# Homework Help: Ferris Wheel question (Motion in 2d)

1. Feb 9, 2013

### jmedina94

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?

2. Feb 9, 2013

3. Feb 9, 2013

### jmedina94

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 .

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"

Last edited: Feb 9, 2013
4. Feb 9, 2013

### Staff: Mentor

Ah, good.

OK, but with respect to what is that angle? How do they want the direction represented?

5. Feb 9, 2013

### jmedina94

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

6. Feb 9, 2013

### Staff: Mentor

The passenger is at the bottom. Draw a picture of the acceleration vector.

You are on the right track, you just have to express it correctly.

7. Feb 9, 2013

### jmedina94

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

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8. Feb 9, 2013

### Staff: Mentor

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.

9. Feb 9, 2013

### haruspex

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.