Sliding Sideways on a Plane

Sliding Sideways on a Plane Problem. A block is placed on a plane inclined at an angle θ. The coefficient of friction between the block and the plane is µ = tan θ. The block initially moves horizontally along the plane at a speed V . In the long-time limit, what is the speed of the block? Solution 1. (Constant of the motion.) Since Ff = µN = tan θ(mg cos θ) = mg sin θ, the magnitudes of the gravitational and friction forces acting on the block are equal. We would like to claim that ay = −at . (t stands for tangent; +y is taken in the direction down the plane; the negative sign is because friction acts oppositely the tangential direction.) Because the two forces are not orthogonal, however, we must thus take both forces into account when considering ay and at . We can show by explicit evaluation or by symmetry arguments that the magnitude of the net acceleration in either direction is indeed the same, so our claim is proven. Direct integration then yields vy = −vt + C = −v + V, where the constant C is obtained from our initial conditions. (The velocity is always in the tangential direction, so vt = v.) The constant C is a constant of the motion, because it is invariant over the entire trajectory. In the long-time limit, by inspection, the motion is down the plane, so v = vy and vy = V /2. Remark. One might be suspicious about the integration: whereas ŷ is a constant vector, t̂ changes with the direction of motion, and it appears that we may need to take that into consideration when integrating at . It turns out that we do not: consider v = v t̂ and a = v̇ = v̇ t̂ + v t̂˙ = v̇ t̂ + v ϕ̇n̂ =: at t̂ + an n̂. (The unit tangent vector t̂ is of unit norm, and its time derivative must therefore be orthogonal to it. This, along with the constraint that t̂, n̂, t̂ × n̂ form a right-handed coordinate system, defines the unit normal vector n̂. ϕ is the angle the tangent vector to the curve makes with the horizontal. The argument for t̂˙ = ϕ̇n̂ is equivalent to that for r̂˙ = θ̇θ̂.) Thus we see that v̇ = at , and our direct integration is acceptable. Remark. Intuitively, why should the tangential deceleration from friction impart the same change in speed as does the vertical acceleration from gravity, the trajectory of the block being curved? The derivation above shows that the crux to this puzzle is that, in both cases, the direction of the acceleration is also the direction of the velocity. In this case the change in speed is linear in time, whereas in the general case we would have to resolve the velocity into orthogonal components and use the Pythagorean theorem. The frictional force imparts a constant deceleration (in t̂) over the same period of time as does the gravitational force (in ŷ), so it is reasonable that both would lead to the same change in speed in vt = v and vy respectively. Solution 2. (Newton’s equations.) We will solve for the trajectory directly. The preferred coordinates here are the intrinsic coordinates s and ϕ, respectively the arclength and angle the tangent to the curve makes with the horizontal. These coordinates are natural here because the frictional force acts opposite the tangential direction, and the tangential direction is defined with respect to the curve (that is, intrinsically). Referring to Solution 1 and recalling ṡ = v, the decomposition of the acceleration in the tangential and normal directions yields ms̈ = mg sin θ (sin ϕ − 1), mṡϕ̇ = mg sin θ cos ϕ. For convenience, substitute u = ṡ/g sin θ and eliminate u, obtaining u̇ = sin ϕ − 1, uϕ̇ = cos ϕ. Eliminating u, some algebra leads to the directly integrable differential equation ϕ̈ = (sec ϕ − 2 tan ϕ)ϕ̇ ϕ̇ ⇐⇒ ϕ̇ = A cos ϕ (1 + sin ϕ). From uϕ̇ = cos ϕ, we get ˙ = cos ϕ(0) = 1 = g sin θ = g sin θ ϕ(0) u(0) u(0) ṡ(0) V ⇐⇒ A= g sin θ , V and, solving for ṡ, we have ṡ = g sin θ u = g sin θ cos ϕ V = ϕ̇ 1 + sin ϕ 2 and lim ṡ = lim ṡ = t→∞ ϕ→π/2 V . 2