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 (the figure (Figure 1) ). The linear speed of a passenger on the rim is constant and equal to 7.76 m/s .

the figure

https://session.masteringphysics.com/problemAsset/1259907/3/YF-03-42.jpg

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?

Part E

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

Explanation / Answer

since it is rotating with constant speed,
a will be same as centripetal acceleration whose magnitude is give by,
a = v^2/r
=(7.76)^2/14
= 4.3 m/s^2

a)
Answer: 4.3 m/s^2
b)
direction is towards the centre
so vertically upward in this case
c)
Answer: 4.3 m/s^2
d)
direction is towards the centre
so vertically downward in this case
e)
v= 7.76 m/s
w = v/r = 7.76 / 14 = 0.554 rad/s
w = change in degree/change in time
0.554 = 2*pi/time
time=11.3 S
Answer: 11.3 S

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