Irodov 1.283: Tilt Cancels Out of Precession – Amit & Deepali Sinha Foundation

Problem Irodov 1.283

A top of mass $m = 0.50\ \mathrm{kg}$ spins about its own symmetry axis with an angular velocity $\omega = 350\ \mathrm{rad/s}$ , its moment of inertia about that axis being $I = 2.0\ \mathrm{g\cdot m^{2}}$ . The top rests on its tip at a fixed point of support, its axis leans at $\theta = 30^{\circ}$ to the vertical, and the distance from the point of support to the center of inertia is $l = 10\ \mathrm{cm}$ . Under gravity the axis sweeps steadily around the vertical at a constant lean. Find:

the angular velocity $\omega'$ of this precession;
the magnitude and the direction of the horizontal component of the force with which the support acts on the top.

The top on its support, its axis leaning θ from the vertical dashed line, the center of inertia a distance l out along it. The top spins about that axis, which sweeps steadily around the vertical. — The top on its support, its axis leaning $\theta$ from the vertical dashed line, the center of inertia a distance $l$ out along it. The top spins about that axis, which sweeps steadily around the vertical.

Given

$m = 0.50\ \mathrm{kg}$
$\theta = 30^{\circ}$
$I = 2.0\ \mathrm{g\cdot m^{2}}$
$\omega = 350\ \mathrm{rad/s}$
$l = 10\ \mathrm{cm}$
$g = 9.8\ \mathrm{m/s^{2}}$

Find

(a)

\omega'

, (b) the horizontal component

F

of the support’s reaction, magnitude and direction

Solution

Step 1The angular momentum lies along the axis

A rapidly spinning symmetric top carries an angular momentum $I\omega$ about its own axis, and the precession adds only a small further component perpendicular to it. Neglecting that extra piece, which is the gyroscopic approximation and is checked at the end, the whole angular momentum about the point of support is directed along the axis.

L = I\omega, \qquad \text{directed along the symmetry axis}

Step 2The torque of the weight about the point of support

The only forces on the top are the weight at the center of inertia and the reaction at the support, and the reaction has no moment about the support because it acts there. The moment arm of the weight is the horizontal offset of the center of inertia from the support.

\begin{align*}\vect\tau &= \vect l\times m\vect g \\ \tau &= mgl\sin\theta\end{align*}

This torque is horizontal and perpendicular to the vertical plane holding the axis, so it is perpendicular to $\vect L$ as well.

Step 3The precession rate from the rate of turning of

\vect L

The angular momentum L lies along the axis, the weight m g at the center of inertia a distance l out, the torque about the support the circled cross. The dashed cone is what the axis sweeps, the loop the sense of ω'. — The angular momentum $\vect L$ lies along the axis, the weight $m\vect g$ at the center of inertia a distance $l$ out, the torque about the support the circled cross. The dashed cone is what the axis sweeps, the loop the sense of $\vect\omega'$ .

Since $\vect L$ keeps a constant length and a constant lean, its tip travels a horizontal circle of radius $L\sin\theta$ at the precession rate, so the magnitude of its rate of change follows from that circular motion. Equating it to the torque gives the rate.

\begin{align*}\left|\frac{d\vect L}{dt}\right| &= \omega' L\sin\theta = \omega' I\omega\sin\theta \\ \omega' I\omega\sin\theta &= mgl\sin\theta\end{align*}

The lean cancels on both sides, because the torque and the swinging part of $\vect L$ carry the same factor $\sin\theta$ , and dividing through by $I\omega\sin\theta$ leaves the rate alone.

\omega' = \frac{mgl}{I\omega} = \frac{0.50\times 9.8\times 0.10}{2.0\times 10^{-3}\times 350} = 0.70\ \mathrm{rad/s}

(1)

Step 4The horizontal reaction is what holds the center of inertia on its circle

The center of inertia is not at rest: it rides a horizontal circle of radius $l\sin\theta$ about the vertical through the support, once per precession turn. Its acceleration is therefore horizontal and center-seeking, and the weight is vertical, so the horizontal part of that acceleration can only be supplied by the reaction at the support.

The same motion from above. The center of inertia C runs counterclockwise on the circle of radius lθ about the vertical at O', with velocity v tangent. The horizontal reaction F points from C at the axis, turning with C at rate ω'. — The same motion from above. The center of inertia $C$ runs counterclockwise on the circle of radius $l\sin\theta$ about the vertical at $O'$ , with velocity $\vect v$ tangent. The horizontal reaction $\vect F$ points from $C$ at the axis, turning with $C$ at rate $\omega'$ .

\begin{align*}a_n &= \omega'^{2}\,(l\sin\theta) \\ F &= m a_n\end{align*}

With the data,

F = 0.50\times(0.70)^{2}\times 0.10\times\sin 30^{\circ}

F = m\omega'^{2}l\sin\theta = 12\ \mathrm{mN}

(2)

Horizontal, from the center of inertia toward the precession axis.

Check

The gyroscopic approximation, $\vect L$ taken along the axis, needs $\omega'\ll\omega$ , and here $\omega'/\omega = 0.70/350 = 2\times 10^{-3}$ . What it drops is the transverse piece of $\vect L$ carried by the precession itself, $I_{\perp}\omega'\sin\theta$ against the retained $I\omega$ , so the neglected part stands to the kept one in the ratio $(I_{\perp}/I)(\omega'/\omega)\sin\theta$ ; with $\sin\theta = 1/2$ and a transverse moment of the same order as $I$ , that is of order $10^{-3}$ and the approximation is excellent. The reaction’s horizontal part is likewise tiny beside its vertical part: $F/mg = 0.01225/4.9 = 2.5\times 10^{-3}$ , as it must be for a center of inertia creeping round a $5\ \mathrm{cm}$ circle once every nine seconds. Dimensionally $mgl/(I\omega)$ is $(\mathrm{kg\cdot m^{2}/s^{2}})/(\mathrm{kg\cdot m^{2}/s})$ , an inverse time.

Remark

The lean drops out of the precession rate but not out of the reaction, and the difference is worth holding onto: $\sin\theta$ enters the torque and the swinging part of $\vect L$ identically, so it cancels from $\omega'$ , while in $F$ it survives as the radius of the circle the center of inertia actually travels. So a given top precesses at one rate whatever angle it is set at, but leans harder on its support the further over it is. That horizontal reaction is not a fixed pull either: it rotates once per precession period, and the contact can only supply a sideways force through friction. Where the table is too smooth to supply it, the tip slips: with no horizontal force acting at all, the center of inertia stays on one fixed vertical line and it is the tip that circles about it.