Two identical bodies of mass m = 0.5 kg are fixed on…

Two identical bodies of mass $m = 0.5\ \mathrm{kg}$ are fixed on opposite sides of a motor shaft rotating at $n = 6000\ \mathrm{rpm}$, each at a distance $R = 0.3\ \mathrm{m}$ from the axis, so the shaft is balanced. Then one body moves 10% farther out. Find the force this unbalanced system exerts on the shaft. Neglect the mass of the rod.

Explanation of Method

Force on a shaft from an unbalanced rotor.

Step 1: Angular velocity

\[ \omega = 6000 \cdot \frac{2\pi}{60} = 628.3\ \mathrm{rad/s}, \]

\[ \omega^2 = 3.948 \cdot 10^5\ \mathrm{s^{-2}}. \]

Step 2: Pull of each body

\[ F_1 = m\omega^2 R = 0.5 \cdot 3.948 \cdot 10^5 \cdot 0.30 \approx 59\,218\ \mathrm{N}, \]

\[ F_2 = m\omega^2 \cdot 1.1R \approx 65\,139\ \mathrm{N}. \]

Step 3: Net force on the shaft

The two pulls are opposite, so only the difference remains: \[F = F_2 - F_1 = m\omega^2 \cdot 0.1R = 0.5 \cdot 3.948 \cdot 10^5 \cdot 0.03.\]

\[F \approx 5.92 \cdot 10^3\ \mathrm{N}\]

$0.30\ \mathrm{m}$ $0.33\ \mathrm{m}$ $59.2\ \mathrm{kN}$ $65.1\ \mathrm{kN}$ $F \approx 5.92\ \mathrm{kN}$ $\omega = 628\ \mathrm{rad/s}$

The pulls needed to keep each body on its circle almost cancel; what is left, about 5.9 kN, rotates with the shaft and loads its bearings.

Result

\[\boxed{\,F \approx 5.92 \cdot 10^3\ \mathrm{N}\,}\]

Remark

About the original solution. The method is right, but $628.32^2$ was evaluated too small, giving $F_1 = 59\,070.5\ \mathrm{N}$ instead of 59 218 N and a net force of 5907 N instead of about 5922 N. The unbalanced force rotates with the shaft, 100 times per second, and is about 1200 times the weight of one body, which is why even small imbalances are dangerous in fast machines.