Stanton D. answered • 02/27/21
Tutor to Pique Your Sciences Interest
So Mike I.,
I'll drop the masses from the equations, they're superfluous here. Suffice it to say, the roller-coaster carries a mixture of kinetic and potential energy. In order to remain on the track (just marginally), it must have "centrifugal" force or acceleration of exactly g at the top of the loop (directed upwards, to balance gravity downwards). You can express the centrifugal (or centripetal, however you think of it) acceleration of a circularly moving object as a = v^2/R . So that must equal g .Work that information backwards to obtain the kinetic energy the train has at the top of the loop, which is (1/2)v^2 = Rg/2. Now, that train also has a potential gravitational energy at the top of the loop: 2gR. Now, you can put the two pieces of energy together: 2.5 Rg. That gives you what you need for your initial gravitational potential energy = gh . So h = 2.5R, n'est-çe pas?
-- Cheers, --Mr. d.
