A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its

Question

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center, as shown in the figure below.(Figure 1)The linear speed of a passenger on the rim is constant and equal to 6.90 m/s .

A) What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion?

a = ____ m/s^2

B) What is the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion?

a = ____ m/s^2

C) How much time does it take the Ferris wheel to make one revolution?

t = ____ s

Explanation / Answer

Let:
r be the radius,
v be the passenger's speed,
a be the acceleration,
t be the time for one revolution.

a = v^2 / r
= 6.9^2 / 14.0
= 3.4m/s^2 to 3 sig. fig.

The acceleration is constant in magnitude, and directed towards the centre of the wheel.

At the bottom, it is 3.4 m/s^2 upwards.
At the top, the acceleration is 3.4 m/s^2 downwards.

t = 2pi r / v
= 2pi * 14.0 / 6.9
= 12.74202898550725 sec.

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