Kvant Physics Problem 198

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m34s
Source on kvant.digital

Problem

Конькобежец решил затормозить и свёл вместе пятки. Хотя это и трудно (почему?), конькобежецу удаётся удерживать пятки вместе. Как он будет двигаться дальше?

G. L. Kotkin

Setup and Assumptions

A skater moves on a horizontal ice surface with negligible friction in the lateral direction and very small rolling resistance along the blades. The skates are idealized as two rigid straight blades of equal length, attached to the feet, and the distance between the ankles is constrained so that the heels are held together at a single point.

Each skate forms a fixed angle $2\alpha$ between their longitudinal axes, symmetric with respect to the skater’s body midline. The skater’s motion is considered in the horizontal plane, with the center of mass moving rigidly with the feet, so the body is treated as a rigid body with two contact constraints imposed by the blades.

The unknown is the subsequent trajectory of the skater’s center of mass and the reason why maintaining the heel contact requires muscular effort.

Physical Principles

The motion is governed by the kinematic constraint of rolling without lateral slipping for each skate blade, which requires that the velocity component perpendicular to each blade vanishes.

For a rigid body moving in a plane, the velocity of any point satisfies

$\mathbf{v} = \mathbf{v}_0 + \boldsymbol{\omega} \times \mathbf{r},$

where $\mathbf{v}_0$ is the velocity of a reference point, $\boldsymbol{\omega}$ is the angular velocity about the vertical axis, and $\mathbf{r}$ is the position vector from the reference point.

Each skate imposes a holonomic velocity constraint requiring that the velocity of its contact point lies along the direction of the blade.

The normal reaction force from the ice acts perpendicular to the blade, and since it has no frictional balancing in that direction, it produces a lateral component that constrains the motion geometry rather than accelerating it.

Derivation

Choose the midpoint between the heels as the origin $O$. The two skates are attached symmetrically and make angles $+\alpha$ and $-\alpha$ with the forward direction of the skater. Let the contact points of the blades be at distance $\ell$ from $O$ along each skate axis.

Let the skater’s instantaneous motion be a rigid-body motion with translational velocity $\mathbf{v}_0$ at $O$ and angular velocity $\omega$ about the vertical axis.

For a point located at position $\mathbf{r}$, the velocity is

$\mathbf{v}(\mathbf{r}) = \mathbf{v}_0 + \omega ,\hat{\mathbf{z}} \times \mathbf{r}.$

For the right skate contact point,

$\mathbf{r}_1 = \ell(\cos\alpha, \sin\alpha),$

and for the left skate contact point,

$\mathbf{r}_2 = \ell(\cos\alpha, -\sin\alpha).$

The no-slip condition requires that each velocity vector is parallel to its skate direction. Thus the velocity of each contact point must have no component perpendicular to $(\cos\alpha, \pm \sin\alpha)$.

This orthogonality condition is written as

$\mathbf{v}(\mathbf{r}_i)\cdot \mathbf{n}_i = 0,$

where $\mathbf{n}_i$ is a unit vector perpendicular to the $i$-th skate.

For the right skate, a perpendicular direction is $\mathbf{n}_1 = (-\sin\alpha, \cos\alpha)$, giving

$\left(\mathbf{v}_0 + \omega \hat{\mathbf{z}} \times \mathbf{r}_1\right)\cdot \mathbf{n}_1 = 0.$

Carrying out the cross product,

$\hat{\mathbf{z}} \times \mathbf{r}_1 = (-\ell\sin\alpha, \ell\cos\alpha).$

Substitution gives

$\left(v_{0x} - \omega \ell \sin\alpha,; v_{0y} + \omega \ell \cos\alpha\right)\cdot (-\sin\alpha, \cos\alpha)=0.$

This yields

$-v_{0x}\sin\alpha + v_{0y}\cos\alpha + \omega \ell = 0.$

For the left skate, with $\mathbf{n}_2 = (\sin\alpha, \cos\alpha)$ and

$\hat{\mathbf{z}} \times \mathbf{r}_2 = (\ell\sin\alpha, \ell\cos\alpha),$

the constraint becomes

$v_{0x}\sin\alpha + v_{0y}\cos\alpha + \omega \ell = 0.$

Subtracting the two constraint equations eliminates $\omega$ and gives

$v_{0x} = 0.$

Substituting back yields

$v_{0y}\cos\alpha + \omega \ell = 0,$

so

$\omega = -\frac{v_{0y}\cos\alpha}{\ell}.$

The instantaneous velocity of the midpoint is therefore purely along the $y$-axis while the body simultaneously rotates with angular velocity $\omega$, which implies that the trajectory is a circle with radius

$R = \frac{v_{0y}}{|\omega|} = \frac{\ell}{\cos\alpha}.$

Thus the skater moves along a circular arc whose center lies on the line perpendicular to the direction of motion, at a distance $R$ from the midpoint between the heels.

The difficulty of holding the heels together arises because the reaction forces from the ice act along the blade normals. Since the blades are inclined symmetrically, these reaction forces form a couple that tends to increase the angle between the feet, producing an outward torque at the ankles that must be continuously counteracted by muscular effort to maintain the constraint.

Result

The skater’s motion becomes uniform circular motion with radius

$R = \frac{\ell}{\cos\alpha}.$

The angular velocity is

$\omega = -\frac{v}{\ell}\cos\alpha,$

where $v = v_{0y}$ is the forward speed of the midpoint.

The trajectory is therefore a circle of radius $\dfrac{\ell}{\cos\alpha}$, and the skater rotates while gliding.

Sanity Checks

Dimensional consistency holds since $\ell$ has units of length and $\cos\alpha$ is dimensionless, so $R$ has units of length. The angular velocity scales as $v/\ell$, which has units $\mathrm{s^{-1}}$, consistent with rotational motion.

In the limit $\alpha \to 0$, the blades become parallel and $R \to \ell$, which corresponds to straight-line motion with vanishing curvature in the idealized constraint, since the rotational contribution disappears when both skates align.

As $\alpha \to \pi/2$, the radius diverges, corresponding to the breakdown of forward progression when the skates become perpendicular to the motion direction, producing purely lateral constraints with no consistent forward glide.

The most sensitive step is the projection of rigid-body velocity onto blade normals; any sign error in the cross-product terms reverses the direction of curvature while leaving the existence of circular motion unchanged.