An 80-ft-diameter Ferris wheel rotates once every 23s. What is the apparent weight of a 80kg passenger at the lowest point of the circle? Express your answer in SI units.
I know to use the equation mg-N=m(v^(2)/r) but I can't figure out how to come away with the apparent weight only the velocity at that point. Please help!!!
To find the apparent weight of a passenger at the lowest point of the Ferris wheel, we need to consider the following:
1. Calculate the gravitational force acting on the passenger. The formula for gravitational force is F = mg, where F is the gravitational force, m is the mass of the passenger, and g is the acceleration due to gravity.
2. Calculate the centripetal force acting on the passenger. The centripetal force required to keep an object moving in a circle is given by F = m(v^2/r), where F is the centripetal force, m is the mass of the passenger, v is the velocity of the passenger, and r is the radius of the circle.
At the lowest point on the Ferris wheel, the velocity of the passenger is at its maximum, and the net force acting on the passenger is the difference between the gravitational force and the centripetal force.
So, the equation becomes:
mg - m(v^2/r) = apparent weight
First, let's calculate the velocity of the passenger at the lowest point. The Ferris wheel completes one rotation every 23 seconds. This means that the passenger's velocity is equal to the circumference of the circle divided by the time taken for one rotation:
v = (2πr) / T
Since the diameter of the Ferris wheel is 80ft, the radius is half of the diameter, which is 40ft. Converting this to meters, we get:
r = 40ft * (0.3048 m/ft) = 12.192 m
T = 23s
Now we can calculate the velocity:
v = (2π * 12.192m) / 23s = 8.366 m/s
Now that we have the velocity, we can substitute this value into the equation mentioned earlier:
mg - m(v^2/r) = apparent weight
Using the given mass, m = 80kg, and the radius, r = 12.192m, and the calculated velocity, v = 8.366 m/s, we can now calculate the apparent weight:
apparent weight = mg - m(v^2/r)
apparent weight = (80kg)(9.8 m/s^2) - (80kg)((8.366 m/s)^2 / 12.192m)
Simplifying this equation will give us the apparent weight of the passenger at the lowest point of the circle in SI units.
To find the apparent weight at the lowest point of the circle, you need to calculate the net force acting on the passenger.
The formula you mentioned, mg - N = m(v^2/r), is used to calculate the net force acting on an object moving in a circular path. Here:
m is the mass of the passenger (80 kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
N is the normal force (the apparent weight),
v is the velocity of the passenger at the lowest point,
and r is the radius of the ferris wheel (half the diameter, so 80 ft = 24.38 m)/2 = 12.19 m).
The net force at the lowest point of the circle is equal to the centripetal force acting on the passenger.
Since the centripetal force is the net force, we can write:
mg - N = mv^2/r
Rearranging the equation to solve for N:
mg - mv^2/r = N
Now, let's calculate the velocity of the passenger at the lowest point of the circle. The circumference of the circle can be calculated as 2πr.
Circumference = 2πr = 2π * 12.19 m = 76.49 m
The time for one complete rotation is given as 23 s. Therefore, the angular velocity is:
Angular velocity (ω) = 2π radians / 23 s ≈ 0.273 rad/s
At the lowest point, the linear velocity (v) is equal to the angular velocity (ω) multiplied by the radius (r):
v = ωr = 0.273 rad/s * 12.19 m ≈ 3.33 m/s
Now we can substitute the values into the previous equation:
80 kg * 9.8 m/s^2 - 80 kg * (3.33 m/s)^2 / 12.19 m ≈ N
Evaluating the equation:
784 N - 69.8 N ≈ N
N ≈ 714.2 N
Therefore, the apparent weight of the 80 kg passenger at the lowest point of the circle is approximately 714.2 N.