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In the game of curling, players "curl" a granite "rock" (of precise size and roughly a flattened cylinder) down a "sheet" of ice towards a target; the "rock" will curve in its path in the direction of the motion of the leading edge. The question here is, what is the mechanism for the curling of the rock?

Standard frictional physics indicates that angular velocity should not have any effect on translational vector; while moving, the bottom surface of the rock is subject to kinetic friction. Due to frictional deceleration, the leading edge of the rock exerts (slightly) more force on the ice than the trailing edge; this would seem to indicate that the rock would curve in the opposite direction from what it does.

Intuition would indicate that the reason for the curling motion would be due to kinetic friction being modified by velocity; that is, along the "left" side of the rock (assuming a rock moving away from the observer) for a rock rotating counterclockwise, the velocity along the ground is slower than for the "right" side; if kinetic friction is decreased as velocity increases, this would be a good explanation for the motion. However, everything I've seen seems to indicate that the coefficient of kinetic friction is constant; is this correct? If so, what is the reason for the (well-observed) phenomenon?

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It seems you must have watched a curling event, so I direct your attention to the sweeping done by the players ahead of the stone. Perhaps I may come back with an answer, but I have some other things to finish right now. –  Mark C Nov 25 '10 at 4:25
@MarcC: the sweeping is relevant to the magnitude of the effect, as it's been proven to have a melting effect on the ice, but it doesn't explain the directionality of the effect, which occurs in the absence of sweeping. –  Paul Sonier Apr 5 '11 at 4:10

3 Answers 3

The coefficient of friction for a curling stone is not constant - motion on ice is different from motion on an unchanging solid surface, thanks to melting, which in turn is proportional to most of the factors that affect friction (e.g. contact area, velocity). So, exactly as you suspect, friction decreases as velocity increases.

Because of the changing CofF, the stone curls more strongly at the end of its trajectory (as it gets slower) than at the beginning.

Additionally, as Mark C says, the team can affect the trajectory after delivery by sweeping - the entire purpose of the sweepers is to melt or at least smooth the ice ahead of the stone (or not, as required), which tends to straighten and lengthen the trajectory by reducing the effect of friction on the leading edge of the stone.

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""...solid surface, thanks to melting, which in turn..."" Are the stones heated? –  Georg Apr 4 '11 at 14:49
@georg: no, friction (and pressure) causes the heating which causes the melting. It's the same melting that occurs with ice skates. –  Paul Sonier Apr 5 '11 at 4:09
@McWafflestix, friction might melt some ice, the rotation would add to this. But pressure melting is a unban legend for skates, in case of those curling stones even more. –  Georg Apr 5 '11 at 8:49
This answer is no good. If the x-y plane is the ice, and x the direction of motion, the motion in y is symmetric to reflections, so that for every spot going down there is a spot going up in y somewhere else So a varying coefficient with speed cannot cause an upward force not the answer. –  Ron Maimon Dec 9 '11 at 9:23
Why do you think the front exerts more pressure due to slowing? There is no reason to think this, it is incorrect. The pressure is even for a symmetric rock. –  Ron Maimon Dec 9 '11 at 9:26

The stone creates a contact patch with the ice. In this case it is roughly circular in shape with approximate radius of $a=0.721\,\sqrt[3]{W\,D\left(\dfrac{1-\nu_{1}}{E_{1}}+\dfrac{1-\nu_{2}}{E_{2}}\right)}$ where $D$ is the diameter of the spherical contact surface (the crown diameter of the stone), $W$ is the weight of the stone, and $\nu$, $E$ are the Poisson's ratio and Youngs modulus for the two materials contacting (ice and stone).

Now because the surface is uneven, the pressure on the contact is not necessarily symmetric, and may have an pressure center of axis from the spin axis of the stone. At most the eccentricity is equal to $a$, yielding the maximum possible torque applied to the stone to be $T=\mu\,W\,a$ where $\mu$ is the coefficient of friction.

The torque the is responsible for altering the spin the stone. Once the stone has spin, then the hydrodynamic contact properties alter the path of the stone.The friction is not constant as it depends on the pressure applied on each spot. Think of the top layer of ice as a lubrication layer, and the stone has to plow through it. As far as the mechanism that this complex interplay creates a side force (tracking force) I am not sure I can explain it. Maybe someone else can.

Ref: "Formulas for Stress & Strain", Roark, 3rd ed. 1954

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I was aware that the reduction in the spin on the stone is due to the torque of the applied coefficient of friction; the real concern is why the vector of the motion changes as it does; yes, the friction is not constant, since it depends on the pressure applied at each spot, but statistically, it evens out, and a pressure-based approach would (as I mention in the original question) indicate a change in vector opposite of the actual effect (due to the front of the rock exerting more pressure due to slowing). –  Paul Sonier Apr 5 '11 at 4:19
I found this in Wikipedia: ""The only part of the stone in contact with the ice is the running surface, a narrow, flat annulus or ring, 0.25 to 0.50 inch (6.3 to 13 mm) wide and about 5 inches (130 mm) in diameter; the sides of the stone bulge convex down to the ring and the inside of the ring is hollowed concave to clear the ice."" –  Georg Apr 5 '11 at 9:00

In order to move up, the back part of the rock must have slightly more friction on it than the front part.

Let x be the direction of motion, and y span the plane of the disk. The easy case to consider is the rolling stone limit, where the stone is moving forward and rotating with angular velocity $R\omega=v$ counterclockwise as you normally draw the x-y plane. In this case, the top point on the edge isn't moving at any point, the back part is going down in y, and the front part is going up. The friction on the back must be slightly higher to give a net up force.

The asymmetry in the friction must be due to the fact that the front edge encounters new ice, while the back edge is going through ice that has been contacted by the front edge already. This must mean that new ice has less friction than ice that has exerted some friction, so that a smooth sheet of ice has a coefficient of friction that increases with the amount of sliding done on it. This is strange, I would have expected the opposite, but it is the only way to explain the sliding.

The leading pressure is the same as the trailing pressure for a symmetric stone, and asymmetries in the stone do not explain the effect, since they average out over a full turn.

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