A mass m = 75 kg slides on a frictionless track that has a drop, followed by a l

Question

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally a flat straight section at the same height as the center of the loop (19.2 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.)

It turns out the engineers designing the loop-the-loop didn

Explanation / Answer

For the mass to stay in contact with the loop at the topmost position,

the gravitational force should account for the centrifeugal force.

Mg = Mv^2/R

Therefore , the velocity should satisfy the following condition at the point of release

at the height of the loop.

v >= sqrt(Rg)

R is radius of loop

g is acceleration due to gravity.

v>= 13.71 m/s

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