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I'm looking for a set of straight ramps, which, under idealised conditions (uniform gravitational acceleration and no friction) would have a point-like body slide down them from the top to the base point in an equal amount of time. Suppose that this set of ramps was arranged such that they all ended at the same base point. What shape would the top ends of the ramps form?

Clearly, the steeper ramps in this set will have to be longer.

Thomas Delaney
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1 Answers1

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Suppose we want a set of ramps which will all take t= sqrt(2/4.9) seconds for a point to roll from the top to the base. This will make calculations easier later on.

For a ramp at an angle, z, to the horizontal; which covers x metres horizontal distance. It therefore has a height of (x tan z) vertical metres.

s (length of the ramp) = x/cos z

a = 9.8 sin z

u (initial velocity) = 0 m/s

t = sqrt (2/4.9)

s = ut + 0.5 at^2

Subbing in, we get: x/cos z = 4.9 sin z * 2/4.9

x = 2 sin z cos z

The vertical distance of the ramp: y = x tan z = 2 sin^2 z

x^2 + (1-y)^2 = 4 sin^2 z cos^2 z + 1 - 4sin^2 z + 4 sin^4 z

= 4 sin^2 z (cos^2 z + sin^2 z - 1) + 1

= 1

Hence, the ends of this set of equi-time straight ramps make up a circle.

In this case, the circle has a radius of 1m. If the time given was different, this radius would also be different.

Thomas Delaney
  • 1,683
  • You seem to be ignoring rotational energy issues, so you'd better say SLIDE rather than ROLL. – Ted Shifrin Jul 14 '15 at 03:55
  • Including rotational inertia will not change the shape of the set of ramps. All the accelerations will be altered by the same factor. – Noah Jul 29 '15 at 14:58