19.6: Motion on a Cycloid, Cusps Down
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We imagine a particle sliding down the outside of an inverted smooth cycloidal bowl, or a bead sliding down a smooth cycloidal wire. We shall suppose that, at time t=0, the particle was at the top of the cycloid and was projected forward with a horizontal velocity v0. See Figure XIX.7.
This time, the equations of motion are
¨s=gsinψ
and
mv2ρ=mgcosϕ−R.
By arguments similar to those made in Section 19.5, we find that
¨s=gs4a
The general solution to this is
s=Aept+Be−pt,
where
p=√g/(2a).
With the initial condition given (at t=0,s=0,˙s=v0 ), we can find A and B and hence:
s=v0√ag(ept−e−pt)
Again proceeding as in Section 19.5, we find for R:
R=m4cosψ(4gacos2ψ−v20).
So – what happens?
If the constraint is two-sided (bead sliding on a wire) R becomes zero when cos2π=v20/(2/ga), and thereafter R is in the opposite direction.
If the constraint is one-sided (particle sliding down the outside of a smooth cycloidal bowl):
- If v20 >4ga, the particle loses contact at the moment of projection.
- If If v20 <4ga the particle loses contact as soon as cos2π=v20/(2ga), is very small (i.e. very much smaller than √(2ga) ), this will happen when ψ=45∘ ; for faster initial speeds, contact is lost sooner.
A particle is projected horizontally with speed v0 = 1 m s−1 from the vertex of the smooth cycloidal hill
x=a(2θ+sin2θ
y=2acos2θ,
where a=2 m. Assuming that g = 9.8 m s−2, how long does it take to get halfway down the hill (i.e. to y=a)?
Solution
We have to use Equation ???. With the numerical data given, this is
s=0.451754(e1.565248t−e−1.565248t).
We can find s from Equation 19.4.12, which gives us s = 2.828427 m. If we let we now have to solve 6.26099 = ξ−1/ξ, or ξ2−6.26099ξ−1=0. From this, ξ = 6.41683 and hence t = 1.19 s.
I leave it to the reader to calculate R at this time – and indeed to see whether the particle loses contact with the hill before then. Perhaps the fact that I got a positive real root for ξ means that we are all right and the particle is still in contact – but I wouldn't be sure of that. I leave it to the reader to investigate further.