Irodov 1.286: Turning the Spin Axis Loads Bearings – Amit & Deepali Sinha Foundation

Problem Irodov 1.286

A uniform sphere of mass $m = 5.0$ kg and radius $R = 6.0$ cm is mounted on a horizontal axle that runs through its center and is carried by two bearings a distance $l = 15$ cm apart, both fixed to a base. The sphere spins about that axle at $\omega = 1250$ rad/s. The base itself is then made to turn about a vertical axis at the much slower rate $\omega' = 5.0$ rad/s, which forces the spin axis to swing round with it. Find the magnitude and the direction of the gyroscopic forces, that is, of the extra forces the spinning sphere and its axle exert on the two bearings because of that forced turning.

The mounting. The sphere spins at ω about the horizontal axle, whose bearings 1 and 2 stand on the base a distance l apart. The base is driven about the vertical dashed axis at ω'. — The mounting. The sphere spins at $\vect\omega$ about the horizontal axle, whose bearings 1 and 2 stand on the base a distance $l$ apart. The base is driven about the vertical dashed axis at $\vect\omega'$ .

Given

$m = 5.0$ kg
$R = 6.0$ cm
$\omega = 1250$ rad/s
$l = 15$ cm
$\omega' = 5.0$ rad/s

Find

the magnitude and direction of the gyroscopic forces on the bearings

Solution

Step 1The angular momentum carried by the spin

Set the $x$ axis along the axle in the sense of $\vect\omega$ and the $z$ axis up, along $\vect\omega'$ . A uniform sphere has the moment of inertia $2mR^2/5$ about any axis through its center, and the spin is $\omega/\omega' = 250$ times faster than the base’s rotation, so the angular momentum is taken to be the spin’s alone and lies along the axle.

\vect L = \frac{2}{5}mR^2\omega\,\vect i

(1)

Step 2Forcing that vector to turn demands a moment

The base carries the axle round the vertical, so the vector of (1) keeps its length and rotates rigidly with the base at $\vect\omega'$ . The rate of change of a vector of fixed length carried by a rotation is the cross product of that rotation with the vector, and here the two are perpendicular, so the moment the bearings must apply to the axle is horizontal, at right angles to the axle itself.

\begin{align*}\vect N = \frac{d\vect L}{dt} &= \vect\omega'\times\vect L \\ &= \omega' L\,(\vect k\times\vect i)\end{align*}

N = \omega' L = \frac{2}{5}mR^2\omega\omega'

(2)

Step 3The moment is shared by the two bearings as a couple

The bearings can act on the axle only where they touch it, at the two points a distance $l$ apart along the $x$ axis. Take the vertical axis of the base’s rotation through the sphere’s center, as the figure shows, so the center travels no circle and needs no net force; the two bearing forces are then equal and opposite, a pure couple. A couple whose forces are separated along $x$ can have its moment along $\vect j$ only if the forces themselves are vertical, and their common magnitude follows from setting the moment of the pair equal to (2).

N = F'\,l

The figure shows that couple, drawn as the pair of forces the axle exerts on the bearings.

The gyroscopic forces on the two bearings, the mounting stripped away. They are vertical, equal and opposite, a pure couple: the bearing that L points toward is pushed up, the other down. — The gyroscopic forces on the two bearings, the mounting stripped away. They are vertical, equal and opposite, a pure couple: the bearing that $\vect L$ points toward is pushed up, the other down.

Numerically,

F' = \frac{2(5.0\ \text{kg})(0.060\ \text{m})^2(1250\ \text{s}^{-1})(5.0\ \text{s}^{-1})}{5(0.15\ \text{m})}

F' = \frac{N}{l} = \frac{2mR^2\omega\omega'}{5l} = 0.30\ \text{kN}

(3)

Vertical, a couple that swings the spin axis towards

\vect\omega'

Check

The units reduce correctly: $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\cdot\text{s}^{-1}/\text{m} = \text{kg}\cdot\text{m}/\text{s}^2$ , a newton. The size is worth reading too: the sphere’s own weight is only $mg = 49$ N, and it sits midway between the bearings, so each bearing carries $mg/2 = 24.5$ N when nothing is turning. The forced turning adds $0.30$ kN to that, about twelve times the static load, which is why the effect cannot be ignored in a real mounting.

Remark

The couple is the mounting’s price for refusing to let the spin axis stay where it points, and its sense always works to fold $\vect L$ into $\vect\omega'$ : reverse either rotation and both forces reverse with it. Two design consequences fall straight out of $F' = 2mR^2\omega\omega'/(5l)$ . Widening the bearing span cuts the load in proportion, since only the couple is fixed and the forces divide it. And the load is linear in both rates, so a rotor that is perfectly quiet on a straight run can wreck its bearings in a slow turn simply because it spins fast.