# Example 3.4: Slider-crank with the mass at the slider
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In the slider-crank mechanism of the figure, crank $OA$ (of length $L$) rotates with constant angular velocity $\omega$ counterclockwise, driven by an unknown input torque $\tau$. The connecting rod $AB$ is $2L$ long. Knowing that the crank and the rod have negligible mass and that the mass of the slider is $M$, compute the instantaneous torque $\tau$ when the crank is at $45^\circ$ from the horizontal. Neglect weight and friction.
(Example solved in {cite:p}`avello2014teoria`.) We follow the three steps of D'Alembert's principle: kinematics, inertia loads and static equilibrium.
:::{dropdown} Step 1: kinematics
**Position.** With $O$ at the origin and $\theta = 45^\circ$:
$$ \vec{r}_{OA} = L\,(\cos 45^\circ, \sin 45^\circ) = (0.7071, 0.7071)\,L $$
The slider is level with $O$: $x_B = L\cos 45^\circ + \sqrt{(2L)^2 - (L \sin 45^\circ)^2} = 2.5779\,L$, and therefore:
$$ \vec{r}_{AB} = (1.8708, -0.7071)\,L, \qquad \sin\beta = \frac{L\sin 45^\circ}{2L} \rightarrow \beta = 20.70^\circ $$
with $\beta$ the angle between the rod and the horizontal (below it).
**Velocities.** $\vec{v}_A = \vec{\omega}\times\vec{r}_{OA} = \omega L\,(-0.7071, 0.7071)$, and $\vec{v}_B = v_B\,\hat{i} = \vec{v}_A + \vec{\omega}_3 \times \vec{r}_{AB}$. Equating components:
$$
\begin{aligned}
v_B &= -0.7071\,\omega L + 0.7071\,\omega_3 L \\
0 &= 0.7071\,\omega L + 1.8708\,\omega_3 L
\end{aligned}
\quad\rightarrow\quad
\omega_3 = -0.3780\,\omega, \qquad v_B = -0.9744\,\omega L
$$
**Accelerations.** With constant $\omega$, $\vec{a}_A = -\omega^2\vec{r}_{OA} = \omega^2 L\,(-0.7071, -0.7071)$, and:
$$ a_B\,\hat{i} = \vec{a}_A - \omega_3^2\, \vec{r}_{AB} + \vec{\alpha}_3 \times \vec{r}_{AB} $$
$$
\begin{aligned}
a_B &= -0.7071\,\omega^2 L - 0.1429 \cdot 1.8708\,\omega^2 L + 0.7071\,\alpha_3 L \\
0 &= -0.7071\,\omega^2 L + 0.1429 \cdot 0.7071\,\omega^2 L + 1.8708\,\alpha_3 L
\end{aligned}
\quad\rightarrow\quad
\alpha_3 = 0.3240\,\omega^2, \qquad a_B = -0.7453\,\omega^2 L
$$
The slider moves to the left ($v_B < 0$) faster and faster ($a_B < 0$, same sign).
:::
:::{dropdown} Step 2: inertia loads
Only the slider has mass, and it translates: it has an inertia force, but no inertia torque. The inertia force is opposite to its acceleration:
$$ \vec{J}_4 = -M\,\vec{a}_B = 0.7453\, M\omega^2 L\;\hat{i} $$
applied at $B$.
:::
:::{dropdown} Step 3: equilibrium (analytical solution)
**Rod (link 3).** It has no mass and no loads: it is a **two-force member** and transmits a force along $AB$. Let $\vec{u} = \vec{r}_{AB}/|\vec{r}_{AB}| = (\cos\beta, -\sin\beta)$ and $\vec{F}_{34} = F_{34}\,\vec{u}$ be the force of the rod on the slider ($F_{34} > 0$ if it pushes, compression).
**Slider (link 4).** It receives $\vec{J}_4$, $\vec{F}_{34}$ and the guide reaction $\vec{F}_{14} = N\,\hat{j}$:
$$
\begin{aligned}
\sum F_x = 0: &\quad 0.7453\,M\omega^2 L + F_{34} \cos\beta = 0 \quad\rightarrow\quad F_{34} = -0.7967\, M\omega^2 L \\
\sum F_y = 0: &\quad N - F_{34} \sin\beta = 0 \quad\rightarrow\quad N = -0.2817\, M\omega^2 L
\end{aligned}
$$
The negative sign of $F_{34}$ means that the rod **pulls** the slider (tension), and that of $N$, that the guide pushes downwards.
