Question 5 of 7: Angular and linear acceleration of a single link

Question 5: Angular and linear acceleration of a single link (20 marks)