Irodov 1.269: A Tilted Rotor Twists Its Bearings – Amit & Deepali Sinha Foundation

Problem Irodov 1.269

A thin uniform rod $AB$ of mass $m$ and length $l$ is clamped by its midpoint $C$ to a straight rotation axle $OO'$ , which turns in fixed bearings at its two ends. The clamp holds the rod rigidly at a fixed angle $\theta$ to the axle, so the point $C$ lies on the axle and the rod pierces it there. The whole assembly turns about $OO'$ with a constant angular velocity $\omega$ . Working in the reference frame that rotates with the axle and the rod, in which every part of the rod is at rest, find the resultant moment about the point $C$ of the centrifugal forces of inertia acting on the rod.

The arrangement. The axle OO' turns at ω in bearings at both ends; the rod AB is clamped to it at the rod's own midpoint C, holding a fixed angle θ between the rod and the axle. — The arrangement. The axle $OO'$ turns at $\vect\omega$ in bearings at both ends; the rod $AB$ is clamped to it at the rod’s own midpoint $C$ , holding a fixed angle $\theta$ between the rod and the axle.

Given

$m$
$l$
$\omega$
$\theta$

Find

the resultant moment

N

of the centrifugal forces about

C

Solution

Step 1The centrifugal force carried by one element of the rod

In the frame rotating with the axle every element of the rod is permanently at rest, so no Coriolis force arises and the only inertial force is the centrifugal one. Let $s$ be the signed distance of an element from $C$ , measured along the rod and counted positive toward $B$ . The element’s mass follows from the uniform linear density, and its offset from the axle follows from the tilt.

dm = \frac{m}{l}\,ds, \qquad r = s\,\sin\theta

The centrifugal force on that element is $\omega^2 r$ per unit mass, directed horizontally away from the axle. Let $dF$ stand for its component along the one fixed horizontal direction that points away from the axle on the $B$ side of the rod; with the sign of $s$ carried along, a single expression then covers both halves of the rod.

dF = \omega^2 r\,dm = \frac{m\omega^2\sin\theta}{l}\,s\,ds

(1)

One element of the rod, a distance s from C. Its distance from the axle is sθ; the centrifugal force d F on it is horizontal, and its perpendicular distance from C is the moment arm sθ. — One element of the rod, a distance $s$ from $C$ . Its distance from the axle is $s\sin\theta$ ; the centrifugal force $d\vect F$ on it is horizontal, and its perpendicular distance from $C$ is the moment arm $s\cos\theta$ .

Step 2The moment of that force about the point

C

The force on the element is horizontal, so its moment arm about $C$ is the vertical separation between $C$ and the element, read off the figure’s right triangle. Multiplying the arm by the force gives the elementary moment.

\begin{align*}dN &= (s\cos\theta)\,dF \\ &= \frac{m\omega^2\sin\theta}{l}\,s\cos\theta\;s\,ds\end{align*}

dN = \frac{m\omega^2\sin\theta\cos\theta}{l}\,s^2\,ds

(2)

Because $s$ enters squared, the two halves of the rod contribute moments of the same sign, and the resultant is a genuine couple rather than a cancellation.

Step 3Summing over the whole rod

The elementary moments are all parallel (each is perpendicular to the plane holding the axle and the rod), so the resultant is their plain integral over the rod, from the end $A$ at $s=-l/2$ to the end $B$ at $s=+l/2$ .

\begin{align*}N &= \int_{-l/2}^{+l/2}\frac{m\omega^2\sin\theta\cos\theta}{l}\,s^2\,ds \\ &= \frac{m\omega^2\sin\theta\cos\theta}{l}\left.\frac{s^3}{3}\right|_{-l/2}^{+l/2} \\ &= \frac{m\omega^2\sin\theta\cos\theta}{l}\cdot\frac{1}{3}\left(\frac{l^3}{8}+\frac{l^3}{8}\right) \\ &= \frac{m\omega^2 l^2}{12}\sin\theta\cos\theta\end{align*}

Folding the product of the two trigonometric factors into the double angle gives the compact form.

N = \frac{1}{24}\,m\omega^2 l^2\sin 2\theta

(3)

Perpendicular to the plane of the axle and the rod, and turning with them.

Check

Dimensionally $m\omega^2l^2$ is a mass times a squared frequency times a squared length, which is a force times a length, as a moment must be. The two degenerate tilts both give zero, and for different reasons that the formula reproduces together: at $\theta=0$ the rod lies along the axle, every element sits on the axis and no centrifugal force exists at all; at $\theta=90^\circ$ the centrifugal forces are large but every one of them is directed straight along the rod through $C$ , so each has zero arm. The moment peaks midway between, at $\theta=45^\circ$ .

Remark

The integral that appeared here, $\sin\theta\cos\theta\int s^2\,dm$ , is exactly the product of inertia of the rod for the axle direction and the direction perpendicular to it through $C$ , so the result reads $N=\omega^2\int xz\,dm$ with $x$ measured from the axle and $z$ along it. That is the standard statement of dynamic unbalance: a rigid body spinning steadily about an axis through its center of mass exerts no couple on its bearings when that axis is a principal axis of the body, and exerts one growing as $\omega^2$ when it is not. Here the rod’s principal axes at $C$ are the rod itself and any perpendicular to it, which is why the couple vanishes at exactly the two tilts where the axle coincides with one of them. Note also that this couple is fixed in the rotating frame, so in the laboratory it sweeps round with the shaft, and the bearings at $O$ and $O'$ feel loads that alternate at the rotation frequency.