Question 3 of 6: Collar on a Rotating Semicircular Rod — Absolute Velocity and Acceleration via a Rotating Frame

Question 3: Collar on a Rotating Semicircular Rod — Absolute Velocity and Acceleration via 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. Semicircular rod $AB$, radius $r=0.2$ m, pivoted at the fixed point $A$ (one end of the diameter; $B$ is the other end, and the arc's centre $O$ is the midpoint of $AB$, also at distance $r$ from every point on the arc). At the instant shown the collar $P$ is at the lowest point of the arc, directly below $O$. The rod rotates about $A$ with $\omega_{AB}=0.8$ rad/s (CCW) and $\alpha_{AB}=0.5$ rad/s² (CW); the collar's speed relative to the rod is held constant at $v_{P/AB}=0.12$ m/s, directed tangent to the arc (horizontal at the lowest point, as drawn).

Find. The absolute velocity $\vec v_P$ and absolute acceleration $\vec a_P$ of the collar at the instant shown.

Approach. Attach the rotating reference frame to the rigid rod $AB$ (angular velocity $\vec\Omega=\omega_{AB}\hat k$, angular acceleration $\dot{\vec\Omega}=\alpha_{AB}\hat k$, both about the fixed point $A$); the collar's absolute motion is then the frame's rigid-body motion of the coincident point plus the collar's motion relative to the frame — and because that relative path is itself the circular arc of radius $r$ centred at $O$ (not at $A$), the relative motion contributes its own centripetal term toward $O$ even though the frame rotates about $A$.

1. Position and the two given rates. With $A$ at the origin and $x$ along $AB$, $O=(r,0)$ and the lowest point $P=(r,-r)$, so $\vec r_{P/A}=(r,-r)=(0.2,-0.2)$ m. Taking CCW as $+\hat k$: $\vec\Omega=0.8\hat k$ rad/s, $\dot{\vec\Omega}=-0.5\hat k$ rad/s² (CW $\Rightarrow$ negative), and $\vec v_{P/AB}=0.12\hat\imath$ m/s (tangent to the arc at its lowest point is horizontal, matching the arrow shown).
2. Absolute velocity. Since $A$ is a fixed pivot ($\vec v_A=0$), $$\vec v_P=\vec\Omega\times\vec r_{P/A}+\vec v_{P/AB} = (0.8\hat k)\times(0.2\hat\imath-0.2\hat\jmath) + 0.12\hat\imath = (0.16\hat\imath+0.16\hat\jmath)+0.12\hat\imath,$$ $$\boxed{\vec v_P = (0.28\hat\imath+0.16\hat\jmath)\ \text{m/s}}, \qquad |\vec v_P|=\sqrt{0.28^2+0.16^2}=0.3225\ \text{m/s},\ \ 29.7^\circ\ \text{above }+x.$$
3. Absolute acceleration — the extra term the geometry hides. The full rotating-frame formula is $$\vec a_P=\dot{\vec\Omega}\times\vec r_{P/A}-\Omega^2\vec r_{P/A}+2\vec\Omega\times\vec v_{P/AB}+\vec a_{P/AB}.$$ The relative acceleration $\vec a_{P/AB}$ is not zero even though the collar's relative speed is constant: as $P$ slides along the arc, the direction of $\vec v_{P/AB}$ continuously turns to stay tangent to a circle of radius $r$ centred at $O$, so $\vec a_{P/AB}$ has no tangential part but a full centripetal part toward $O$, $$\vec a_{P/AB}=\frac{v_{P/AB}^2}{r}\,\hat u_{P\to O} = \frac{0.12^2}{0.2}\,\hat\jmath = 0.072\hat\jmath\ \text{m/s}^2\qquad(\hat u_{P\to O}=+\hat\jmath\text{ at the lowest point}).$$ The other three terms, with $\vec r_{P/A}=(0.2,-0.2)$: $$\dot{\vec\Omega}\times\vec r_{P/A}=(-0.5\hat k)\times(0.2\hat\imath-0.2\hat\jmath)=(-0.1,-0.1),\qquad -\Omega^2\vec r_{P/A}=-0.64(0.2,-0.2)=(-0.128,0.128),$$ $$2\vec\Omega\times\vec v_{P/AB}=2(0.8\hat k)\times(0.12\hat\imath)=(0,0.192).$$
4. Sum the four terms. $$\vec a_P=(-0.1,-0.1)+(-0.128,0.128)+(0,0.192)+(0,0.072),$$ $$\boxed{\vec a_P=(-0.228\hat\imath+0.292\hat\jmath)\ \text{m/s}^2}, \qquad |\vec a_P|=\sqrt{0.228^2+0.292^2}=0.3705\ \text{m/s}^2,\ \ 128.0^\circ\ \text{from }+x.$$