Linear Vibrations Lab

Worked examples › Example 5

Example 5Critical damping of a pinned bar

rotational EOMcritical damping

Problem

A homogeneous bar of mass $m$ and length $b$ is pinned to ground at O and moves in a horizontal plane. A spring $k$ is attached at distance $a$ from O and a damper $c$ at the end, distance $b$. $\theta$ is the bar's angle, counterclockwise positive. Assume small oscillations ($\sin\theta\approx\theta$).

Find: the value of $c$ for critical damping.

InteractiveThe system in motion

Displacements in the drawing are scaled up so the motion is visible; the plot shows the true values.

Presets:
θ(t)

Solution

Every solution follows the same five steps: coordinates, free-body diagrams, kinetics, kinematics, solve. Try the problem first, then open one step at a time to check your work.

Step 1 · Choose a coordinate system

Let $\theta$ be the bar's rotation about O, counterclockwise positive, measured from the equilibrium where the spring is unstretched.

Step 2 · Draw the free-body diagrams
Bar, top view (horizontal plane)OxOyk(aθ)c(bθ)ab+θweight acts out of the page
Free-body diagram of the bar for a small counterclockwise rotation θ. The spring and damper resist the motion.
Step 3 · Kinetics (Newton–Euler equations)

Moments about the fixed pin O ($I_O=\tfrac13mb^2$ for a uniform bar about its end):

$$\textstyle\sum M_O=I_O\ddot\theta:\qquad -(k\,a\theta)\,a-(c\,b\dot\theta)\,b=\tfrac13mb^2\,\ddot\theta$$

The pin reactions pass through O and produce no moment. The bar moves in a horizontal plane, so gravity has no moment about O either.

Step 4 · Kinematics (constraints)

For small rotations the attachment points move along arcs that are nearly straight:

$$\text{spring point: } a\theta,\qquad \text{damper point: } b\theta,\ \ \text{velocity } b\dot\theta$$
Step 5 · Solve

Rearranged:

$$\tfrac13mb^2\,\ddot\theta+cb^2\,\dot\theta+ka^2\,\theta=0\qquad(m_{eq}=\tfrac13mb^2,\ c_{eq}=cb^2,\ k_{eq}=ka^2)$$

Critical damping requires $c_{eq}=2\sqrt{k_{eq}m_{eq}}$:

$$c\,b^2=2\sqrt{ka^2\cdot\tfrac13mb^2}\qquad\Longrightarrow\qquad c_{cr}=\frac{2a}{b}\sqrt{\frac{km}{3}}$$
Answer

$$c_{cr}=\frac{2a}{b}\sqrt{\frac{km}{3}}$$

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