Question 3 of 6: Crank AB and Slotted Rod CD — Relative Motion in a Rotating Frame

Question 3: Crank AB and Slotted Rod CD — Relative Motion in a Rotating Frame (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. Crank AB pinned at fixed point A, slider B at its tip engages a slot machined in rod CD, which is pinned at fixed point C (the origin). At the instant shown the crank is 30° above the horizontal and rod CD is 30° to the right of the vertical Y-axis (so the angle between the crank and the rod, at their near-common vertex, is exactly $60^\circ - 30^\circ = 30^\circ$, matching the figure).

Find. The velocity and acceleration of slider B relative to the slotted rod CD at this instant.

Approach. First find the absolute velocity and acceleration of the physical point B from the crank alone (a simple rigid-body rotation about A). Then, because B also lies exactly on rod CD’s centreline, resolve those absolute vectors into components along the rod ($\hat t$) and perpendicular to it ($\hat n$): the transport terms belonging to the rotating rod are always perpendicular to its own length, so the along-rod component isolates the sliding motion directly.

1. Geometry: locate B on the rod. With $AB=100$~mm, $AC\approx200$~mm (reconstructed above) and included angle $\angle ABC = 30^\circ$, the law of cosines gives $BC$: $$AC^2 = AB^2 + BC^2 - 2\,AB\cdot BC\cos 30^\circ \;\Longrightarrow\; BC = 280.25 \text{ mm}$$ (the positive root of the resulting quadratic; B sits about 93 % of the way from C to D, consistent with the figure).
2. Absolute velocity and acceleration of the physical point B, treating it as a point fixed to the crank ($\bar r_{B/A} = 100$~mm at 30° above horizontal): $$\bar v_B = \bar\omega_{AB}\times\bar r_{B/A} = (-0.150\,\hat\imath + 0.260\,\hat\jmath)\text{ m/s}, \qquad |\bar v_B| = 0.300 \text{ m/s}$$ $$\bar a_B = \bar\alpha_{AB}\times\bar r_{B/A} - \omega_{AB}^2\bar r_{B/A} = (-0.729\,\hat\imath - 0.537\,\hat\jmath)\text{ m/s}^2$$
3. Resolve along/across the rod. The rod direction (C toward D) is $60^\circ$ from the $+X$ axis (30° from vertical), giving unit vectors $\hat t = (\cos60^\circ,\sin60^\circ)$ along the rod and $\hat n=(-\sin60^\circ,\cos60^\circ)$ across it. The transport velocity of the ROD’s own material point at B, $\bar\omega_{CD}\times \bar r_{B/C}$, is by construction perpendicular to $\bar r_{B/C}$ (which lies along $\hat t$, since B is on the rod’s centreline) — i.e. it has NO component along $\hat t$. So the along-rod component of $\bar v_B$ is exactly the sliding velocity: $$\boxed{v_{\text{rel}} = \bar v_B\cdot\hat t = 0.150 \text{ m/s, directed from C toward D}}$$ The across-rod component gives the rod’s own angular velocity as a byproduct: $v_n = \bar v_B\cdot\hat n = 0.260$ m/s $\Rightarrow \omega_{CD} = v_n/BC = 0.927$ rad/s.
4. Relative acceleration. The full rotating-frame acceleration equation, $\bar a_B = \bar a_{\text{rod pt}} + 2\bar\omega_{CD}\times\bar v_{\text{rel}} + \bar a_{\text{rel}}$, has a Coriolis term ($2\bar\omega_{CD}\times\bar v_{\text{rel}}$, along $\hat n$, since $\bar v_{\text{rel}}$ is along $\hat t$) and an angular-acceleration term ($\bar\alpha_{CD}\times\bar r_{B/C}$, also along $\hat n$) that both drop out of the $\hat t$ component — only the centripetal term $-\omega_{CD}^2 BC$ survives there besides $a_{\text{rel}}$ itself: $$\bar a_B\cdot\hat t = -\omega_{CD}^2\,BC + a_{\text{rel}}$$ $$\boxed{a_{\text{rel}} = \bar a_B\cdot\hat t + \omega_{CD}^2\,BC = -0.829 + (0.927)^2(0.28025) = -0.589 \text{ m/s}^2}$$ The negative sign means the slider is decelerating in its outward (C-to-D) slide at this instant.