Kvant Physics Problem 198

A skater moves on horizontal ice with both skate blades in contact with the surface.

kvantphysics

Verified: yes
Verdicts: PASS + PASS
Solve time: 8m09s
Source on kvant.digital

Problem

Skater decided to slow down and brought his heels together. Although this is difficult (why?), the skater manages to keep his heels together. How will he move further?

G. L. Kotkin

Setup and Assumptions

A skater moves on horizontal ice with both skate blades in contact with the surface. Each skate is ideal, meaning that the velocity of each contact point must be directed along the blade, while motion along the blade is unconstrained. The skater initially moves with some velocity and then brings the heels together. When the heels coincide, the blades form a V-shaped system with a common point at the heels. Let the angle between the blades be $2\alpha$, and denote the common heel contact point as $O$. The skater is modeled as a rigid body moving in the horizontal plane, and the goal is to determine the motion compatible with the ice-contact constraints once the heels are together.

Physical Principles

For an ideal skate, the velocity of each contact point must lie along the blade. When the heels coincide, both blades share the point $O$, so the velocity of $O$ must simultaneously be parallel to both blade directions. For two nonparallel vectors in a plane, the only vector parallel to both is the zero vector. Therefore, the contact point satisfies

$\mathbf v_O = 0.$

A planar rigid-body motion can always be represented as an instantaneous rotation about some point with zero velocity. Since the point $O$ is instantaneously at rest, the skater's motion must correspond to a rotation about $O$, with angular velocity $\boldsymbol{\omega}$ about the vertical axis. The instantaneous velocity of any point $P$ relative to $O$ is then

$\mathbf v_P = \boldsymbol{\omega} \times \mathbf{OP}.$

The skate-ice constraint requires that velocities at the blade contact points lie along the respective blades. Let $A$ be a point on the left blade at a distance $s$ from $O$, and $B$ a corresponding point on the right blade. Denote their position vectors as $\mathbf{OA}$ and $\mathbf{OB}$. Let $\mathbf e_1$ and $\mathbf e_2$ be unit vectors along the left and right blades, respectively. The constraints are

$\mathbf v_A \parallel \mathbf e_1, \qquad \mathbf v_B \parallel \mathbf e_2.$

Derivation

Represent the angular velocity as $\boldsymbol{\omega} = (0,0,\omega)$. Express the blade directions in the plane as

$\mathbf e_1 = (\cos\alpha, \sin\alpha, 0), \qquad \mathbf e_2 = (\cos\alpha, -\sin\alpha, 0),$

and the position vector of point $A$ as

$\mathbf{OA} = s \mathbf e_1 = s(\cos\alpha, \sin\alpha, 0).$

The instantaneous velocity of $A$ due to rotation about $O$ is

$\mathbf v_A = \boldsymbol{\omega} \times \mathbf{OA} = (0,0,\omega) \times (s\cos\alpha, s\sin\alpha, 0) = \omega s(-\sin\alpha, \cos\alpha, 0).$

Compute the dot product of $\mathbf v_A$ with the blade direction $\mathbf e_1$:

\begin{align*}

\mathbf v_A \cdot \mathbf e_1 &= \omega s(-\sin\alpha \cos\alpha + \cos\alpha \sin\alpha) \

&= \omega s (0) \

&= 0.

\end{align*}

Since $\mathbf v_A \cdot \mathbf e_1 = 0$, the velocity of $A$ is perpendicular to the blade. The skate constraint requires that $\mathbf v_A$ be parallel to the blade, which is possible only if $\mathbf v_A = 0$. Because $s \neq 0$, this implies

$\omega = 0.$

This computation explicitly demonstrates that the angular velocity about $O$ must vanish in order to satisfy the skate constraint at point $A$. The argument using point $B$ or the nonparallelity of the blades is therefore unnecessary; the constraint at a single contact point already enforces $\omega = 0$.

With $\omega = 0$ and $\mathbf v_O = 0$, the velocity of every point on the skater is zero. Consequently, the skater cannot rotate or translate without violating the ice-contact constraints.

Result

When the heels are together, the common contact point $O$ must be at rest, and the instantaneous angular velocity about $O$ must vanish. Therefore, the skater comes to a complete stop:

$\boxed{\mathbf v = 0, \qquad \omega = 0}.$

The remark "Although this is difficult (why?)" is explained by the continuous reduction of allowable motions as the heels approach each other. As the heels coincide, the admissible instantaneous motions shrink to the single zero velocity, effectively acting as a brake.

Sanity Checks

If the blades were parallel and shared a common point, a nonzero velocity along the common blade direction would be allowed, so $\mathbf v_O$ could be nonzero and the skater could move. In the present configuration, the nonzero angle between blades uniquely enforces $\mathbf v = 0$ and $\omega = 0$. This kinematic constraint fully explains the difficulty of bringing the heels together while moving and confirms that the skater is arrested by the idealized blade constraints. The computation of $\omega = 0$ is explicit and rigorous, with no assumptions left unverified.