Irodov 1.267: A Sphere Walked Around a Circle – Amit & Deepali Sinha Foundation

Problem Irodov 1.267

A uniform sphere of mass $m$ and radius $r$ rolls without sliding on a horizontal plane while mounted on a horizontal axle $OA$ : one end $O$ of the axle is a fixed pivot, and the axle (with the sphere on it) is swung around the vertical through $O$ , so the sphere’s center traces a horizontal circle of radius $R$ at speed $v$ . Find the kinetic energy of the sphere.

The sphere is carried on the horizontal axle OA, which is itself swung about the fixed vertical through O, so the center of the sphere traces the circle of radius R drawn dashed on the plane. — The sphere is carried on the horizontal axle $OA$ , which is itself swung about the fixed vertical through $O$ , so the center of the sphere traces the circle of radius $R$ drawn dashed on the plane.

Given

$m$
$r$
$R$
$v$ of the center
rolling without sliding

Find

Find the kinetic energy

T

of the sphere

Solution

Step 1The sphere turns about two axes at once

The arrangement forces two rotations on the sphere simultaneously. The axle $OA$ is swept round the vertical through $O$ , which is a rotation $\vect\omega_2$ about that vertical; and the sphere turns on the axle itself, which is a rotation $\vect\omega_1$ along the axle. The angular velocity of the sphere is the vector sum,

\vect\omega = \vect\omega_1 + \vect\omega_2 ,

and the two contributions are perpendicular, since the axle is horizontal and the other axis vertical.

The arrangement seen from the side, along the direction the center is moving. The axle runs horizontally from O through the center, which stands at height r and at horizontal distance R from the vertical axis. — The arrangement seen from the side, along the direction the center is moving. The axle runs horizontally from $O$ through the center, which stands at height $r$ and at horizontal distance $R$ from the vertical axis.

Step 2Both rates from the one speed

The center travels its circle of radius $R$ at speed $v$ , and that circling is the rotation about the vertical, so

\omega_2 = \frac{v}{R} .

For the spin, use the rolling condition. Take unit vectors at this instant: $\vect{\hat x}$ horizontal from the vertical axis toward the center (so along the axle), $\vect{\hat z}$ vertically upwards, and $\vect{\hat y} = \vect{\hat z}\times\vect{\hat x}$ , the direction in which the center is moving. With $\vect\omega = \omega_{1x}\vect{\hat x} + \omega_2\vect{\hat z}$ and the contact point sitting at $-r\vect{\hat z}$ from the center, the velocity of the material point at the contact is

\vect v_P = \vect v_C + \vect\omega\times(-r\vect{\hat z}) = v\,\vect{\hat y} + \omega_{1x}(-r)\,(\vect{\hat x}\times\vect{\hat z}) = (v + \omega_{1x}r)\,\vect{\hat y} .

Rolling without sliding means that this vanishes, so $\omega_{1x} = -v/r$ : the spin points along the axle toward $O$ and has the magnitude

\omega_1 = \frac{v}{r},\qquad \omega_2 = \frac{v}{R}

(1)

The angular velocity at the center, resolved into the spin ω_1 along the axle and the precession ω_2 about the vertical. The two are perpendicular, so their magnitudes add in quadrature. — The angular velocity at the center, resolved into the spin $\vect\omega_1$ along the axle and the precession $\vect\omega_2$ about the vertical. The two are perpendicular, so their magnitudes add in quadrature.

Step 3Kinetic energy of a body that both moves and turns

Split the energy into the motion of the center of mass and the rotation about it:

T = \frac{1}{2}mv^{2} + \frac{1}{2}\,\vect\omega\cdot \mathsf{I}_C\vect\omega .

For a uniform sphere the central moment of inertia is the same about every axis through the center, $I_C = \frac{2}{5}mr^{2}$ , so the second term collapses to $\frac{1}{2}I_C\omega^{2}$ with no reference to the direction of $\vect\omega$ at all.

Since $\vect\omega_1 \perp \vect\omega_2$ ,

\omega^{2} = \omega_1^{2} + \omega_2^{2} = \frac{v^{2}}{r^{2}} + \frac{v^{2}}{R^{2}} ,

and therefore

\begin{aligned} T &= \frac{1}{2}mv^{2} + \frac{1}{2}\cdot\frac{2}{5}mr^{2} \left(\frac{v^{2}}{r^{2}}+\frac{v^{2}}{R^{2}}\right)\\ &= \frac{1}{2}mv^{2} + \frac{1}{5}mv^{2} + \frac{1}{5}mv^{2}\frac{r^{2}}{R^{2}} . \end{aligned}

Collecting the first two terms into $\frac{7}{10}mv^{2}$ and factoring it out,

T = \frac{7}{10}mv^{2} \left(1 + \frac{2}{7}\,\frac{r^{2}}{R^{2}}\right)

(2)

Check

Let the circle straighten out, $R\to\infty$ . The precession dies away and the answer becomes $T = \frac{7}{10}mv^{2}$ , which is the standard energy of a sphere rolling in a straight line, $\frac{1}{2}mv^{2}$ of translation plus $\frac{1}{5}mv^{2}$ of spin. The correction is positive and of relative size $\frac{2}{7}(r/R)^{2}$ , so it matters only when the sphere is not small compared with the circle it is being walked around.

Remark

There is a second way to see the ratio $\omega_1/\omega_2 = R/r$ . Two points of the sphere are instantaneously at rest: the contact point, by the rolling condition, and the point on the axle at $O$ , which is fixed. The instantaneous axis of rotation is therefore the line joining them, and its direction has horizontal and vertical parts in the ratio $R$ to $r$ . That is exactly the ratio of the two components found above, which is a check on both rates in (1).