The small mass m sliding without friction along a looped track is to remain on t

Question

The small mass m sliding without friction along a looped track is to remain on the track at all times, even at the very top of the loop of radius r.

For the following answers, use only the symbols given.

Now assume the actual release height is 1.7h. Give the following forces in terms of given parameters m, r, and g.

(a) Calculate the magnitude of the normal force exerted by the track at the bottom of the loop.

(b)Calculate the magnitude of the normal force exerted by the track at the top of the loop.

At one point for part a I calculated that the normal force is 19/2 of mg but now I cannot figure out how I got that answer....

Explanation / Answer

at the bottom most point ,normal force(reaction) will be mv^2/r (centrifugal force)+mg(weight)...where v can be calculated from newtons laws - v^2 = 2as,s-vertical displacement between point of release and the point where you want to find normal force...at top most point ,normal force= mv^2/r - mg(weight)

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