Irodov 1.54: How Sharply the Rim Points Turn – Amit & Deepali Sinha Foundation

Problem Irodov 1.54

A cylinder of radius $r$ rolls without slipping over a horizontal plane. Consider the rim point $O$ currently touching the plane, the point $A$ diametrically opposite $O$ , and the point $B$ at the end of the diameter perpendicular to $OA$ , on the leading side (ahead of the center, in the sense of the cylinder’s motion). Each of $A$ and $B$ traces a cycloid as the cylinder rolls. Find the radius of curvature of the path traced by $A$ , and the radius of curvature of the path traced by $B$ , at the instants each occupies that named position.

The cylinder at one instant: O the point on the ground, A diametrically opposite it, B at the end of the perpendicular diameter on the leading side, v the center's velocity. Each point is drawn where the problem names it. — The cylinder at one instant: $O$ the point on the ground, $A$ diametrically opposite it, $B$ at the end of the perpendicular diameter on the leading side, $\vect v$ the center’s velocity. Each point is drawn where the problem names it.

Given

radius $r$
rolling without slipping

Find

radius of curvature of the trajectory at

A

; radius of curvature of the trajectory at

B

Solution

Step 1The cycloid traced by a rim point

With the point starting at the origin on the ground, and the center’s position $x_C=r\varphi$ from the no-slip condition,

x(\varphi) = r(\varphi-\sin\varphi), \qquad y(\varphi) = r(1-\cos\varphi).

Step 2Curvature radius as a function of the turned angle

The radius of curvature of a plane curve given parametrically is $R_c=(x'^2+y'^2)^{3/2}/|x'y''-y'x''|$ , with primes denoting $d/d\varphi$ .

x' = r(1-\cos\varphi), \quad y' = r\sin\varphi, \quad x'' = r\sin\varphi, \quad y'' = r\cos\varphi

\begin{align*}x'y''-y'x'' &= r^2(1-\cos\varphi)\cos\varphi - r^2\sin^2\varphi \\ &= r^2(\cos\varphi-1) \\ x'^2+y'^2 &= r^2(1-\cos\varphi)^2+r^2\sin^2\varphi \\ &= 2r^2(1-\cos\varphi)\end{align*}

Combining, and using $1-\cos\varphi=2\sin^2(\varphi/2)$ ,

\begin{align*}R_c(\varphi) &= \frac{\bigl[2r^2(1-\cos\varphi)\bigr]^{3/2}}{r^2(1-\cos\varphi)} \\ &= 2r\sqrt{2(1-\cos\varphi)} \\ &= 4r\sin(\varphi/2).\end{align*}

R_c(\varphi) = 4r\sin(\varphi/2)

(1)

Step 3Evaluating at

A

and at

B

Point $A$ , diametrically opposite the ground, is reached after a half turn, $\varphi=\pi$ , where $x-x_C=-r\sin\pi=0$ places it directly above the center, as required. Point $B$ sits at the same height as the center ( $y=r$ , so $\cos\varphi=0$ ) on the leading side, $x-x_C=-r\sin\varphi>0$ , which selects $\varphi=3\pi/2$ (three-quarter turn) over $\varphi=\pi/2$ , since $\sin(3\pi/2)=-1<0$ . Substituting into (1),

R_A = 4r\sin(\pi/2), \qquad R_B = 4r\sin(3\pi/4) = 4r\sin(\pi/4).

R_A = 4r

(2)

R_B = 2\sqrt2\,r

(3)

Check

At $\varphi\to0$ (a point just leaving the ground) $R_c\to0$ : the cycloid has a cusp at the ground, an infinitely sharp turn, which is the well known feature of rolling motion that this formula reproduces automatically. And $R_A=4r>R_B=2\sqrt2\,r\approx2.83r$ , consistent with $A$ (the fastest-moving rim point, speed $2v_C$ ) sweeping the gentlest curve of the three.

Remark

The curvature radius $4r\sin(\varphi/2)$ is exactly twice the chord $O$ -to-point-at- $\varphi$ , which has length $2r\sin(\varphi/2)$ (a standard circle chord): the trajectory curves exactly as sharply as a circle of radius equal to twice the rim point’s instantaneous distance from the contact would, which is the same relation that makes the contact point itself, at $\varphi=0$ , a genuine cusp rather than merely a tight bend.