Irodov 1.285: Precession Follows the Effective Vertical

Problem Irodov 1.285

A top of mass $m = 1.0\ \mathrm{kg}$ spins about its own symmetry axis with an angular velocity $\omega = 310\ \mathrm{rad/s}$ , its moment of inertia about that axis being $I = 4.0\ \mathrm{g\cdot m^{2}}$ . It stands on its tip at a point of support that is fixed to a block, and the block is driven horizontally with a constant acceleration $w = 1.0\ \mathrm{m/s^{2}}$ . The distance from the point of support to the top’s center of inertia is $l = 10\ \mathrm{cm}$ . Find the magnitude of the angular velocity of precession $\vect\omega'$ of the top, and the direction of that vector.

The top on its tip on the block, driven horizontally at w. Its axle leans arbitrarily, the problem stating none. The spin is ω, the center of inertia is a distance l out, and the dashed line is the true vertical. — The top on its tip on the block, driven horizontally at $\vect w$ . Its axle leans arbitrarily, the problem stating none. The spin is $\vect\omega$ , the center of inertia is a distance $l$ out, and the dashed line is the true vertical.

Given

$m = 1.0\ \mathrm{kg}$
$I = 4.0\ \mathrm{g\cdot m^{2}}$
$\omega = 310\ \mathrm{rad/s}$
$w = 1.0\ \mathrm{m/s^{2}}$
$l = 10\ \mathrm{cm}$
$g = 9.8\ \mathrm{m/s^{2}}$

Find

the magnitude of

\vect\omega'

and its direction

Solution

Step 1Move into the frame of the block

The point of support is not at rest in the laboratory, so the convenient frame is the block’s own. It translates with a constant acceleration and does not rotate, so the only price of using it is one uniform inertial force $-m\vect w$ on every body, added at the center of inertia alongside the weight. The two constant forces combine into a single uniform field, an effective gravity.

The effective gravity in the block's frame: the true g downward and the inertial term - w opposite to the block's acceleration add to g_eff, which leans by θ from the vertical. The horizontal leg is drawn much longer than it is. — The effective gravity in the block’s frame: the true $\vect g$ downward and the inertial term $-\vect w$ opposite to the block’s acceleration add to $\vect g_{\mathrm{eff}}$ , which leans by $\theta$ from the vertical. The horizontal leg is drawn much longer than it is.

\begin{align*}\vect g_{\mathrm{eff}} &= \vect g-\vect w \\ g_{\mathrm{eff}} &= \sqrt{g^{2}+w^{2}}\end{align*}

g_{\mathrm{eff}} = \sqrt{g^{2}+w^{2}}, \qquad \tan\theta = \frac{w}{g}

(1)

Step 2The torque about the point of support

In this frame the point of support is at rest, so the angular momentum equation may be written about it, and the reaction acting there has no moment about it. The only moment left is that of the effective weight, applied at the center of inertia a distance $l$ out along the axis. Writing $\vartheta$ for the angle between the top’s axis and the effective vertical, which the problem never states,

\begin{align*} \vect\tau &= \vect l\times m\vect g_{\mathrm{eff}} \\ \tau &= m g_{\mathrm{eff}}\,l\sin\vartheta \tag{2}\end{align*}

This torque is horizontal in the effective sense, meaning perpendicular to the plane holding the axis and the effective vertical, so it swings $\vect L$ round that vertical instead of tipping the top over.

Step 3The precession rate under the effective gravity

The angular momentum is $L = I\omega$ along the axis, to the accuracy of the gyroscopic approximation, and its tip runs a circle of radius $L\sin\vartheta$ about the effective vertical at the precession rate. Equating that rate of change to the torque (2) is the whole of the dynamics.

\begin{align*}\left|\frac{d\vect L}{dt}\right| &= \omega' L\sin\vartheta = \omega' I\omega\sin\vartheta \\ \omega' I\omega\sin\vartheta &= m g_{\mathrm{eff}}\,l\sin\vartheta\end{align*}

The unstated lean cancels, which is why the problem can be answered without it. Dividing through by $I\omega\sin\vartheta$ and substituting the effective gravity from (1) gives the rate.

\omega' = \frac{m g_{\mathrm{eff}}\,l}{I\omega} = \frac{ml\sqrt{g^{2}+w^{2}}}{I\omega}

With the data,

\omega' = \frac{1.0\times 0.10\times\sqrt{9.8^{2}+1.0^{2}}}{4.0\times 10^{-3}\times 310}

\omega' = \frac{ml\sqrt{g^{2}+w^{2}}}{I\omega} = 0.79\ \mathrm{rad/s}

(3)

Step 4The direction of the precession vector

L lies along the axle, and the torque τ about the support is the circled cross. The dotted line is the true vertical, the dashed one the precession axis, tilted by θ toward w and carrying the ω' loop. — $\vect L$ lies along the axle, and the torque $\vect\tau$ about the support is the circled cross. The dotted line is the true vertical, the dashed one the precession axis, tilted by $\theta$ toward $\vect w$ and carrying the $\vect\omega'$ loop.

A top precesses about the axis its effective gravity defines, so $\vect\omega'$ lies along the effective vertical rather than the true one. With the spin directed away from the point of support, as it is drawn in the figure, the precession vector points along the upward effective vertical, and the opposite spin would reverse it.

\vect\omega' = -\,\omega'\,\frac{\vect g_{\mathrm{eff}}}{g_{\mathrm{eff}}}

That direction has a vertical part $g$ and a horizontal part $w$ pointing the way the block accelerates, so its tilt from the true vertical is the angle already fixed in (1).

\theta = \arctan\frac{w}{g} = \arctan\frac{1.0}{9.8} = 5.8^{\circ}\approx 6^{\circ}

(4)

The tilt is from the vertical, toward the block’s acceleration.

Check

Switching the block off should return the ordinary top. Setting $w=0$ gives $\theta = 0$ and the plain rate $\omega' = mgl/(I\omega)$ , which is $0.790\ \mathrm{rad/s}$ against $0.794\ \mathrm{rad/s}$ here, a difference of half a percent. That is exactly the factor $\sqrt{1+(w/g)^{2}} = 1.005$ by which the effective field exceeds $g$ . So the acceleration barely changes how fast the top precesses, while turning its axis through six degrees. The gyroscopic approximation holds comfortably as well, $\omega'/\omega = 0.79/310 = 2.5\times 10^{-3}$ . Dimensionally $ml\sqrt{g^{2}+w^{2}}/(I\omega)$ is $(\mathrm{kg\cdot m}\cdot\mathrm{m/s^{2}})/(\mathrm{kg\cdot m^{2}/s})$ , an inverse time.

Remark

The rate reads the magnitude of the effective field and the axis reads its direction, so a gyroscope carried on an accelerating vehicle borrows its reference direction from the plumb line: the direction a hanging weight in the same block would take is exactly the direction the spinning top precesses about, though the top keeps whatever lean it was given rather than settling onto that direction as the plumb line does. That is the practical trouble with a gyrocompass or an artificial horizon aboard a maneuvering craft, and it is the Newtonian face of the statement that a uniformly accelerated frame and a uniform gravitational field are indistinguishable from the inside. Note that the first-order effect is entirely in the direction: the tilt grows like $w/g$ for small $w$ , while the rate grows only like $(w/g)^{2}/2$ .