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

Question 5: Inverted crank-slider — 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. $r_2=0.20$ m, $r_1=G_2G_4=0.50$ m, $\dot\theta_2=\omega_2=2000$ rpm $=209.44$ rad/s (const, $\ddot\theta_2=0$). Only link 3 has mass: $m_3=0.25$ kg at $G_3=A$, $I_{G3}=0.125$ kg·m$^2$.

Find. Kinematics $s,\theta_4,\omega_4,\alpha_4(\theta_2)$; shaking force $F_s(\theta_2,\dot\theta_2)$ at $45^\circ,90^\circ$; a balancing scheme and its residual.

Approach. Close the position loop for $s(\theta_2)$ and $\theta_4(\theta_2)$, differentiate for $\omega_4,\alpha_4$; then note the coupler mass centre coincides with the crank pin, so its acceleration is pure centripetal — giving a constant-magnitude shaking force that a single crank counterweight cancels exactly.

1. Position. With $G_2=(0,0)$, $G_4=(r_1,0)$, $A=(r_2\cos\theta_2,\,r_2\sin\theta_2)$: slider length and follower angle are $$s=|G_4A|=\sqrt{r_1^2+r_2^2-2r_1r_2\cos\theta_2},\qquad \theta_4=\operatorname{atan2}(r_2\sin\theta_2,\ r_2\cos\theta_2-r_1)$$ At $\theta_2=45^\circ$: $s=0.386$ m, $\theta_4=158.5^\circ$; at $90^\circ$: $s=0.539$ m, $\theta_4=158.2^\circ$.
2. Velocities. Differentiating the loop, $\dot s=\dfrac{r_1r_2\dot\theta_2\sin\theta_2}{s}$ and $\omega_4=\dfrac{r_2\dot\theta_2\,(r_2-r_1\cos\theta_2)}{s^2}$. At $45^\circ$: $\dot s=38.4$ m/s, $\omega_4=-43.3$ rad/s; at $90^\circ$: $\dot s=38.9$ m/s, $\omega_4=+28.9$ rad/s.
3. Accelerations. A second differentiation ($\ddot\theta_2=0$) gives $\alpha_4$ and $\ddot s$. At $45^\circ$: $\alpha_4=2.95\times10^{4}$ rad/s$^2$, $\ddot s=4.22\times10^{3}$ m/s$^2$; at $90^\circ$: $\alpha_4=1.10\times10^{4}$ rad/s$^2$, $\ddot s=-2.81\times10^{3}$ m/s$^2$. The coupler mass centre is the crank pin, so its acceleration is simply centripetal, $\mathbf a_{G3}=-\omega_2^2 r_2\,\hat r$ (toward $G_2$), of constant magnitude $\omega_2^2 r_2=8773$ m/s$^2$.
4. Shaking force. Only link 3 has mass, so $\mathbf F_s=-m_3\mathbf a_{G3}=m_3\omega_2^2 r_2\,\hat r$ — it points radially outward along the crank and has constant magnitude: $$|F_s|=m_3\,\dot\theta_2^{\,2}\,r_2=(0.25)(209.44)^2(0.20)=\boxed{2193\ \text{N}}$$ Therefore $|F_s|=2193$ N at both $\theta_2=45^\circ$ and $90^\circ$ (only its direction rotates with the crank).
5. Balancing scheme. Because $F_s$ is exactly that of a rotating unbalance $m_3$ at radius $r_2$, one counterweight on the crank (opposite $A$) with $$m_{cw}r_{cw}=m_3 r_2=(0.25)(0.20)=0.050\ \text{kg}\cdot\text{m}$$ cancels it completely: $\boxed{F_s=0}$ at every $\theta_2$ (including $45^\circ,90^\circ$). The scheme is fully effective for the shaking force; a small shaking couple from the coupler inertia $I_{G3}\alpha_3$ remains, which would need a second (moment) balance if required.