Question 5 of 7: Inverted crank-slider — kinematics, shaking force, balancing

Question 5: Inverted crank-slider — kinematics, shaking force, balancing (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. $r_2=0.25\ \text{m}$, $r_1=0.50\ \text{m}$, $\omega_2=1750\ \text{rpm}=183.3\ \text{rad/s}$, $\alpha_2=0$; coupler (slider 3) $m_3=0.3\ \text{kg}$ with $G_3$ at the crank pin $A$, $I_{G3}=0.15\ \text{kg}\cdot\text{m}^2$; crank 2 and follower 4 massless.

Find. (i) coupler/follower kinematics vs $\theta_2$; (ii) shaking force at $\theta_2=0,45^\circ,90^\circ$; (iii) a balancing scheme.

Approach. Write the loop $G_2A$ + (block slip along 4) closing on $G_4$: the pin $A$ lies on follower 4, so $\theta_4$ and the slip $b=|G_4A|$ come straight from $A$’s position. Since $G_3$ sits at $A$ and the crank turns at constant speed, $\mathbf a_{G3}=\mathbf a_A$ is purely centripetal — that makes the shaking force constant.

(i) Position, velocity, acceleration relations

1. Position. With $G_2$ at the origin and $G_4=(r_1,0)$, the crank pin is $A=(r_2\cos\theta_2,\ r_2\sin\theta_2)$. The follower passes through $G_4$ and $A$, so $$\theta_4=\operatorname{atan2}\!\big(r_2\sin\theta_2,\ r_2\cos\theta_2-r_1\big),\qquad b=|G_4A|=\sqrt{r_1^2+r_2^2-2r_1r_2\cos\theta_2}.$$
2. Velocity. $\mathbf v_A=\omega_2 r_2$ ($\perp$ crank). Resolving along and across follower 4 gives the follower angular speed and the block slip speed $$\omega_4=\frac{\mathbf v_A\cdot\hat{\mathbf n}_4}{b},\qquad \dot b=\mathbf v_A\cdot\hat{\mathbf e}_4,$$ with $\hat{\mathbf e}_4$ along $G_4A$ and $\hat{\mathbf n}_4\perp$ to it. The block (coupler 3) shares the follower’s rotation, $\omega_3=\omega_4$.
3. Acceleration. $\alpha_2=0\Rightarrow \mathbf a_A=-\omega_2^2\mathbf r_{A/G_2}$ (centripetal, toward $G_2$). Projecting $\mathbf a_A=\mathbf a_{\text{coincident on }4}+2\boldsymbol\omega_4\times\dot{\mathbf b}+\alpha_4\times\mathbf r+\dots$ yields $\alpha_4=\alpha_3$ and the slip acceleration $\ddot b$ (Coriolis term $2\omega_4\dot b$ retained). Numerically:

   $\theta_2$	$b=|G_4A|$ (m)	$\theta_4$	$\omega_4=\omega_3$ (rad/s)
   $0^\circ$	0.250	$180^\circ$	$-183.3$
   $45^\circ$	0.368	$151^\circ$	$-35.0$
   $90^\circ$	0.559	$153^\circ$	$+36.7$

(ii) Shaking force

1. Only the coupler has mass, and its centre is at the crank pin. Because $G_3\equiv A$ and $\alpha_2=0$, the mass-centre acceleration is purely centripetal with constant magnitude: $$a_{G3}=\omega_2^2 r_2=(183.3)^2(0.25)=8.40\times10^{3}\ \text{m/s}^2,\quad \text{directed }A\!\to\!G_2.$$
2. Shaking force. The net inertia force on the frame is $$\mathbf F_s=m_3\,\mathbf a_{G3}=-m_3\,\omega_2^2 r_2\,\hat{\mathbf r}_{A/G_2},\qquad |\mathbf F_s|=m_3\,\omega_2^2 r_2=0.3(183.3)^2(0.25)=\boxed{2.52\times10^{3}\ \text{N}}.$$ Its magnitude is independent of $\theta_2$, so $$|\mathbf F_s|_{0^\circ}=|\mathbf F_s|_{45^\circ}=|\mathbf F_s|_{90^\circ}=2519\ \text{N};$$ only its direction changes — it rotates with the crank (a pure rotating force). (The inertia couple $I_{G3}\alpha_3$ produces a shaking moment, not a force.)

(iii) Balancing scheme

Since $\mathbf F_s$ is a constant-magnitude force rotating with the crank, it is completely cancelled by a single rotating counterweight on the crank, placed diametrically opposite $A$, sized so its centrifugal force equals $\mathbf F_s$: $$m_{cw}\,r_{cw}\,\omega_2^2=m_3\,r_2\,\omega_2^2\;\Rightarrow\;\boxed{m_{cw}\,r_{cw}=m_3\,r_2=0.075\ \text{kg}\cdot\text{m}}$$ — e.g. $m_{cw}=0.75\ \text{kg}$ at $r_{cw}=0.10\ \text{m}$. Reason: the shaking force is entirely the centripetal reaction of the coupler mass concentrated at the rotating pin $A$; an equal, opposite rotating mass produces an equal, opposite centrifugal force at every instant, so the resultant on the frame is zero. The only residual is the small inertia couple $I_{G3}\alpha_3$, which a counter-rotating balance shaft (or an added moment counterweight) could further reduce if required.