Ferris Wheel: Find Direction of Passenger's Acceleration

[rsala]

Homework Statement

The Ferris wheel in the figure View Figure , 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 \;{\rm m}/{\rm s}^{2} .

other data: radius is 14m

find the direction of the passenger's acceleration at this instant.
the answer must be in degrees from the right of the vertical axis. edited.

Homework Equations

a$$^{ }_{centripetal}$$ = $$\omega$$$$^{2}_{}$$ x R
or
a$$^{ }_{centripetal}$$ = v$$^{2}_{}$$ / R

The Attempt at a Solution

since the instant they are referring to is the the lowest point of the ferris wheel (i think)
the centripetal acceleration there is .643 m/s^2 direction is directly upwards...so there's a magnitude of .643 in the y direction with 0 in the x direction there.

since the ferris wheel is accelerating counterclockwise @ .5 m/s^2 the tangential acceleration is .5. with direction to the direct right on the x axis, with 0 magnitude in the y.

this is the point at which i do not understand.
my normal approach to this problem would be to acknowledge that I am finding A, and A = Ax + Ay...Ax = the tangential, and Ay = the centripetal.

and proceed by just pluging the calculator ARCTAN Ay / Ax (which is arctan.643/.5 = arctan1.286)= .90...impossible IT CANT be .90 degrees (nor .9 radians because that's still 52 degrees and not the correct answer)

the correct answer is in fact 37.9 degrees,, how can i find the solution? thanks and sorry for the long read.

the answer must be in degrees from the right of the vertical axis. edited.

 

[Dick]

You never quite got around to stating the actual question. What is it exactly? I do note that the 'incorrect answer' of 52.1 degrees and the 'correct answer' of 37.9 degrees add up to 90. Interesting coincidence, yes?

 

[rsala]

i forgot to state that the answer must be in degrees from the right of the vertical axis. edited.

is the 52 degrees that I am getting from the positive x axis?

ive drawn up a photo of what I am thinking, i am not sure if how I am picturing the "degrees to the right of vertical" correctly as the problem wants.

http://img225.imageshack.us/img225/4088/rightfromverticalfromhofz4.jpg

 

[Dick]

i forgot to state that the answer must be in degrees from the right of the vertical axis. edited.

is the 52 degrees that I am getting from the positive x axis?

Yes. Exactly.

 

[rsala]

ah i get it now, disregard my picture i see where i messed up

 

[Dick]

i forgot to state that the answer must be in degrees from the right of the vertical axis. edited.

is the 52 degrees that I am getting from the positive x axis?

ive drawn up a photo of what I am thinking, i am not sure if how I am picturing the "degrees to the right of vertical" correctly as the problem wants.

http://img225.imageshack.us/img225/4088/rightfromverticalfromhofz4.jpg

[/URL]

The 37.9 degree angle you should be picturing is the one between the acceleration vector (the one that the 52.1 degree angle is measuring and the vertical axis. I'm not sure why you drew it between the vertical axis and some downward pointing vector.