Irodov 1.287: The Bearings Feel the Wobble – Amit & Deepali Sinha Foundation

Problem Irodov 1.287

A gyroscope’s cylindrical disc, of mass $m = 15$ kg and radius $r = 5.0$ cm, spins about its own axis at angular velocity $\omega = 330$ rad/s. The disc’s axle is carried in two bearings a distance $l = 15$ cm apart. A driving mechanism swings the whole axle back and forth about a horizontal axis (perpendicular to itself), one full oscillation taking $T = 1.0$ s and carrying the axle $\varphi_m = 20^\circ$ away from its mean position at each extreme. Find the largest gyroscopic force the axle exerts on a bearing.

The machine seen from the side. The disc spins about the axle, which is carried in two bearings a distance l apart and is itself forced to swing in this plane, its extreme position drawn dashed at φ_m from the mean one. — The machine seen from the side. The disc spins about the axle, which is carried in two bearings a distance $l$ apart and is itself forced to swing in this plane, its extreme position drawn dashed at $\varphi_m$ from the mean one.

Given

$m = 15$ kg
$r = 5.0$ cm
$\omega = 330$ rad/s
$l = 15$ cm
$T = 1.0$ s
$\varphi_m = 20^\circ$

Find

Find the maximum gyroscopic force

F

on a bearing

Solution

Step 1The angular momentum and the rate at which it is turned

The disc spins fast about its own axle, so its angular momentum lies along the axle and has the magnitude

L = I\omega, \qquad I = \frac{1}{2}mr^{2},

$I$ being the moment of inertia of a uniform disc about its axis. The forced oscillation swings that axle, and with it the vector $\vect L$ , about a horizontal axis perpendicular to it. Writing the swing as $\varphi = \varphi_m\sin(2\pi t/T)$ , its angular velocity is

\Omega = \frac{d\varphi}{dt} = \varphi_m\,\frac{2\pi}{T}\cos\!\left(\frac{2\pi t}{T}\right),

whose greatest magnitude occurs as the axle passes through the middle of its swing:

\Omega_{\max} = \frac{2\pi\varphi_m}{T}

(1)

Step 2The moment needed to turn the angular momentum

Turning a vector of fixed length $L$ at the angular rate $\Omega$ about an axis perpendicular to it changes it at the rate

\left|\frac{d\vect L}{dt}\right| = \Omega L ,

and by the equation of moments that rate must be supplied as a moment $N$ about the third axis, perpendicular to both the axle and the oscillation axis. So

N = \Omega L = \Omega\,\frac{1}{2}mr^{2}\omega .

The same machine seen from above. The bearings can only supply that moment as a couple: two equal and opposite forces perpendicular to the axle, a distance l apart. By the third law the axle presses back on the bearings with the same pair. — The same machine seen from above. The bearings can only supply that moment as a couple: two equal and opposite forces perpendicular to the axle, a distance $l$ apart. By the third law the axle presses back on the bearings with the same pair.

Step 3The couple at the bearings

The bearings are the only things holding the axle, and each can push only at its own point. Two equal and opposite forces $F$ a distance $l$ apart make a couple of moment $Fl$ , and this couple is what the bearings apply to the axle, so

Fl = N = \frac{1}{2}mr^{2}\omega\,\Omega .

The forces the axle exerts on the bearings are their reactions, equal in magnitude. Taking $\Omega$ at its maximum from (1),

F_{\max} = \frac{1}{l}\cdot\frac{1}{2}mr^{2}\omega\cdot \frac{2\pi\varphi_m}{T} ,

and the two twos cancel:

F_{\max} = \frac{\pi m r^{2}\varphi_m\,\omega}{lT}

(2)

Step 4Numbers

The amplitude must go in as a pure number of radians before it can be substituted.

\begin{aligned} \varphi_m &= 20^\circ = 0.35\ \text{rad}, \\ F_{\max} &= \frac{\pi \times 15 \times (5.0\times10^{-2})^{2} \times 0.35 \times 330}{0.15 \times 1.0} \approx 0.9\times10^{2}\ \mathrm{N} = 0.09\ \mathrm{kN}. \end{aligned}

Check

The gyroscopic treatment assumes the angular momentum stays along the axle, which is safe only if the axle is swung far more slowly than the disc spins. Here $\Omega_{\max} = 2\pi\times0.35/1.0 = 2.2$ rad/s against $\omega = 330$ rad/s, a ratio of about $150$ , so the assumption holds comfortably. The dimensions also check: $mr^{2}\omega$ is an angular momentum, dividing by $T$ and multiplying by the dimensionless $\varphi_m$ gives a moment, and dividing by $l$ gives a force.

Remark

Forgetting to convert $\varphi_m$ into radians is the standard way to lose this problem: $20$ in place of $0.35$ inflates the answer by a factor of $57$ . The reason radians are compulsory is that $\varphi_m$ enters only through $d\varphi/dt$ , and the derivative of $\sin\varphi$ is $\cos\varphi$ only when $\varphi$ is in radians.