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Re: Block sliding on curvilinear surface



A block with mass m, originally at rest slides downward on a
curvilinear surface given by y = (9x)^1/2 and r = 9cot(angle)
cos(angle). This is a parabola on its side. Does the block leaves
the surface before reaching the ground (y = 0)?

Unless I am misinterpreting something, it seems to me that the answer
has to be an unequivocal "yes" since the surface is vertical at the
point that it touches the ground.

What is the minimum value for n in such way that the block will
leave the surface? (y = (nx)^1/2)

My guess is that the block leaves the surface for any positive value of n.

The first thing I noticed is that, in my opinion, you cannot use
centripetal dynamics.

I'm not sure what you mean by this.

Two questions:
1. Is this an original problem? I highly doubt it (There is nothing
new under the Sun) but I have never seen this type of problem before.

As you noted, it is similar to the classic problem that has an object
sliding from the top of a hemispherical dome. The only real
difference here is that the radius of curvature is not constant.
That makes it harder and possibly more interesting as a result. IMO
the main point of a problem like this might be to emphasize the
utility of the instantaneous radius of curvature.

2. Is it solvable, even with simple calculus? Is it solvable algebraically?

I would expect it to be readily solvable. Calculus ideas would be
necessary to find the radius of curvature.

3. Are there other simple curves where this type of problem could be solved?

A sine curve might be more interesting for the type of question you
pose because it doesn't go vertical.

--
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm