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?
To find the apparent weight of the student at the bottom of the loop-the-loop, we need to consider the forces acting on the student.
At the top of the loop-the-loop, the only force acting on the student is the force due to the seat, FN, which is vertically down. We can find the magnitude of this force using the student's weight, which is given by:
Weight = mass × gravity
W = M × g
where M is the mass of the student and g is the acceleration due to gravity.
At the top of the loop, the student experiences an apparent weight, which is the force exerted on the student by the seat. This force is given by:
Apparent weight = Weight - Normal force
At the top of the loop, the Normal force is equal to the force due to the seat, FN. Therefore, at the top of the loop:
Apparent weight = Weight - FN
Now, to find the apparent weight at the bottom of the loop-the-loop, we need to consider the forces acting on the student.
At the bottom of the loop-the-loop, the student is moving in a circular path, which requires a centripetal force pointing towards the center of the loop. This centripetal force is provided by the normal force, FN.
Since the student is moving in a vertical loop, the net force at the bottom of the loop consists of the normal force, FN, and the student's weight, W, both pointing downwards. Therefore, the net force can be expressed as:
Net force = FN + W
To find the apparent weight at the bottom of the loop, we subtract the normal force, FN, from the net force:
Apparent weight = Net force - FN
Now we can substitute the values given in the problem:
Mass of the student, M = 82 kg
Radius of the loop-the-loop, R = 15 m
Force due to the seat at the top of the loop, FN = 696 N
First, calculate the weight of the student:
Weight = M × g
Weight = 82 kg × 9.8 m/s²
Weight ≈ 803.6 N
Next, find the apparent weight at the top of the loop:
Apparent weight at the top = Weight - FN
Apparent weight at the top = 803.6 N - 696 N
Apparent weight at the top ≈ 107.6 N
Finally, find the apparent weight at the bottom of the loop:
Apparent weight at the bottom = (FN + Weight) - FN
Apparent weight at the bottom = FN + Weight - FN
Apparent weight at the bottom ≈ 803.6 N
Therefore, the apparent weight of the student at the bottom of the loop-the-loop is approximately 803.6 N.
what is the momentum of a 1500-kg car traveling with a speed of 20 m/s (45MPH)
To find the apparent weight of the student at the bottom of the loop-the-loop, we need to consider the forces acting on the student at that point.
At the bottom of the loop-the-loop, the student experiences two forces: the force due to gravity (weight) and the normal force from the seat.
1. Weight (W): The weight of the student acts vertically downwards and is given by the formula W = mg, where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s^2).
W = mg
W = 82 kg * 9.8 m/s^2
W = 803.6 N
2. Normal Force (FN): The normal force acts perpendicular to the surface of contact (the seat). At the bottom of the loop-the-loop, the normal force is directed upwards and is equal in magnitude to the centripetal force required to keep the student moving in a circular path.
FN = mv^2 / R
Given that the normal force at the top of the loop-the-loop is 696 N, we can use this information to find the velocity of the student at the bottom of the loop-the-loop.
FN = mv^2 / R
696 N = 82 kg * v^2 / 15 m
v^2 = (696 N * 15 m) / 82 kg
v^2 ≈ 127.073 m^2/s^2
v ≈ √(127.073 m^2/s^2)
v ≈ 11.28 m/s
Now that we have the velocity, we can calculate the apparent weight at the bottom of the loop-the-loop.
Apparent Weight = FN + W
Apparent Weight = 696 N + 803.6 N
Apparent Weight ≈ 1499.6 N
Therefore, the apparent weight of the student at the bottom of the loop-the-loop is approximately 1499.6 N.
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)