Irodov 1.264: The Speed That Clears the Edge – Amit & Deepali Sinha Foundation

Problem Irodov 1.264

A uniform solid cylinder of radius $R = 15$ cm rolls without sliding along a horizontal floor that meets, at a sharp edge, an inclined section descending at angle $\alpha = 30^\circ$ to the horizontal. The direction of the center’s velocity must change abruptly at that corner. Find the largest approach speed $v_0$ of the cylinder’s center for which it still stays on the surface all the way round the edge, without leaving it (without jumping off the corner).

The horizontal surface breaks at a sharp edge into an incline falling away at α; the dashed line continues the horizontal so the angle can be read off. — The horizontal surface breaks at a sharp edge into an incline falling away at $\alpha$ ; the dashed line continues the horizontal so the angle can be read off.

Given

$R = 15$ cm
$\alpha = 30^\circ$
rolling without sliding

Find

Find the largest

v_0

for which the cylinder stays in contact all the way round the edge

Solution

Step 1What the cylinder actually does at the edge

The break in the surface is a sharp corner, not a rounded fillet. While the cylinder is on the horizontal plane it touches along a line directly below its axis; once that line reaches the corner there is nothing ahead of it to touch, because the ground has fallen away. From that moment until the cylinder comes down flat on the incline, the only point of the body in contact with anything is the one resting on the corner itself. The cylinder therefore turns about the edge, and its center swings on a circle of radius $R$ centered at the edge.

Let $\theta$ be the angle between the vertical and the line from the edge to the center. It runs from $\theta = 0$ , the instant the contact line arrives at the corner, to $\theta = \alpha$ , at which point the line from the edge to the center is normal to the incline and the cylinder settles onto it.

Turning about the edge. The cylinder touches only the corner, so the normal part N of the reaction lies along the radius, and the center travels on a circle of radius R about the corner while θ grows from 0 to α. — Turning about the edge. The cylinder touches only the corner, so the normal part $\vect N$ of the reaction lies along the radius, and the center travels on a circle of radius $R$ about the corner while $\theta$ grows from $0$ to $\alpha$ .

Step 2No energy is lost when the turn begins

A rolling cylinder turns instantaneously about its contact line, and the contact line at the moment of arrival is the corner. The instantaneous axis therefore does not jump when the turn begins: the body’s velocity field is the same just before and just after, so no impulsive reaction acts and no kinetic energy is lost. Throughout the turn the contact point stays fixed at the edge, so the reaction does no work either.

Step 3Speed as a function of the angle turned through

Take the edge as the axis. For a uniform solid cylinder $I_c = mR^2/2$ about its own axis, and the parallel axis theorem gives its moment of inertia about the edge:

I_O = I_c + mR^2 = \frac{mR^2}{2} + mR^2 = \frac{3mR^2}{2}.

Rolling without sliding fixes the angular velocity as $\omega = v/R$ , where $v$ is the speed of the center. Now write conservation of energy between $\theta = 0$ , where the speed is $v_0$ and the center sits a height $R$ above the edge, and a general angle $\theta$ , where the center is at height $R\cos\theta$ . The center has therefore descended by $R(1 - \cos\theta)$ , and

\begin{aligned} \frac{1}{2}I_O\omega^2 &= \frac{1}{2}I_O\omega_0^2 + mgR(1 - \cos\theta), \\ \frac{1}{2}\cdot\frac{3mR^2}{2}\cdot\frac{v^2}{R^2} &= \frac{1}{2}\cdot\frac{3mR^2}{2}\cdot\frac{v_0^2}{R^2} + mgR(1 - \cos\theta), \\ \frac{3}{4}v^2 &= \frac{3}{4}v_0^2 + gR(1 - \cos\theta). \end{aligned}

Multiplying through by $4/3$ gives the speed at any angle of the turn:

v^2 = v_0^2 + \dfrac{4}{3}gR\,(1 - \cos\theta)

(1)

Step 4The condition for staying on the edge

The center travels on a circle of radius $R$ about the edge, so its acceleration has a component $v^2/R$ directed at the edge. Two things act on the cylinder: the weight $mg$ straight down, and the reaction of the edge. That reaction is not purely radial, because the cylinder turns about the corner without sliding and it is static friction at the corner that keeps it turning. Moments about the edge settle the tangential motion: only the weight has a moment there, at the arm $R\sin\theta$ , and the moment of inertia about the edge is the $I_O$ computed above.

