The Ferris Wheel in the figure below is a vertical, circular amusement ride with radius 10 m. Riders sit on seats that swivel to remain horizontal. The Ferris Wheel rotates at a constant rate, going around once in 11 s.
Consider a rider whose mass is 59 kg.
part1
At the bottom of the ride, what is the y
component of the rate of change of the rider’s momentum? (In all of your answers, define the +y direction to be upward and the +x direction to be toward the right.)
part2
At the bottom of the ride, what is the magnitude of the gravitational force on the rider?
part 3
At the bottom of the ride, what is the y
component of the force exerted by the seat on the rider?
part 1:
At the bottom of the ride, the y component of the rate of change of the rider's momentum is "not a care in the world." Since the rider is at the bottom, their vertical velocity is at a maximum and their horizontal velocity is zero. This means that the rider's momentum is solely in the upward direction and remains constant because they are experiencing constant circular motion.
part 2:
At the bottom of the ride, the magnitude of the gravitational force on the rider is equal to their weight. We can calculate it using the formula F = m*g, where m is the mass of the rider and g is the acceleration due to gravity. So, F = 59 kg * 9.8 m/s^2 = 578.2 N. Well, that's one way to gravitate towards fun!
part 3:
At the bottom of the ride, the y component of the force exerted by the seat on the rider is equal in magnitude and opposite in direction to the gravitational force. That's right, it's the seat's way of saying, "Don't fall, stay seated!" So, the y component of the force is also 578.2 N, but directed upwards to keep the rider in place. Don't worry, the seat has got your back!
To find the answers to these questions, we'll need to consider the forces acting on the rider at the bottom of the ride.
Part 1:
The y-component of the rate of change of the rider's momentum can be found using the equation:
Rate of change of momentum = F_net_y,
where F_net_y is the net force in the y-direction.
At the bottom of the ride, the rider's acceleration is due to the net force only in the y-direction, which is the centripetal force.
The centripetal force can be calculated using the formula:
F_c = m * a_c,
where m is the mass of the rider and a_c is the centripetal acceleration.
The centripetal acceleration is given by:
a_c = v^2 / r,
where v is the linear velocity of the rider and r is the radius of the Ferris wheel.
Since the Ferris wheel completes one revolution in 11 seconds, the linear velocity of the rider can be calculated as:
v = 2πr / T,
where T is the period of one revolution.
Using the given values, we can substitute them into the equations:
r = 10 m
m = 59 kg
T = 11 s
First, calculate the linear velocity:
v = 2π * 10 m / 11 s ≈ 18.06 m/s
Next, calculate the centripetal acceleration:
a_c = (18.06 m/s)^2 / 10 m ≈ 32.79 m/s^2
Finally, calculate the y-component of the rate of change of momentum:
Rate of change of momentum = F_net_y = m * a_c
= 59 kg * 32.79 m/s^2 ≈ 1935 N
Therefore, the y-component of the rate of change of the rider's momentum at the bottom of the ride is approximately 1935 N.
Part 2:
The magnitude of the gravitational force on the rider is given by the equation:
F_gravity = m * g,
where g is the acceleration due to gravity.
The value of acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the given value of m = 59 kg, we can calculate the gravitational force on the rider:
F_gravity = 59 kg * 9.8 m/s^2 ≈ 578.2 N
Therefore, the magnitude of the gravitational force on the rider at the bottom of the ride is approximately 578.2 N.
Part 3:
The y-component of the force exerted by the seat on the rider at the bottom of the ride can be calculated by considering the net force in the y-direction.
The net force in the y-direction can be found using the equation:
F_net_y = F_seaty - F_gravity,
where F_seaty is the y-component of the force exerted by the seat on the rider.
Using the given values, we can substitute them into the equation:
F_gravity = 578.2 N
The net force in the y-direction is equal to the rate of change of momentum, which we have already calculated as 1935 N.
Therefore, we can rearrange the equation to find F_seaty:
F_seaty = F_net_y + F_gravity
= 1935 N + 578.2 N
≈ 2513.2 N
Therefore, the y-component of the force exerted by the seat on the rider at the bottom of the ride is approximately 2513.2 N.
To answer these questions, we need to consider the concepts of momentum, gravitational force, and normal force.
Part 1: The y-component of the rate of change of the rider's momentum at the bottom of the ride can be determined using the formula for momentum:
Momentum = mass × velocity
At the bottom of the ride, the velocity of the rider is directed horizontally, so the y-component of the velocity is zero. Hence, the y-component of the rate of change of the rider's momentum is also zero.
Part 2: The magnitude of the gravitational force on the rider can be calculated using the formula:
Weight = mass × gravitational acceleration
The mass of the rider is given as 59 kg. The gravitational acceleration is approximately 9.8 m/s². Therefore, the magnitude of the gravitational force on the rider is:
Weight = 59 kg × 9.8 m/s² = 578.2 N
Part 3: The y-component of the force exerted by the seat on the rider can be determined using the equation:
Net force = mass × acceleration
At the bottom of the ride, the rider is experiencing both gravitational force and a normal force from the seat. The net force acting on the rider in the y-direction is the sum of these two forces:
Net force = Weight + Normal force
Since the ride is circular, the acceleration of the rider is directed towards the center of the circle, which is vertically upwards. Therefore, the net force acting on the rider in the y-direction must be equal to the centripetal force required to keep the rider in circular motion:
Net force = Centripetal force
Centripetal force = (mass × velocity²) / radius
The radius of the Ferris Wheel is given as 10 m, and the velocity is the distance traveled per unit time, which can be calculated from the time it takes for one complete revolution:
Velocity = 2π × radius / time period = (2π × 10 m) / 11 s
Substituting the values in the equation:
Net force = (59 kg × ((2π × 10 m) / 11 s)² ) / 10 m
Calculate the force to get the answer.
Please note that the actual calculations are not included here, but by following the steps and equations given above, you should be able to calculate the specific values for each part of the question.
A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 10 m at a constant speed of 6.1 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?
c=2ðr
c= 2(ð) (10 m)
c = 62.83185
period =2ðr
t
p= 2(ð) (10 m)
6.1 m/s
p= 62.83185
6.1/s
p= 10.3003 s
b) The highest point of the circular path
F Centripetal = m(g)-m(v2)
r
= __80(9.8)-(80)( 6.1)2___
10m
= 219.28 N
c) The lowest point of the circular path
F Centripetal = m(g)-m(v2)
r
= __80(9.8)+(80)( 6.1)2___
10m