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 6.30 m/s .

Part A

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

Part B

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

Part C

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

Part D

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

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

Part E

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

Express your answer in seconds to three significant figures.

Explanation / Answer

Let us take the following:
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
= 6.30^2 / 14.0
a = 2.84 m/s^2

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

B)

So, a) Upward.

C)At the top, the acceleration is 4.13 m/s^2

D) Downwards.

E)

t = 2pi r / v
= 2pi * 14.0 / 6.30 = 13.955 s
t = 14.0 sec three significant figures

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