A student of mass M = 62 kg takes a ride in a frictionless loop-the-loop at an amusement park. The radius of the loop-the-loop is R = 19 m. The normal force by the track on the student at the top of the loop-the-loop is N1 = 100 N
Find the speed of M at the top of the loop.
Determine the normal force on M by the track at the bottom of the loop-the-loop, assuming the speed at the bottom is the same as at the top.
To find the speed of the student at the top of the loop, we can use the principle of conservation of mechanical energy. At the top of the loop, the student is at the highest point, so all of their kinetic energy is converted to potential energy.
The potential energy at the top of the loop is given by:
PE = mgh
where m is the mass of the student (62 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the loop. In this case, the height is equal to twice the radius of the loop, since the student is at the top.
h = 2R = 2 * 19 m = 38 m
So, the potential energy at the top is:
PE = m * g * h = 62 kg * 9.8 m/s^2 * 38 m = 22756 J
Since all of the kinetic energy is converted to potential energy, we can equate them:
KE = PE
where KE is the kinetic energy of the student. The kinetic energy is given by:
KE = (1/2) * m * v^2
where v is the velocity of the student. Rearranging the equation, we get:
v = sqrt(2 * KE / m)
Substituting the value of potential energy calculated earlier, we have:
v = sqrt(2 * 22756 J / 62 kg) = sqrt(732.645 m^2/s^2) = 27.04 m/s
So, the speed of the student at the top of the loop is approximately 27.04 m/s.
To determine the normal force on the student by the track at the bottom of the loop, assuming the speed at the bottom is the same as at the top, we can use the equation of motion for circular motion.
At the bottom of the loop, the centripetal force is provided by the normal force and the weight of the student:
F = N2 + mg
where N2 is the normal force at the bottom of the loop and g is the acceleration due to gravity.
Since the centripetal force is given by:
F = (mv^2) / R
where R is the radius of the loop, we can equate the two equations:
N2 + mg = (mv^2) / R
Rearranging the equation, we get:
N2 = (mv^2) / R - mg
Substituting the values, we have:
N2 = (62 kg * (27.04 m/s)^2) / 19 m - 62 kg * 9.8 m/s^2 = 1990.06 N - 607.6 N = 1382.46 N
So, the normal force on the student by the track at the bottom of the loop is approximately 1382.46 N.