A. We know that the linear speed (v) of the passenger on the rim is constant and equal to 6.00 m/s, and the radius (r) of the Ferris wheel is 14.0 m. The centripetal acceleration (a_c) at any point in the circular motion can be calculated using the formula:
a_c = v^2 / r
Plugging in the values, we get:
a_c = (6.00 m/s)^2 / 14.0 m
a_c = 36.00 m^2/s^2 / 14.0 m
a_c = 2.57 m/s^2
So, the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion is 2.57 m/s^2.
B. The magnitude of the centripetal acceleration is the same at any point in the circular motion. Therefore, the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion is also 2.57 m/s^2.
C. To find the time it takes the Ferris wheel to make one revolution, we can use the formula:
t = 2πr/v
Plugging in the values, we get:
t = 2π(14.0 m) / 6.00 m/s
t = 28π m / 6.00 m/s
t ≈ 14.67 s
So, it takes the Ferris wheel approximately 14.67 seconds to make one revolution.
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?
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