- #1

rsala

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## 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[tex]^{ }_{centripetal}[/tex] = [tex]\omega[/tex][tex]^{2}_{}[/tex] x R

or

a[tex]^{ }_{centripetal}[/tex] = v[tex]^{2}_{}[/tex] / 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.

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