Ferris Wheel question (Motion in 2d)

AI Thread Summary
The discussion revolves around calculating the direction of a passenger's acceleration on a counterclockwise Ferris wheel at the lowest point of its motion. The passenger moves at a speed of 3.00 m/s with a tangential acceleration of 0.500 m/s² and a centripetal acceleration of 0.64 m/s². Participants clarify that the angle should be expressed as degrees north of east, and emphasize the importance of accurately representing the acceleration vectors. The correct approach involves drawing the tangential and centripetal (or radial) acceleration vectors to determine their resultant direction. The conversation highlights the need for clarity in vector representation and calculation methods.
jmedina94
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1. The Ferris wheel [...] rotates counterclockwise, is just starting up. At a given instant, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00 m/s and is gaining speed at a rate of 0.500 m/s^2 .
2. Find the direction of the passenger's acceleration at this instant.
3. My attempt was using: the arctan 0.64/0.5, and that yielded ~52.0° but no dice so, could someone help me?
 
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Show how you made your calculation. (Some information is missing from your problem statement.)
 
Doc Al said:
Show how you made your calculation. (Some information is missing from your problem statement.)

The problem: The Ferris wheel in the figure (Figure 1) , which rotates counterclockwise, is just starting up. At a given instant, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00 m/s and is gaining speed at a rate of 0.500 m/s^2 .

The ferris wheel's radius is 14m**** Sorry about that!

Centripetal Acc = v^2/r
= 9/14 m/s^2 = 0.64

Tangential Acc = 0.5 m/s^2

Magnitude of both vectors = 0.81 m/s

θ = tan^-1 (0.64/0.5) *this was wrong*"find the direction of the passenger's acceleration at this instant"
 
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jmedina94 said:
The ferris wheel's radius is 14m**** Sorry about that!
Ah, good.

Centripetal Acc = v^2/r
= 9/14 m/2^2 = 0.64

Tangential Acc = 0.5 m/s^2

Magnitude of both vectors = 0.81 m/s

θ = tan^-1 (0.64/0.5) *this was wrong*
OK, but with respect to what is that angle? How do they want the direction represented?
 
OK, but with respect to what is that angle? How do they want the direction represented?

Not sure I was just testing out a method I had already practiced, probably the center of the ferris wheel?

They want it as:

θ= __° north of east
 
jmedina94 said:
Not sure I was just testing out a method I had already practiced, probably the center of the ferris wheel?
The passenger is at the bottom. Draw a picture of the acceleration vector.

They want it as:

θ= __° north of east
You are on the right track, you just have to express it correctly.
 
Doc Al said:
The passenger is at the bottom. Draw a picture of the acceleration vector.

This is my rendition of the question, not sure if it's the correct way to solve the direction
 

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jmedina94 said:
This is my rendition of the question, not sure if it's the correct way to solve the direction
What I have in mind is this: Put a dot where the passenger is. Then draw the vectors representing the tangential and centripetal acceleration of the passenger. Then draw their sum and see where it points.
 
Doc Al said:
the tangential and centripetal acceleration
Technically, I think you mean tangential and radial acceleration, though in this case, since r is constant, radial and centripetal will be the same, just opposite sign.
 
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