# Example 3.2: Disk driven by a piston

In the mechanism of the figure, a horizontally guided piston $P$ drives a disk through a connecting rod $PQ$. The disk rotates about its center $O$, and at the instant considered the pin $Q$ is right above $O$ and the rod is at $45^\circ$ from the horizontal. What force $F$ must the piston exert for the disk to have an angular acceleration of $1000$ rpm/s clockwise?

Data: piston and rod of negligible mass; disk of mass $M = 1$ kg and diameter $20$ cm; rod length $L = 40\sqrt{2}$ cm. Neglect friction.

<img src="figs/L03_piston_disc.png" alt="Disk driven by a piston" width="480px">

:::{dropdown} Solution
**Kinematics.** The angular acceleration, in SI units, is:

$$ \alpha = 1000~\frac{\text{rev}}{\text{min}\cdot\text{s}} \cdot \frac{2\pi~\text{rad}}{1~\text{rev}} \cdot \frac{1~\text{min}}{60~\text{s}} = 104.72~\text{rad/s}^2 \quad (\text{clockwise}) $$

**Inertia loads.** Only the disk has mass. Since it rotates about its center of mass, $\vec{a}_G = \vec{0}$ and **there is no inertia force**: just an inertia torque, opposite to $\alpha$ (counterclockwise):

$$ I_G = \frac{1}{2} M R^2 = \frac{1}{2} \cdot 1 \cdot 0.1^2 = 0.005~\text{kg·m}^2, \qquad |M_{in}| = I_G\, \alpha = 0.005 \cdot 104.72 = 0.5236~\text{N·m} $$

Note that this result **does not depend on the angular velocity** of the disk: the disk rotates about its center of mass, and the rod and piston are massless.

**Rod.** With no mass and no loads, it is a two-force member: the force $F_b$ it transmits acts along $PQ$, at $45^\circ$.

**Disk.** Taking moments about $O$ (so that the reaction at the shaft does not appear), the moment of the rod force must balance the inertia torque. The distance from $O$ to line $PQ$ is $d = R \sin 45^\circ$ (the angle between $OQ$, vertical, and the rod is $45^\circ$):

$$ F_b\, R \sin 45^\circ = I_G\, \alpha \quad\rightarrow\quad F_b = \frac{0.5236}{0.1 \cdot \sin 45^\circ} = 7.405~\text{N} $$

The rod is in compression: it pushes pin $Q$ up and to the right, which gives a clockwise moment about $O$, as needed to accelerate the disk clockwise.

**Piston.** It receives the force $F$, the guide reaction $N$ (vertical) and the rod force (at $45^\circ$, downwards and to the left). Its force equilibrium gives:

$$
\begin{aligned}
\sum F_x = 0: &\quad F - F_b \cos 45^\circ = 0 \quad\rightarrow\quad F = 7.405 \cdot \cos 45^\circ = 5.24~\text{N} \\
\sum F_y = 0: &\quad N - F_b \sin 45^\circ = 0 \quad\rightarrow\quad N = 5.24~\text{N}
\end{aligned}
$$

Combining the expressions, $F = I_G\,\alpha / R$: with the rod at $45^\circ$ and the pin right above $O$, the piston force is the same as if it acted tangentially on the rim of the disk. The rod length plays no role.

**Reaction at the disk shaft.** Since the disk has no inertia force, $\vec{F}_{O} = -\vec{F}_b = -(5.24\,\hat{i} + 5.24\,\hat{j})$ N.
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