- #1
Yae Miteo
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Homework Statement
A mass m = 77 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.1 m and finally a flat straight section at the same height as the center of the loop (15.1 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.)
part 2) What height above the ground must the mass begin to make it around the loop-the-loop?
Homework Equations
[tex]PE=mgh[/tex]
[tex]KE = \frac {mv^2}{2} [/tex]
and perhaps
[tex]a_c = \frac {v^2} {r}[/tex]
The Attempt at a Solution
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I began by setting KE = PE
[tex]KE = PE[/tex]
[tex]mgh = \frac {mv^2}{2}[/tex]
cancel out
[tex]\frac {v^2}{r} = gh[/tex]
solve for h
[tex]h = \frac{v^2}{2g}[/tex]
plug in and find h
[tex]h=7.54999[/tex]
add to radius of loop, and answer is
[tex]22.65[/tex]
However, this is not the correct answer. I feel that I am close, but I am not sure what I am missing.