# Example 2.9: Bicycle suspension

The mechanism in the photo (dimensions in centimeters) shows the rear suspension of a bicycle (adapted from the January 2023 exam). Find:

1. The number of degrees of freedom.
2. The velocity of point $A$, using the analytical method.
3. The angular accelerations of all links and the acceleration of the rear wheel center, $C$.

::::{grid} 2
:::{grid-item}
![photo](figs/problemas_cin_an/examen2023_ene_4_foto.jpg)
:::
:::{grid-item}
![diagram](figs/problemas_cin_an/examen2023_ene_4_enunciado.png)
:::
::::

Data: $\omega_4= 1~rad/s$ counterclockwise; $\alpha_4= 2~rad/s^2$ clockwise. Any other dimension needed can be measured from the figure.

:::{dropdown} Solution

**1. Degrees of freedom**

The spring-damper unit $O_5$-$D$ neither adds nor removes degrees of freedom, so it may or may not be counted. If it is not counted:

- $N=4$ bodies (counting the ground as a body).
- $p_I=4$ revolute joints ($O_2$: 1-2, $O_4$: 1-4, $A$: 2-3 and $B$: 3-4).
- $p_{II}=0$, since there are no two-d.o.f. joints.

Applying Grübler's formula: $G=3(N-1)-2 p_I - p_{II} = 3 \cdot 3 - 2 \cdot 4 = 1$ d.o.f.

If the spring-damper unit is counted, it consists of two bodies sliding on each other: Body 5 (lower half) and Body 6 (upper half, the one attached to Body 2). Then:

- $N=6$ bodies.
- $p_I=7$: six revolute joints (1-2, 1-4, 2-3, 3-4, 2-6 and 1-5) and one prismatic joint (5-6).
- $p_{II}=0$.

This gives $G = 3 \cdot 5 - 2 \cdot 7 = 1$ d.o.f., matching the previous result, as expected.

**2. Velocity of point $A$**

Since this is a one-d.o.f. mechanism, a single velocity (and acceleration) input is enough to solve for all the others.
The strategy is to find a loop that covers all the bodies of the kinematic chain to be analyzed, write the relative-motion equations of the points along that loop, and solve the resulting system.
The loop will be $O_4$-$B$-$A$-$O_2$, with the following position vectors (in cm, measured from the figure):

$$
\vec{r}_{O_2A} = (5.529,~2.499), \quad
\vec{r}_{O_4B} = (22.246,~-2.259), \quad
\vec{r}_{AB} = (20.609,~-17.248), \quad
\vec{r}_{BC} = (4.6,~0.6)
$$

The velocity equations along the loop are:

$$
\begin{aligned}
\vec{v}_B &= \vec{\omega}_4 \times \vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{v}_A &= \vec{\omega}_2 \times \vec{r}_{O_2A} && \text{(Body 2)} \\
\vec{v}_B &= \vec{v}_A + \vec{\omega}_3 \times \vec{r}_{AB} && \text{(Body 3)}
\end{aligned}
$$

The first one can be evaluated directly: $\vec{v}_B = (1\,\hat{k}) \times (22.246\,\hat{i} - 2.259\,\hat{j}) = 2.259\,\hat{i} + 22.246\,\hat{j}~(cm/s)$.
Substituting the other two into the third:

$$
2.259\,\hat{i} + 22.246\,\hat{j} = \omega_2 \left(-2.499\,\hat{i} + 5.529\,\hat{j}\right) + \omega_3 \left(17.248\,\hat{i} + 20.609\,\hat{j}\right)
$$

which is a system of two equations with two unknowns, whose solution is:

$$
\omega_2 = 2.296~\text{rad/s}~(\circlearrowleft), \qquad \omega_3 = 0.4636~\text{rad/s}~(\circlearrowleft)
$$

The requested velocity is therefore:

$$
\vec{v}_A = \vec{\omega}_2 \times \vec{r}_{O_2A} = -5.74\,\hat{i} + 12.69\,\hat{j}~(cm/s)
$$

**3. Accelerations**

We write the equivalent system with the rigid-body acceleration equations along the same loop:

$$
\begin{aligned}
\vec{a}_B &= \vec{\alpha}_4 \times \vec{r}_{O_4B} - \omega_4^2\,\vec{r}_{O_4B} && \text{(Body 4)} \\
\vec{a}_A &= \vec{\alpha}_2 \times \vec{r}_{O_2A} - \omega_2^2\,\vec{r}_{O_2A} && \text{(Body 2)} \\
\vec{a}_B &= \vec{a}_A + \vec{\alpha}_3 \times \vec{r}_{AB} - \omega_3^2\,\vec{r}_{AB} && \text{(Body 3)}
\end{aligned}
$$

where $\vec{a}_B = -26.76\,\hat{i} - 42.23\,\hat{j}~(cm/s^2)$ is known and the unknowns are $\alpha_2$ and $\alpha_3$. Solving:

$$
\alpha_2 = -4.803~\text{rad/s}^2~(\circlearrowright), \qquad
\alpha_3 = -0.3016~\text{rad/s}^2~(\circlearrowright), \qquad
\vec{a}_A = -17.13\,\hat{i} - 39.72\,\hat{j}~(cm/s^2)
$$

Point $C$ is not part of the loop used to set up the system, but once $\omega_3$ and $\alpha_3$ are known, its acceleration follows directly from the equation of rigid body 3:

$$
\vec{a}_C = \vec{a}_B + \vec{\alpha}_3 \times \vec{r}_{BC} - \omega_3^2\,\vec{r}_{BC} = -27.57\,\hat{i} - 43.75\,\hat{j}~(cm/s^2)
$$

The following MATLAB code solves the problem using the Symbolic Math Toolbox:

```matlab
clear, clc, close all

% Geometric data (cm):
rO2A=[5.529 2.499 0];
rO4B=[22.246 -2.259 0];
rAB=[20.609 -17.248 0];
rBC=[4.6 0.6 0];

% Known inputs:
w4=[0 0 1];
af4=[0 0 -2];

% Symbolic unknowns:
syms w2M w3M real

% Velocities:
w3=[0 0 w3M];
w2=[0 0 w2M];
vA=cross(w2,rO2A);
vB=cross(w4,rO4B);

eq1= vB==vA+cross(w3,rAB);

sol1=solve(eq1,[w3M,w2M]);
w2M=eval(sol1.w2M);
w3M=eval(sol1.w3M);
w3=[0 0 w3M]; w2=[0 0 w2M];

fprintf('\nVelocity results:\n')
fprintf('v_A='); disp(eval(vA));
fprintf('v_B='); disp(vB);
fprintf('w_2='); disp(w2M);
fprintf('w_3='); disp(w3M);

% Accelerations:
syms af2M af3M real

af2=[0 0 af2M];
aA=cross(af2,rO2A) + cross(w2,cross(w2,rO2A));
aB=cross(af4,rO4B) + cross(w4,cross(w4,rO4B));

af3=[0 0 af3M];
aBA=cross(af3,rAB) + cross(w3,cross(w3,rAB));

eq2= aB == aA + aBA;
sol2=solve(eq2,[af2M,af3M]);

af2M=eval(sol2.af2M)
af3M=eval(sol2.af3M)
af2=[0 0 af2M]; af3=[0 0 af3M];

% Acceleration of C:
aC = aB + cross(af3, rBC) + cross(w3, cross(w3, rBC));

fprintf('\nAcceleration results:\n')
fprintf('a_A='); disp(eval(aA));
fprintf('a_B='); disp(aB);
fprintf('a_C='); disp(eval(aC));
```
:::