# physics

A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.5 m and finally a flat straight section at the same height as the center of the loop (15.5 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

3) If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

6) It turns out the engineers designing the loop-the-loop didn’t really know physics – when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning.
How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?

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1. when it reaches the top of the loop it must have centripetal acceleration = gravitational acceleration of it will drop off the track.

v^2/R = g = 9.8 m/s^2 approx

v^2 / R = g
so
v^2 = g R at top
so how far above the top of the loop must it start?
call that distance h above the top
m g h = (1/2) m v^2
so
g h =(1/2) g R
h = (1/2) R

If it starts at the level of the top of the loop, it better be going at speed v there :)

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2. oh, and at ground level of course
(1/2) m V^2 = m g (h + 2R)

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