Irodov 1.52: Rolling Points Always Accelerate Toward Center

Problem Irodov 1.52

A wheel of radius $R=0.50$ m rolls without slipping along a horizontal surface, the center of the wheel moving at a constant speed $v=1.00$ m/s. A point $A$ is marked on the rim. Find:

the modulus and the direction of the acceleration vector of $A$ , at a general instant;
the total path length $s$ traced out by $A$ between two successive instants at which it touches the ground.

A rim point A at a general instant. The wheel rolls without slipping on the hatched surface below, its center traveling at the constant velocity v, which is drawn beside the center because motion indicators go off the body, not on it. — A rim point $A$ at a general instant. The wheel rolls without slipping on the hatched surface below, its center traveling at the constant velocity $\vect v$ , which is drawn beside the center because motion indicators go off the body, not on it.

Given

$R=0.50$ m
$v=1.00$ m/s, constant

Find

(a)

|\vect w_A|

and its direction, (b)

s

between successive contacts

Solution

Step 1The rim point’s acceleration, from the rigid body formula

With $\vect w_C=\vect 0$ and $\vect\beta=\vect 0$ , only the centripetal term survives for any rim point $P$ , in particular for $A$ .

\vect w_A = -\omega^2\vect r_{A/C}

Substituting $\omega=v/R$ and taking the modulus, since $|\vect r_{A/C}|=R$ ,

|\vect w_A| = \omega^2 R = \frac{v^2}{R^2}\cdot R = \frac{v^2}{R}.

w_A = \dfrac{v^2}{R} = 2.0\ \text{m/s}^2

(1)

directed from $A$ straight toward the center $C$ at every instant, since $\vect w_A=-\omega^2\vect r_{A/C}$ is antiparallel to $\vect r_{A/C}$ for any position of $A$ on the rim.

The same wheel with the direction this step establishes: w_A runs from the rim point A straight to the center, and does so wherever A happens to sit on the rim. — The same wheel with the direction this step establishes: $\vect w_A$ runs from the rim point $A$ straight to the center, and does so wherever $A$ happens to sit on the rim.

Step 2The rim point’s speed, as a function of the turned angle

Let $\varphi=\omega t$ be the angle turned since $A$ left the ground. With the contact point instantaneously at rest, $A$ ’s velocity has modulus $2R\omega\sin(\varphi/2)$ (the classic cycloid speed, reached by resolving $\vect v_A=\vect v_C+\vect\omega\times\vect r_{A/C}$ at general $\varphi$ ), so the element of arc length is

ds = |\vect v_A|\,dt = 2R\omega\sin(\varphi/2)\,dt = 2R\sin(\varphi/2)\,d\varphi,

using $d\varphi=\omega\,dt$ .

Step 3Integrating over one full turn

Two successive contacts with the ground are exactly one full turn apart, $\varphi:0\to2\pi$ .

\begin{align*}s &= \int_0^{2\pi} 2R\sin(\varphi/2)\,d\varphi \\ &= 2R\Bigl[-2\cos(\varphi/2)\Bigr]_0^{2\pi} \\ &= 2R\bigl[(-2\cos\pi)-(-2\cos0)\bigr] \\ &= 2R(2+2)\end{align*}

s = 8R = 4.0\ \text{m}

(2)

Check

Dimensionally $v^2/R$ is (m/s) $^2$ /m = m/s $^2$ , correct for an acceleration. And $s=8R=4.0$ m exceeds the straight-line distance between the two contact points, $2\pi R=3.1$ m, exactly as a curved path must exceed the chord it spans.

Remark

The length $s=8R$ , four rim diameters, is a purely geometric property of the cycloid: it never depended on $v$ at all, only on $R$ , even though the whole derivation ran through a time integral with $\omega$ in it. Every rim point traces a congruent arch, so this same $8R$ is the answer for any marked point on any uniformly rolling wheel, not just the one on the ground at $t=0$ .