Irodov 1.146: Friction Steers a Precessing Rolling Cone

Problem Irodov 1.146

A solid cone $A$ of mass $m=3.2$ kg and half-angle $\alpha=10^\circ$ rolls, without slipping, on the inside of a second, fixed conical surface $B$ , in such a way that cone $A$ ’s apex, at a point $O$ shared with surface $B$ ’s own apex, stays motionless throughout the motion. Cone $A$ ’s own axis is horizontal (its center of gravity sits at the same height as $O$ ), at a distance $l=17$ cm from $O$ along that axis, and this axis sweeps steadily around the vertical through $O$ at constant angular velocity $\omega$ , carrying the whole cone around with it while it spins about its own axis fast enough to keep the rolling contact slip-free.

Find the static friction force that surface $B$ exerts on cone $A$ when $\omega=1.0$ rad/s.
Given a coefficient of friction $k=0.25$ between the two surfaces, find the values of $\omega$ for which cone $A$ can still roll without sliding.

Meridian section through the vertical axis and cone A's horizontal axis. The lower generatrix, at angle α below that axis, contacts the hatched surface B; the center of gravity (dot) sits a distance l from O, and the assembly precesses at ω. — Meridian section through the vertical axis and cone $A$ ’s horizontal axis. The lower generatrix, at angle $\alpha$ below that axis, contacts the hatched surface $B$ ; the center of gravity (dot) sits a distance $l$ from $O$ , and the assembly precesses at $\omega$ .

Given

$m=3.2\text{ kg}$
$\alpha=10^\circ$
$l=17\text{ cm}$
$\omega=1.0\text{ rad/s}$ (part a)
$k=0.25$ (part b)

Find

(a)

F_{fr}

; (b) the range of

\omega

for rolling without slipping

Solution

Step 1The contact line’s direction is fixed by cone

A

’s own half-angle

Cone $A$ ’s axis is horizontal, and its lateral surface makes angle $\alpha$ with that axis all the way around; only the lower generatrix, at angle $\alpha$ below the horizontal axis, can touch the surrounding surface $B$ from inside. So, in the vertical plane containing the instantaneous axis (the meridian plane), the contact line points away from $O$ at angle $\alpha$ below horizontal, and the reaction normal to it points at angle $(90^\circ-\alpha)$ above horizontal, away from $B$ ’s material.

Step 2Newton’s second law for the center of mass

The same section with the three forces on cone A: the normal reaction N and the static friction F_fr at the contact generatrix, and the weight m g at the center of gravity. — The same section with the three forces on cone $A$ : the normal reaction $N$ and the static friction $F_{fr}$ at the contact generatrix, and the weight $m\vect g$ at the center of gravity.

The center of gravity moves on a horizontal circle of radius $l$ at constant $\omega$ , so its acceleration is purely centripetal, magnitude $\omega^2 l$ , horizontal, toward the vertical axis; it has no vertical component. Resolving the three forces (weight $m\vect g$ , normal $N$ , friction $F_{fr}$ ) along the vertical and along the horizontal radial direction,

\begin{aligned} N\cos\alpha - F_{fr}\sin\alpha - mg &= 0\\ N\sin\alpha + F_{fr}\cos\alpha &= -m\omega^2 l \end{aligned} \tag{1}

where the horizontal equation carries a minus sign on the right because the centripetal acceleration points inward while the stated positive sense for $N$ and $F_{fr}$ ’s components was taken outward.

Step 3Solve for

N

and

F_{fr}

The coefficient matrix is a rotation by $\alpha$ , so it inverts by the same rotation through $-\alpha$ , giving $N$ and the friction’s outward-sense component (the sense (1) fixed as positive) directly; the normal reaction is boxed for reuse in Step 5.

\begin{aligned} F_{fr,\text{outward}} &= -\bigl(mg\sin\alpha + m\omega^2 l\cos\alpha\bigr) \end{aligned} \tag{2}

N = mg\cos\alpha - m\omega^2 l\sin\alpha

(3)

Step 4Evaluate part (a)

Taking the magnitude in that true, up-slope sense and substituting $m=3.2\text{ kg}$ , $g=9.8\text{ m/s}^2$ , $\alpha=10^\circ$ , $l=0.17\text{ m}$ , $\omega=1.0\text{ rad/s}$ ,

\begin{align*}F_{fr} &= mg\left(\sin\alpha+\frac{\omega^2 l}{g}\cos\alpha\right)\\ &= (3.2\text{ kg})(9.8\text{ m/s}^2)\left[\sin10^\circ+\frac{(1.0\text{ rad/s})^2(0.17\text{ m})}{9.8\text{ m/s}^2}\cos10^\circ\right]\end{align*}

F_{fr} = mg\left(\sin\alpha+\dfrac{\omega^2 l}{g}\cos\alpha\right) = 6.0\text{ N}

(4)

Step 5The no-slip condition bounds

\omega

Setting $F_{fr}=kN$ at the threshold and dividing through by $\cos\alpha$ ,

\begin{align*}mg\sin\alpha+m\omega^2 l\cos\alpha &= k\bigl(mg\cos\alpha-m\omega^2 l\sin\alpha\bigr)\\ g\tan\alpha+\omega^2 l &= kg-k\omega^2 l\tan\alpha\\ \omega^2 l(1+k\tan\alpha) &= g(k-\tan\alpha)\end{align*}

Solving for $\omega$ gives the bound; Step 6 evaluates it numerically.

Step 6Evaluate part (b)

With $k=0.25$ , $\alpha=10^\circ$ ( $\tan\alpha=0.176$ ), $g=9.8\text{ m/s}^2$ , $l=0.17\text{ m}$ ,

\begin{align*}\omega_{\text{max}} &= \sqrt{\frac{g(k-\tan\alpha)}{l(1+k\tan\alpha)}}\\ &= \sqrt{\frac{(9.8\text{ m/s}^2)(0.25-0.176)}{(0.17\text{ m})(1+0.25\times0.176)}}\end{align*}

\omega < \sqrt{\dfrac{g(k-\tan\alpha)}{l(1+k\tan\alpha)}} \approx 2.0\text{ rad/s}

(5)

Check

Dimensionally, $[g][\tan\alpha]/[l]=(\text{m/s}^2)/\text{m}=\text{s}^{-2}$ , and its square root is an angular velocity, as required. If $k\le\tan\alpha$ the bound under the square root is zero or negative and no $\omega$ at all permits rolling without slipping; here $k=0.25>\tan10^\circ=0.176$ , so a nonzero range does exist, consistent with the numeric answer.

Remark

Because $N$ decreases and $F_{fr}$ increases together as $\omega$ grows, the cone is always closest to slipping at the fastest precession it is asked to sustain, never at the slowest: a governor or a centrifugal clutch built on this same rolling-cone-on-cone geometry would fail by slipping at high speed, not by stalling at low speed.