## Friday, April 16, 2010

### Fox 272 - Solution Yu claims the following:
Without the parabolic track a particle (initially at rest) at the vertex will be in free fall and drops at maximum acceleration of g. With the parabolic track the motion is restricted by the track and the particle is no longer in free fall; it slides down the track with increasing speed. Will it leave the track at some point?

All parabolic tracks are similar (See Fox 270). Consider the track y=-x^2 which ends at the point (x,-x^2). When a particle slides from rest at the vertex (0,0) to the point (x,-x^2), it acquires speed v=sqrt(2gx) in the direction given by tanθ=dy/dx=-2x, where θ is the angle with the x-axis.
.: cosθ = 1/sqrt(1+4x^2), sinθ=-2/sqrt(1+4x^2).

At point (x,-x^2) the particle is in free fall,
v_H=sqrt(2gx)/sqrt(1+4x^2),
v_y=-2sqrt(2gx^2)/sqrt(1+4x^2).

After a further time Δt, Δx=sqrt(2gx)Δt/sqrt(1+4x^2),
Δy=-2sqrt(2gx^2)Δt/sqrt(1+4x^2) - (1/2)g(Δt)^2
= -2xΔx-(1+(1/4x^2))(Δx)^2,
and the particle is at y=-x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.

If the track did not end at (x,-x^2) but continued to infinity,
y=-(x+Δx)^2
= -x^2-2xΔx-(Δx)^2 > -x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.
.: The particle stays on the parabolic track.

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