**Crank (link 2).** The rod exerts $\vec{F}_{32} = -\vec{F}_{34}$ on it at $A$. Taking moments about $O$:
$$ \tau + \left( \vec{r}_{OA} \times \vec{F}_{32} \right)_z = 0 \quad\rightarrow\quad \tau = \left( \vec{r}_{OA} \times \vec{F}_{34} \right)_z = F_{34}\, L \left( -0.7071 \sin\beta - 0.7071 \cos\beta \right) $$
$$ \boxed{\tau = 0.7262\, M\omega^2 L^2} \quad (\text{counterclockwise}) $$
**Power check.** The power of the input torque plus that of the inertia force is zero: $\tau\,\omega + \vec{J}_4\cdot\vec{v}_B = 0$, so $\tau = -J_4\, v_B / \omega = 0.7453 \cdot 0.9744\, M\omega^2L^2 = 0.7262\, M\omega^2L^2$. ✓
:::
:::{dropdown} Step 3: equilibrium (graphical solution)
With the rod as a two-force member, the slider is a **three-force member**: $\vec{J}_4$ (known), $\vec{F}_{34}$ (along $AB$) and $\vec{F}_{14}$ (perpendicular to the guide). All three go through $B$, so closing the force triangle is enough to get the magnitudes and senses of $\vec{F}_{34}$ and $\vec{F}_{14}$. Then the force that the rod transmits to the crank has an arm $d$ about $O$, measured on the drawing ($d = 0.9114\,L$), and $\tau = |\vec{F}_{32}|\, d = 0.7967 \cdot 0.9114\, M\omega^2L^2 = 0.7262\, M\omega^2L^2$.
:::
## Interactive construction
Go through the construction step by step. Dragging the crank shows the solution at any other position (magnitudes are given in units of $M\omega^2L$ and $M\omega^2L^2$):
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```
## Torque over a revolution
Repeating the computation at every position gives the curve of the torque needed to keep $\omega$ constant. The torque changes sign: in some regions the motor has to accelerate the slider, and in others it must brake it. Its mean value over a revolution is zero, because the kinetic energy of the mechanism is the same at the beginning and at the end of the revolution (lesson 4). This fluctuating torque is what a flywheel smooths out.
(Example solved in {cite:p}`avello2014teoria`.) We follow the three steps of D'Alembert's principle: kinematics, inertia loads and static equilibrium.
:::{dropdown} Step 1: kinematics
**Position.** With $O$ at the origin and $\theta = 45^\circ$:
$$ \vec{r}_{OA} = L\,(\cos 45^\circ, \sin 45^\circ) = (0.7071, 0.7071)\,L $$
The slider is level with $O$: $x_B = L\cos 45^\circ + \sqrt{(2L)^2 - (L \sin 45^\circ)^2} = 2.5779\,L$, and therefore:
$$ \vec{r}_{AB} = (1.8708, -0.7071)\,L, \qquad \sin\beta = \frac{L\sin 45^\circ}{2L} \rightarrow \beta = 20.70^\circ $$
with $\beta$ the angle between the rod and the horizontal (below it).
**Velocities.** $\vec{v}_A = \vec{\omega}\times\vec{r}_{OA} = \omega L\,(-0.7071, 0.7071)$, and $\vec{v}_B = v_B\,\hat{i} = \vec{v}_A + \vec{\omega}_3 \times \vec{r}_{AB}$. Equating components:
$$
\begin{aligned}
v_B &= -0.7071\,\omega L + 0.7071\,\omega_3 L \\
0 &= 0.7071\,\omega L + 1.8708\,\omega_3 L
\end{aligned}
\quad\rightarrow\quad
\omega_3 = -0.3780\,\omega, \qquad v_B = -0.9744\,\omega L
$$
**Accelerations.** With constant $\omega$, $\vec{a}_A = -\omega^2\vec{r}_{OA} = \omega^2 L\,(-0.7071, -0.7071)$, and:
$$ a_B\,\hat{i} = \vec{a}_A - \omega_3^2\, \vec{r}_{AB} + \vec{\alpha}_3 \times \vec{r}_{AB} $$
$$
\begin{aligned}
a_B &= -0.7071\,\omega^2 L - 0.1429 \cdot 1.8708\,\omega^2 L + 0.7071\,\alpha_3 L \\
0 &= -0.7071\,\omega^2 L + 0.1429 \cdot 0.7071\,\omega^2 L + 1.8708\,\alpha_3 L
\end{aligned}
\quad\rightarrow\quad
\alpha_3 = 0.3240\,\omega^2, \qquad a_B = -0.7453\,\omega^2 L
$$
The slider moves to the left ($v_B < 0$) faster and faster ($a_B < 0$, same sign).