\begin{align*}I_O\,\dot\omega &= mgR\sin\theta, \\ \frac{3mR^2}{2}\,\dot\omega &= mgR\sin\theta, \\ R\dot\omega &= \frac{2}{3}\,g\sin\theta.\end{align*}

The tangential component of Newton’s second law for the center, whose tangential acceleration is that same $R\dot\omega$ , then leaves the friction the corner has to supply:

\begin{align*}m\,R\dot\omega &= mg\sin\theta - f, \\ \frac{2}{3}\,mg\sin\theta &= mg\sin\theta - f, \\ f &= \frac{1}{3}\,mg\sin\theta .\end{align*}

What matters for the contact condition is that this friction is tangential, so it drops out of the radial equation entirely: only the normal component $\vect N$ , along the line from the edge to the center, appears below.

The weight resolved on the axes the turn defines: n points from the edge out to the center and τ along the direction of travel. The inward radial pull available is mgθ. — The weight resolved on the axes the turn defines: $n$ points from the edge out to the center and $\tau$ along the direction of travel. The inward radial pull available is $mg\cos\theta$ .

Resolving along the inward radial direction $n$ , the weight contributes $mg\cos\theta$ inward and the reaction $N$ contributes $N$ outward, so Newton’s second law along that direction reads

mg\cos\theta - N = \frac{mv^2}{R}. \tag{2}

The corner can push the cylinder away but cannot pull it back, so contact survives only while $N \ge 0$ . Rearranging (2) for the normal force,

N = m\left(g\cos\theta - \frac{v^2}{R}\right),

so the contact condition, multiplied through by $R/m$ , is a ceiling on the speed at each angle of the turn:

v^2 \le gR\cos\theta.

Step 5Combining the two, and finding where the condition bites

Substitute the speed from (1) into the contact condition, so that everything is expressed in $v_0$ and $\theta$ :

\begin{aligned} v_0^2 + \frac{4}{3}gR(1 - \cos\theta) &\le gR\cos\theta, \\ v_0^2 &\le gR\cos\theta - \frac{4}{3}gR + \frac{4}{3}gR\cos\theta, \\ v_0^2 &\le \frac{1}{3}gR\,(7\cos\theta - 4). \end{aligned}

This has to hold for every $\theta$ from $0$ to $\alpha$ . The right-hand side falls monotonically as $\theta$ grows, because $\cos\theta$ does, so the tightest demand is the last instant of the turn, $\theta = \alpha$ :

v_0^2 \le \dfrac{1}{3}gR\,(7\cos\alpha - 4)

(3)

Taking the equality gives the largest speed that still gets the cylinder round. With $g = 9.8$ m/s $^2$ , $R = 0.15$ m and $\cos30^\circ = 0.866$ , the bracket evaluates to

7\cos\alpha - 4 = 7(0.866) - 4 = 2.06,

and substituting it back gives the speed:

v_{0,\max} = \sqrt{\frac{(9.8)(0.15)(2.06)}{3}} = \sqrt{1.01}\ \text{m/s}.

v_{0,\max} = \sqrt{\frac{1}{3}gR\,(7\cos\alpha - 4)} = 1.0\ \text{m/s}

(4)

Check

Send $\alpha \to 0$ , a surface with the faintest possible kink. The formula gives $v_0^2 = \frac{1}{3}gR(7 - 4) = gR$ , and that is exactly what (2) demands at $\theta = 0$ read on its own: with the center directly above the edge, $mg - N = mv_0^2/R$ , so $N \ge 0$ needs $v_0^2 \le gR$ . The two routes agree, which they must, since $\theta = 0$ is the first instant of every turn. Dimensions also close: $gR$ has units m $^2$ /s $^2$ , so its square root is a speed.

Remark

Two things worth noticing. First, the answer contains no mass, and $R$ enters only as an overall scale: the shape has entered solely through $I_O/mR^2 = 3/2$ , so every uniform solid cylinder behaves the same way and a sphere or a hoop would carry a different numerical factor. Second, the bracket $7\cos\alpha - 4$ vanishes at $\alpha = \arccos(4/7) = 55^\circ$ . On an incline steeper than that no speed works at all, not even $v_0 = 0$ : a cylinder set down gently on such an edge tips over it and loses contact before it lands.