Irodov 1.270: The Rod's Cone Needs More Spin – Amit & Deepali Sinha Foundation

Problem Irodov 1.270

A thin uniform rod of length $l$ and mass $m$ hangs from a hinge at its upper end, free to take any inclination. The rod is kept turning about the vertical axis through the hinge at a constant angular velocity $\omega$ , so its lower end sweeps out a horizontal circle and the rod itself sweeps out a cone. Find the steady angle $\theta$ between the rod and the vertical.

The rod as a conical pendulum. Hinged at the top, it leans at a fixed angle θ to the vertical axis and turns about that axis at ω, its lower end running round the dashed horizontal circle. — The rod as a conical pendulum. Hinged at the top, it leans at a fixed angle $\theta$ to the vertical axis and turns about that axis at $\vect\omega$ , its lower end running round the dashed horizontal circle.

Given

$l$
$m$
$\omega$
$g$

Find

the cone angle

\theta

Solution

Step 1Choose the frame in which the rod stands still

Work in the reference frame that rotates with the rod at $\omega$ . Every element of the rod is permanently at rest in that frame, so no Coriolis force appears, and the rod is in equilibrium under three things: its weight, the hinge reaction, and the centrifugal forces of inertia. Taking moments about the hinge kills the reaction, whose line of action passes through it, and leaves a balance between the other two.

N_{\mathrm{cf}} = N_{\mathrm{grav}}

One element of the rod a distance s from the hinge, circling the axis at radius sθ under the horizontal centrifugal force d F, whose arm about the hinge is sθ. The weight m g acts at the rod's midpoint. — One element of the rod a distance $s$ from the hinge, circling the axis at radius $s\sin\theta$ under the horizontal centrifugal force $d\vect F$ , whose arm about the hinge is $s\cos\theta$ . The weight $m\vect g$ acts at the rod’s midpoint.

Step 2The moment of the centrifugal forces about the hinge

Let $s$ measure distance along the rod from the hinge. The element there has mass $dm$ fixed by the uniform linear density and circles the axis at the radius the figure dimensions, which sets the size of its centrifugal force.

dm = \frac{m}{l}\,ds, \qquad dF = \omega^2 (s\sin\theta)\,dm

That force is horizontal, so its arm about the hinge is the vertical drop $s\cos\theta$ , and the moments of all the elements are parallel and simply add.

\begin{align*}N_{\mathrm{cf}} &= \int_0^l (s\cos\theta)\,\omega^2 (s\sin\theta)\frac{m}{l}\,ds \\ &= \frac{m\omega^2\sin\theta\cos\theta}{l}\int_0^l s^2\,ds \\ &= \frac{m\omega^2\sin\theta\cos\theta}{l}\cdot\frac{l^3}{3}\end{align*}

N_{\mathrm{cf}} = \frac{1}{3}\,m\omega^2 l^2\sin\theta\cos\theta

(1)

Step 3The moment of the weight about the hinge

The weight of a uniform rod acts at its midpoint, a distance $l/2$ from the hinge along the rod. It is vertical, so its arm about the hinge is the horizontal offset of that midpoint from the axis.

d_{\mathrm{grav}} = \frac{l}{2}\sin\theta

N_{\mathrm{grav}} = mg\,d_{\mathrm{grav}} = \frac{1}{2}\,mgl\sin\theta

(2)

Step 4Setting the two moments equal

Equating (1) and (2) and canceling the common factor $\sin\theta$ , which is legitimate for any tilted position, leaves a relation for $\cos\theta$ alone.

\begin{aligned} \frac{1}{3}\,m\omega^2 l^2\sin\theta\cos\theta &= \frac{1}{2}\,mgl\sin\theta \\ \frac{1}{3}\,\omega^2 l\cos\theta &= \frac{1}{2}\,g \end{aligned} \tag{3}

\cos\theta = \frac{3g}{2\omega^2 l}

(4)

Check

The right-hand side is a ratio of an acceleration to (angular velocity squared times length), so it is dimensionless as a cosine must be. As $\omega$ grows without bound the cosine tends to zero and the rod approaches the horizontal without ever reaching it, which is right: at exactly horizontal the centrifugal moment vanishes, its arm $s\cos\theta$ going to zero, while the weight’s moment is at its largest, $mgl/2$ , so no balance is possible there. At the other end a tilted solution exists only while $3g/(2\omega^2l)\le1$ ; for slower rotation the equation has no tilted root and the rod simply hangs vertically along the axis, where both moments vanish separately.

Remark

The balance (3) can be read as $I\omega^2\cos\theta = mgd$ , with $I$ the rod’s moment of inertia about the hinge and $d$ the hinge to center of mass distance. A point bob on a string of length $L$ instead satisfies $\cos\theta = g/(\omega^2L)$ , and the two statements agree exactly when $L$ is the combination $I/md$ , which for this rod is $2l/3$ . That combination is the equivalent length of a physical pendulum, the same one that fixes its small-oscillation period, so a single length characterizes the rod in both the swinging and the coning problem. It also explains why the rod is harder to lift off the vertical than a bob at the rod’s own tip: its equivalent length is shorter, so the threshold angular velocity is higher by a factor $(3/2)^{1/2}$ .