:::
:::{dropdown} Step 2: inertia loads
Only the slider has mass, and it translates: it has an inertia force, but no inertia torque. The inertia force is opposite to its acceleration:
$$ \vec{J}_4 = -M\,\vec{a}_B = 0.7453\, M\omega^2 L\;\hat{i} $$
applied at $B$.
:::
:::{dropdown} Step 3: equilibrium (analytical solution)
**Rod (link 3).** It has no mass and no loads: it is a **two-force member** and transmits a force along $AB$. Let $\vec{u} = \vec{r}_{AB}/|\vec{r}_{AB}| = (\cos\beta, -\sin\beta)$ and $\vec{F}_{34} = F_{34}\,\vec{u}$ be the force of the rod on the slider ($F_{34} > 0$ if it pushes, compression).
**Slider (link 4).** It receives $\vec{J}_4$, $\vec{F}_{34}$ and the guide reaction $\vec{F}_{14} = N\,\hat{j}$:
$$
\begin{aligned}
\sum F_x = 0: &\quad 0.7453\,M\omega^2 L + F_{34} \cos\beta = 0 \quad\rightarrow\quad F_{34} = -0.7967\, M\omega^2 L \\
\sum F_y = 0: &\quad N - F_{34} \sin\beta = 0 \quad\rightarrow\quad N = -0.2817\, M\omega^2 L
\end{aligned}
$$
The negative sign of $F_{34}$ means that the rod **pulls** the slider (tension), and that of $N$, that the guide pushes downwards.
**Crank (link 2).** The rod exerts $\vec{F}_{32} = -\vec{F}_{34}$ on it at $A$. Taking moments about $O$:
$$ \tau + \left( \vec{r}_{OA} \times \vec{F}_{32} \right)_z = 0 \quad\rightarrow\quad \tau = \left( \vec{r}_{OA} \times \vec{F}_{34} \right)_z = F_{34}\, L \left( -0.7071 \sin\beta - 0.7071 \cos\beta \right) $$
$$ \boxed{\tau = 0.7262\, M\omega^2 L^2} \quad (\text{counterclockwise}) $$
**Power check.** The power of the input torque plus that of the inertia force is zero: $\tau\,\omega + \vec{J}_4\cdot\vec{v}_B = 0$, so $\tau = -J_4\, v_B / \omega = 0.7453 \cdot 0.9744\, M\omega^2L^2 = 0.7262\, M\omega^2L^2$. ✓
:::
:::{dropdown} Step 3: equilibrium (graphical solution)
With the rod as a two-force member, the slider is a **three-force member**: $\vec{J}_4$ (known), $\vec{F}_{34}$ (along $AB$) and $\vec{F}_{14}$ (perpendicular to the guide). All three go through $B$, so closing the force triangle is enough to get the magnitudes and senses of $\vec{F}_{34}$ and $\vec{F}_{14}$. Then the force that the rod transmits to the crank has an arm $d$ about $O$, measured on the drawing ($d = 0.9114\,L$), and $\tau = |\vec{F}_{32}|\, d = 0.7967 \cdot 0.9114\, M\omega^2L^2 = 0.7262\, M\omega^2L^2$.
:::
## Interactive construction
Go through the construction step by step. Dragging the crank shows the solution at any other position (magnitudes are given in units of $M\omega^2L$ and $M\omega^2L^2$):
```{raw} html
```
## Torque over a revolution
Repeating the computation at every position gives the curve of the torque needed to keep $\omega$ constant. The torque changes sign: in some regions the motor has to accelerate the slider, and in others it must brake it. Its mean value over a revolution is zero, because the kinetic energy of the mechanism is the same at the beginning and at the end of the revolution (lesson 4). This fluctuating torque is what a flywheel smooths out.
