Irodov 1.284: The Elevator Sets the Effective Gravity

Problem Irodov 1.284

A gyroscope stands on the floor of a lift car that is moving upward with the constant acceleration $w = 2.0$ m/s $^2$ . It consists of a uniform disc of radius $R = 5.0$ cm carried at one end of a rod of length $l = 10$ cm, the disc spinning about the rod’s own direction; the rod’s other end is held at a point $O$ by a hinge that lets the rod swing freely in any direction. The rod is horizontal, so that the disc’s center is a horizontal distance $l$ from the hinge, and the rod sweeps about the vertical through $O$ at the steady rate $n = 0.5$ revolutions per second. The rod’s mass and all friction are to be neglected. Find the angular velocity of the disc about its own axis.

The gyroscope on the car's floor. The rod runs from the hinge O to the disc's center a distance l away; the disc spins at ω and the axle's tip circles at Ω. The car accelerates at w. — The gyroscope on the car’s floor. The rod runs from the hinge $O$ to the disc’s center a distance $l$ away; the disc spins at $\vect\omega$ and the axle’s tip circles at $\vect\Omega$ . The car accelerates at $\vect w$ .

Given

$R = 5.0$ cm
$l = 10$ cm
$w = 2.0$ m/s $^2$
$n = 0.5$ rps
$g = 9.8$ m/s $^2$

Find

the disc’s own angular velocity

\omega

Solution

Step 1Work in the car, where gravity is stronger

The hinge $O$ is at rest in the car, not in the ground, so take the car’s own frame. It translates with the constant acceleration $\vect w$ upward, so every mass in it carries the uniform inertial force $-m\vect w$ in addition to its weight, and the two combine into a single effective weight pointing straight down.

m\vect g_{\text{eff}} = m\vect g - m\vect w, \qquad g_{\text{eff}} = g + w

Step 2The moment the effective weight exerts about the hinge

The rod is horizontal, so the effective weight of the disc acts a horizontal distance $l$ from $O$ and its moment arm about the hinge is that full length. The rod itself is massless and contributes nothing.

N = m(g+w)\,l

(1)

Step 3The angular momentum stored in the spin

The disc turns about its own axis at $\omega$ , and for a uniform disc of mass $m$ and radius $R$ the moment of inertia about that axis is $mR^2/2$ . Since the spin is far faster than the precession, the angular momentum of the precessional motion is neglected and $\vect L$ is taken to lie along the rod.

L = \frac{1}{2}mR^2\omega

(2)

The same mounting, with L of along the rod. The arrow at the disc's center is the true weight m g; the inertial force -m w is not drawn. The moment N of is the circled cross. — The same mounting, with $\vect L$ of (2) along the rod. The arrow at the disc’s center is the true weight $m\vect g$ ; the inertial force $-m\vect w$ is not drawn. The moment $\vect N$ of (1) is the circled cross.

Step 4Precession turns that vector without changing its length

The rod stays horizontal, so $\vect L$ keeps its magnitude and simply swings round the vertical through $O$ , sweeping the angle $d\varphi = \Omega\,dt$ in the time $dt$ . The figure is that sweep seen from above: the increment $d\vect L$ closes the narrow isosceles triangle, so it is perpendicular to $\vect L$ and its length is the arc $L\,d\varphi$ .

The tip of L seen from above during a time dt. The vector turns through dφ = Ω dt without changing its length, so the increment d L that closes the triangle is perpendicular to L and has the length L dφ. — The tip of $\vect L$ seen from above during a time $dt$ . The vector turns through $d\varphi = \Omega\,dt$ without changing its length, so the increment $d\vect L$ that closes the triangle is perpendicular to $\vect L$ and has the length $L\,d\varphi$ .

Dividing that increment by $dt$ and using the equation of moments $d\vect L/dt = \vect N$ gives the precession relation.

\begin{align*}|d\vect L| &= L\,d\varphi \\ N = \frac{|d\vect L|}{dt} &= L\,\frac{d\varphi}{dt}\end{align*}

N = \Omega L

(3)

where the precession rate in radians per second follows from the stated $n$ revolutions per second as $\Omega = 2\pi n$ .

Step 5Solving for the spin

Putting the moment (1) and the angular momentum (2) into the precession relation (3) gives one equation in $\omega$ , and the disc’s mass drops out of it because it appears on both sides.

\begin{align*}m(g+w)\,l &= 2\pi n\cdot\frac{1}{2}mR^2\omega \\ (g+w)\,l &= \pi n R^2\omega\end{align*}

With the data,

\omega = \frac{(9.8+2.0)\ \text{m/s}^2\cdot 0.10\ \text{m}}{\pi(0.5\ \text{s}^{-1})(0.050\ \text{m})^2}

\omega = \frac{(g+w)l}{\pi n R^2} = 3.0\times 10^2\ \text{rad/s}

(4)

Check

The units come out right: $(\text{m/s}^2)(\text{m})/(\text{s}^{-1}\,\text{m}^2) = \text{s}^{-1}$ . More important, the answer justifies the approximation behind (2). What was dropped is the angular momentum of the precession itself, $(ml^2+mR^2/4)\Omega$ about the vertical through $O$ , and the right comparison is against the spin angular momentum $mR^2\omega/2$ that was kept, not against the two rates. The mass cancels between them, leaving $(l^2+R^2/4)\Omega\big/(R^2\omega/2) = (0.010625)(3.14)/(0.376) = 0.089$ , so the kept part is the larger by about eleven to one. The margin is narrower than the ratio of the rates suggests, because $l^2$ dwarfs $R^2/2$ , but it goes the right way. Here the dropped part in fact costs nothing at all: with the rod horizontal and the precession steady it is a constant vertical vector, so it contributes nothing to $d\vect L/dt$ and (3) is exact.

Remark

The car enters the answer only through the combination $g+w$ , which is the whole content of the problem: a gyroscope is a gravimeter. Watching the precession rate of a known top is a way of measuring the effective gravity of whatever vehicle carries it, and in a freely falling car, $w = -g$ , the effective weight vanishes, the moment about the hinge vanishes with it, and the axle simply keeps pointing wherever it was left. The disc’s mass never appears either, since it multiplies the moment and the angular momentum alike, so a heavier disc of the same shape precesses at exactly the same rate.