Irodov 1.263: Rolling Delays the Ball's Departure – Amit & Deepali Sinha Foundation

Problem Irodov 1.263

A uniform solid ball of radius $r$ starts, with negligible speed, from the highest point of a fixed sphere of radius $R$ and rolls down the sphere’s outer surface without slipping. At some point of the descent the ball leaves the surface. Find the ball’s angular velocity at that instant.

The ball of radius r on the outside of the fixed sphere of radius R. Its center stays a distance R+r from the sphere's center O, and θ is the angle from the upward vertical (dashed) to the radius through the contact. — The ball of radius $r$ on the outside of the fixed sphere of radius $R$ . Its center stays a distance $R+r$ from the sphere’s center $O$ , and $\theta$ is the angle from the upward vertical (dashed) to the radius through the contact.

The ball alone on the sphere's local tangent plane at the same instant, showing the center's velocity v, the spin ω, the marked contact point, and the radius line of length r from the center to the rim. — The ball alone on the sphere’s local tangent plane at the same instant, showing the center’s velocity $\vect v$ , the spin $\omega$ , the marked contact point, and the radius line of length $r$ from the center to the rim.

Given

$R$
$r$
rolls without slipping
negligible initial speed

Find

the angular velocity

\omega

at the moment the ball breaks off

Solution

Step 1The kinetic energy of a ball that rolls

Let $m$ be the ball’s mass, $v$ the speed of its center and $\omega$ its angular velocity. Rolling without slipping means the contact point is momentarily at rest, so the center’s speed and the spin are locked together, and the moment of inertia is that of a uniform solid ball about a central axis.

v = \omega r, \qquad I = \frac{2}{5}mr^2 \tag{1}

Adding the translational and rotational parts and eliminating $\omega$ ,

\begin{align*}K &= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \\ &= \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mr^2\cdot\frac{v^2}{r^2} \\ &= \frac{1}{2}mv^2 + \frac{1}{5}mv^2\end{align*}

K = \dfrac{7}{10}mv^2

(2)

Step 2Energy conservation fixes the speed at every angle

Both surface forces, the normal force and the static friction, are applied at the contact point, which rolling without slipping holds instantaneously at rest, so neither of them does any work and all the potential energy released goes into the kinetic energy (2). The center starts at height $R+r$ above $O$ and at angle $\theta$ it stands at height $(R+r)\cos\theta$ , so the drop is

h = (R+r)(1 - \cos\theta)

and equating the energy released to the kinetic energy gained,

\begin{align*}\frac{7}{10}mv^2 &= mgh \\ &= mg(R+r)(1-\cos\theta)\end{align*}

v^2 = \dfrac{10}{7}g(R+r)(1-\cos\theta)

(3)

Step 3The break-off condition

While the ball stays on the sphere its center travels on a circle of radius $R+r$ about $O$ , so the resultant force along that radius must supply the centripetal acceleration. The two forces are the sphere’s push, which can only be outward along the radius, and the weight, which makes the angle $\theta$ with that radius.

The two forces on the ball at the angle θ: the normal force N outward along the radius through the contact, away from O, and the weight m g vertical at the ball's center. — The two forces on the ball at the angle $\theta$ : the normal force $\vect N$ outward along the radius through the contact, away from $O$ , and the weight $m\vect g$ vertical at the ball’s center.

Resolving the weight along the inward radius and subtracting the outward normal force,

\frac{mv^2}{R+r} = mg\cos\theta - N

The ball breaks off at the instant the surface stops pushing, that is when $N$ falls to zero, and putting $N=0$ into the line above leaves

v^2 = g(R+r)\cos\theta

(4)

Step 4The angle at which the ball leaves

Both (3) and (4) give the same $v^2$ , and they can only agree at one angle. Setting them equal and canceling the common factor $g(R+r)$ ,

\begin{align*}\frac{10}{7}g(R+r)(1-\cos\theta) &= g(R+r)\cos\theta \\ 10(1-\cos\theta) &= 7\cos\theta \\ 10 &= 17\cos\theta\end{align*}

\cos\theta = \dfrac{10}{17}

(5)

Neither $R$ nor $r$ survives, so a rolling ball always leaves at this same angle.

Step 5The angular velocity at that instant

Substituting the break-off angle (5) into (4) gives the speed of the center there, and the rolling condition (1) converts it into the spin.

\begin{align*}v^2 &= g(R+r)\cdot\frac{10}{17} \\ \omega^2 &= \frac{v^2}{r^2} = \frac{10g(R+r)}{17r^2}\end{align*}

\omega = \sqrt{\dfrac{10g(R+r)}{17r^2}}

(6)

Check

Rebuild the speed (3) without invoking energy at all. Along the surface Newton’s second law for the center reads $m\,dv/dt = mg\sin\theta - f$ , while friction is the only force with a moment about the ball’s own center, $fr = \frac{2}{5}mr^{2}\,d\omega/dt$ ; rolling keeps $v = \omega r$ at every instant, so $f = \frac{2}{5}m\,dv/dt$ and the two equations together leave $dv/dt = \frac{5}{7}g\sin\theta$ . Writing $dv/dt = v\,dv/ds$ with $ds = (R+r)\,d\theta$ and integrating from rest at $\theta = 0$ returns $v^{2} = \frac{10}{7}g(R+r)(1-\cos\theta)$ , which is (3) term for term, reached by forces rather than by energy. For a fixed path radius $R+r$ the boxed answer scales as $1/r$ , as it must, since a smaller ball has to spin faster to roll at the same speed. Dimensionally $g(R+r)/r^2$ carries units of $\mathrm{s}^{-2}$ , whose square root is indeed an angular velocity. The ball’s mass canceled from every equation, as it must when gravity is the only force doing work.

Remark

Compare the same descent with the friction switched off, so the ball slides instead of rolling. Then all the released energy goes into the center’s motion, the energy balance reads $v^2 = 2g(R+r)(1-\cos\theta)$ , and the same break-off condition gives $\cos\theta = 2/3$ . Rolling therefore delays the departure, because a rolling ball has to spend part of the released energy on spin and so arrives at any given angle moving more slowly, which lets the surface keep curving away underneath it for longer. Both cases are one formula: writing the moment of inertia as $I = kmr^2$ , the ball leaves at $\cos\theta = 2/(3+k)$ , which is $2/3$ for the sliding particle, $10/17$ for the uniform ball, and $1/2$ for a thin hoop rolling on its rim.