Question
A student of mass M = 82 kg takes a ride in a frictionless loop-the-loop at an amusement park. The radius of the loop-the-loop is R = 15 m. The force due to the seat on the student at the top of the loop-the-loop is FN = 696 N and is vertically down. What is the apparent weight of the student at the bottom of the loop-the-loop?
Answers
bobpursley
at top:
force downward on student:
mg+696-mv^2/r=0
mg=mv^2/r-696 but
mv^2/r=mg+696
at bottom, apparent weight
weight=mg+mv^2/r=mg+mg+696
apparent weight= 82(2g)+696
for force at bottom, reverse v^2/r
force=m(g+v^2/r)
force downward on student:
mg+696-mv^2/r=0
mg=mv^2/r-696 but
mv^2/r=mg+696
at bottom, apparent weight
weight=mg+mv^2/r=mg+mg+696
apparent weight= 82(2g)+696
for force at bottom, reverse v^2/r
force=m(g+v^2/r)
what is the momentum of a 1500-kg car traveling with a speed of 20 m/s (45MPH)