Question 5 of 7: Equations of motion of the PR manipulator (20 marks)

Question 5: Equations of motion of the PR manipulator (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. A two-degree-of-freedom arm: link 1 slides along $\hat{Z}_{0}$ and link 2 turns on link 1. The base triad is drawn with $\hat{X}_{0}$ vertical, so the slide is horizontal. Both inertia tensors are diagonal in their own link frames and the mass centres are placed by the two given vectors.

Find. The two equations of motion, i.e. the actuator force $\tau_{1}$ on the slide and the actuator torque $\tau_{2}$ at the revolute joint, as functions of $q,\dot q,\ddot q$.

[Figure not reproduced: Figure 5.1 — the PR manipulator of Figure 2, redrawn with the frames used below. Link 1 (mass $m_{1}$) slides horizontally along $\hat{Z}_{0}$ through two fixed bearings; its mass centre $C_{1}$ lies a distance $l_{1}$ behind the joint. Link 2 (mass $m_{2}$) turns about $\hat{Y}_{1}$ with its . See the official exam paper.]

Approach. Use the Lagrangian formulation: write the two mass-centre positions in the base frame, differentiate for their velocities, assemble kinetic and potential energy, and apply $\tau_{i}=\frac{d}{dt}\frac{\partial L}{\partial\dot q_{i}} -\frac{\partial L}{\partial q_{i}}$. For a two-joint arm this is far quicker than a Newton-Euler outward-inward recursion and it exposes the structure of the answer directly.

1. Fix the frames and the generalized coordinates. Take $q=[q_{1}\;\;q_{2}]^{T}$ with $q_{1}=d_{1}$ the slide displacement and $q_{2}=\theta_{2}$ the joint angle. Frame $\{1\}$ rides on the slide, parallel to $\{0\}$ at all times, with its origin at $$^{0}P_{1ORG}=\begin{bmatrix}0\\0\\q_{1}\end{bmatrix},\qquad{}^{0}_{1}R=I,$$ and frame $\{2\}$ shares that origin but is rolled about $\hat{Y}_{1}$, so $^{0}_{2}R=R_{Y}(q_{2})$. Reading these two facts off Figure 2 correctly is worth more than any subsequent algebra.
2. Locate the two mass centres in the base frame. Carrying the given vectors across, $$^{0}P_{C1}=\begin{bmatrix}0\\0\\q_{1}-l_{1}\end{bmatrix},\qquad{}^{0}P_{C2}={}^{0}P_{1ORG}+{}^{0}_{2}R\,{}^{2}P_{C2} =\begin{bmatrix}0\\0\\q_{1}\end{bmatrix}.$$ The second result is the key structural fact of this problem: $^{2}P_{C2}$ is the zero vector, so the mass centre of link 2 sits exactly on the joint axis and rotating the link does not move it.
3. Differentiate for the velocities. Both mass centres translate with the slide alone, $$^{0}v_{C1}={}^{0}v_{C2}=\begin{bmatrix}0\\0\\\dot q_{1}\end{bmatrix},$$ while the angular velocities are $\omega_{1}=0$ (link 1 only translates) and $^{2}\omega_{2}=[0\;\;\dot q_{2}\;\;0]^{T}$, written in frame $\{2\}$ where the inertia tensor is diagonal.
4. Assemble the kinetic energy. Summing translational and rotational parts, $$k=\tfrac12 m_{1}\dot q_{1}^{2}+\tfrac12 m_{2}\dot q_{1}^{2} +\tfrac12\,{}^{2}\omega_{2}^{T}\,{}^{C_2}I_{2}\,{}^{2}\omega_{2} =\tfrac12(m_{1}+m_{2})\dot q_{1}^{2}+\tfrac12 I_{yy2}\dot q_{2}^{2},$$ because the diagonal tensor picks out only $I_{yy2}$ when $\omega$ lies along $\hat{Y}_{2}$. Hence the mass matrix is $$\boxed{M(q)=\begin{bmatrix}m_{1}+m_{2}&0\\0&I_{yy2}\end{bmatrix},}$$ constant and diagonal - the two axes are completely decoupled.
5. Assemble the potential energy. Height is measured along $\hat{X}_{0}$, and both mass centres have zero $\hat{X}_{0}$ component for every $q$, so $$u=g\left(m_{1}\,{}^{0}P_{C1}\cdot\hat{X}_{0}+m_{2}\,{}^{0}P_{C2}\cdot\hat{X}_{0}\right) =0\quad\text{(constant)},\qquad G(q)=\frac{\partial u}{\partial q}=\begin{bmatrix}0\\0\end{bmatrix}.$$ Gravity acts perpendicular to every direction in which the mechanism can move, so it does no work and produces no generalized force.
6. Collect the velocity-product terms. The Coriolis and centrifugal vector is built from derivatives of the mass matrix, $$V(q,\dot q)_{i}=\sum_{j,k}\left(\frac{\partial M_{ij}}{\partial q_{k}} -\tfrac12\frac{\partial M_{jk}}{\partial q_{i}}\right)\dot q_{j}\dot q_{k},$$ and every partial derivative of a constant matrix vanishes, so $V(q,\dot q)=0$. There is no centrifugal stiffening of the slide and no Coriolis coupling into the joint.
7. State the equations of motion. With $\tau=M(q)\ddot q+V(q,\dot q)+G(q)$, $$\boxed{\tau_{1}=(m_{1}+m_{2})\,\ddot q_{1},\qquad \tau_{2}=I_{yy2}\,\ddot q_{2}.}$$ The slide actuator sees the total mass of everything it carries and nothing else; the rotary actuator sees only the second link's moment of inertia about its own rotation axis. The inertia components $I_{xx1},I_{yy1},I_{zz1},I_{xx2},I_{zz2}$ and the length $l_{1}$ appear nowhere, which is a legitimate result rather than an omission: link 1 never rotates, and the other two axes of link 2 are never spun about.

It is worth saying explicitly why the answer is this clean, because an examiner is testing whether the candidate can see it. Two independent circumstances collapse the coupling. First, the slide is horizontal while gravity is vertical, so no term $m g \sin q_{2}$ can appear. Second, the mass centre of link 2 lies on its own rotation axis, so the rotation contributes no translational kinetic energy and the off-diagonal entry $M_{12}$ vanishes. Move that mass centre off the axis by even a small offset $l_{2}$ and the answer immediately acquires $M_{12}=-m_{2}l_{2}\sin q_{2}$, a centrifugal term $-m_{2}l_{2}\cos q_{2}\,\dot q_{2}^{2}$ in the slide equation and a gravity torque $m_{2}gl_{2}\cos q_{2}$ at the joint.