A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center, as shown in the figure below. The linear speed of a passenger on the rim is constant and equal to 6.00 m/s.
A.What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion?
B. What is the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion?
C. How much time does it take the Ferris wheel to make one revolution?
As for A and B
you have to use the circular acceleration rule
ac = v^2 / R
A) acL = 6^2 / 14
acH = 2.57 m/s^2
B) acH = 6^2 / 14
acH = 2.57 m/s^2
To find the magnitude of the passenger's acceleration as she passes through different points on the Ferris wheel, we can use the following equations:
1. Linear speed (v) = radius (r) × angular velocity (ω)
2. Centripetal acceleration (a) = radius (r) × angular velocity (ω)^2
By using these equations, we can calculate the answers to each part of the question.
A. To find the magnitude of the passenger's acceleration at the lowest point, we need to consider that the linear speed is constant and equal to 6.00 m/s. Since the lowest point is on the side of the Ferris wheel with the greatest radius, the radius (r) is equal to 14.0 m.
Using equation 1, we can rearrange the equation to solve for angular velocity (ω):
ω = v / r = 6.00 m/s / 14.0 m = 0.429 rad/s
Now, using equation 2, we can find the centripetal acceleration (a):
a = r × ω^2 = 14.0 m × (0.429 rad/s)^2 = 3.41 m/s^2
So, the magnitude of the passenger's acceleration at the lowest point is 3.41 m/s^2.
B. To find the magnitude of the passenger's acceleration at the highest point, we can still use the same linear speed of 6.00 m/s, but now the radius (r) is equal to the difference between the radius of the Ferris wheel and the passenger's height from the center (which is also the highest point), which is 14.0 m - 0 m = 14.0 m.
Following the same steps as in part A, we find:
ω = v / r = 6.00 m/s / 14.0 m = 0.429 rad/s
a = r × ω^2 = 14.0 m × (0.429 rad/s)^2 = 3.41 m/s^2
Therefore, the magnitude of the passenger's acceleration at the highest point is also 3.41 m/s^2.
C. The time it takes for the Ferris wheel to make one revolution is equal to the period (T) of its motion. The period can be calculated using the formula:
T = 2π / ω,
where ω is the angular velocity.
Using the angular velocity we calculated earlier (0.429 rad/s), we can solve for the period:
T = 2π / 0.429 rad/s ≈ 14.66 s
Therefore, it takes the Ferris wheel approximately 14.66 seconds to make one revolution.
To solve these questions, we'll need to use the formulas for centripetal acceleration, linear velocity, and time.
A. The magnitude of the passenger's acceleration when passing through the lowest point can be found using the formula for centripetal acceleration:
ac = v^2 / r
where ac is the centripetal acceleration, v is the linear speed of the passenger, and r is the radius of the Ferris wheel.
Substituting the given values: v = 6.00 m/s and r = 14.0 m, we have:
ac = (6.00 m/s)^2 / 14.0 m
ac = 36.0 m^2/s^2 / 14.0 m
ac ≈ 2.57 m/s^2
Therefore, the magnitude of the passenger's acceleration as she passes through the lowest point is approximately 2.57 m/s^2.
B. To find the magnitude of the passenger's acceleration when passing through the highest point, we can use the same formula:
ac = v^2 / r
Substituting the given values: v = 6.00 m/s and r = 14.0 m, we have:
ac = (6.00 m/s)^2 / 14.0 m
ac = 36.0 m^2/s^2 / 14.0 m
ac ≈ 2.57 m/s^2
Therefore, the magnitude of the passenger's acceleration as she passes through the highest point is also approximately 2.57 m/s^2.
C. The time it takes for the Ferris wheel to make one revolution can be calculated using the formula:
T = 2πr / v
where T is the period of time, r is the radius of the Ferris wheel, and v is the linear velocity of the passenger.
Substituting the given values: r = 14.0 m and v = 6.00 m/s, we have:
T = 2π(14.0 m) / 6.00 m/s
T = 28π / 6.00 s
Calculating this value, we find:
T ≈ 14.66 s
Therefore, it takes approximately 14.66 seconds for the Ferris wheel to make one revolution.