A student of mass M = 85 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 = 542 N and is vertically down. What is the apparent weight of the student at the bottom of the loop-the-loop?
At the top of the loop-the-loop, the student experiences a downward gravitational force (mg) and a downward force due to the seat (FN). These forces add up to provide the centripetal force necessary for the circular motion.
Hence, we have:
mg + FN = (mv^2)/R
To find the velocity of the student at the top of the loop, we can use the equation:
v = √(gR)
Plugging this value of velocity into the previous equation, we can solve for mg:
mg + FN = (m(gR))/R
mg = m(gR)/R - FN
g = (gR)/R - (FN/m)
Substituting the given values:
g = (9.8 m/s^2)(15 m)/15 m - (542 N)/(85 kg)
g ≈ 98.0 - 6.38
g ≈ 91.6 N
Thus, the apparent weight of the student at the bottom of the loop-the-loop is approximately 91.6 N.