Irodov 1.57: The Line Joining Two Rest Points – Amit & Deepali Sinha Foundation

Problem Irodov 1.57

A right circular cone of half-angle $\alpha=30^\circ$ and base radius $R=5.0$ cm rolls without slipping over a horizontal plane, its apex fixed by a hinge at the point $O$ . The cone’s own axis stays horizontal throughout, with the base’s center $C$ always at the same height as $O$ ; at each instant exactly one point of the base rim touches the plane. Point $C$ moves at a constant speed $v=10.0$ cm/s along its circular path around the vertical through $O$ . Find:

the modulus of the cone’s angular velocity, and the angle it makes with the vertical;
the modulus of the cone’s angular acceleration.

The cone in side elevation: apex O hinged at height R, the lower generator touching the plane at the marked point, the axis OC horizontal, the apex angle 2α between the slant lines. C's velocity v points out of the page. — The cone in side elevation: apex $O$ hinged at height $R$ , the lower generator touching the plane at the marked point, the axis $OC$ horizontal, the apex angle $2\alpha$ between the slant lines. $C$ ’s velocity $\vect v$ points out of the page.

Given

$\alpha=30^\circ$
$R=5.0$ cm
$v=10.0$ cm/s, constant

Find

(a)

\omega

and its angle to the vertical, (b)

\beta

Solution

Step 1The line through

O

and the rolling contact is the instantaneous axis

So the cone’s instantaneous angular velocity $\vect\omega$ is directed along the lower generator, the line from $O$ through the contact point. Since the axis $OC$ makes angle $\alpha$ (the half-angle) with every generator, $\vect\omega$ makes angle $\alpha$ with the horizontal axis $OC$ , and therefore angle $90^\circ-\alpha$ with the vertical.

The same elevation with what this step establishes: ω lies along the lower generator, the line joining the hinge O to the rolling contact, and so stands 90°-α from the dashed vertical through O. — The same elevation with what this step establishes: $\vect\omega$ lies along the lower generator, the line joining the hinge $O$ to the rolling contact, and so stands $90^\circ-\alpha$ from the dashed vertical through $O$ .

Step 2The magnitude of

\vect\omega

, from the speed of

C

The axis $OC$ , the base radius $R$ and a generator form a right triangle at $C$ , whose angle at $O$ is the half-angle.

\tan\alpha = \frac{R}{OC}

Solving that for the axis length, a result Step 3 needs again,

OC = \dfrac{R}{\tan\alpha}

(1)

The perpendicular distance from $C$ to the generator line is the leg $OC\sin\alpha$ of the right triangle dropped from $C$ onto that line; substituting (1),

\begin{align*}d &= OC\sin\alpha \\ &= \frac{R}{\tan\alpha}\sin\alpha \\ &= R\cos\alpha\end{align*}

Since $C$ travels on a circle of radius $d$ about the instantaneous axis at speed $v$ ,

v = \omega d = \omega R\cos\alpha.

\omega = \dfrac{v}{R\cos\alpha} = 2.3\ \text{rad/s}

(2)

\text{angle to the vertical} = 90^\circ-\alpha = 60^\circ

(3)

Step 3The precession rate of the whole picture

$C$ travels a horizontal circle of radius $OC$ about the vertical, at speed $v$ , so the precession rate is that speed divided by that radius; substituting (1) for the radius,

\begin{align*}\Omega &= \frac{v}{OC} \\ &= \frac{v}{R/\tan\alpha}\end{align*}

\Omega = \dfrac{v\tan\alpha}{R}

(4)

Step 4The angular acceleration, from

\vect\omega

’s constant-magnitude precession

Since $|\vect\omega|$ is constant and $\vect\omega$ only turns, at rate $\Omega$ , about the vertical, its rate of change is the standard precession formula.

\vect\beta = \frac{d\vect\omega}{dt} = \vect\Omega\times\vect\omega

Taking moduli, with $\theta$ the angle between the vertical $\vect\Omega$ and $\vect\omega$ ,

\beta = \Omega\,\omega\sin\theta.

Step 1 fixed that angle at $90^\circ-\alpha$ , whose sine is $\cos\alpha$ . Substituting it, then (4) for $\Omega$ and (2) for $\omega$ ,

\begin{align*}\beta &= \Omega\,\omega\cos\alpha \\ &= \left(\frac{v\tan\alpha}{R}\right)\left(\frac{v}{R\cos\alpha}\right)\cos\alpha\end{align*}

\beta = \dfrac{v^2\tan\alpha}{R^2} = 2.3\ \text{rad/s}^2

(5)

Check

Dimensionally $v/(R\cos\alpha)$ is a speed over a length, rad/s, and $v^2\tan\alpha/R^2$ is a speed squared over a length squared, rad/s $^2$ , both correct. As a limiting case, a slender cone ( $\alpha\to0$ ) has $OC\to\infty$ at fixed $R$ , so $C$ barely turns and $\Omega\to0$ even as $\omega\to v/R$ stays finite: a needle-thin cone spins almost like a cylinder rolling in a straight line, with vanishing precession, consistent with $\beta=\Omega\omega\cos\alpha\to0$ .

Remark

$\vect\beta$ here is not caused by the cone speeding up: $v$ and hence $\omega$ are both constant. It exists purely because $\vect\omega$ ’s direction sweeps around with the rolling, exactly the mechanism behind a gyroscope’s precession, $\vect\beta=\vect\Omega\times\vect\omega$ , even though nothing here is spinning about its own axis independently of the rolling.