# Ferris wheel Rotational Kinematics Question

fleabass123

## Homework Statement

The given information is:
A Ferris wheel with a radius of 9.0 m rotates at a constant rate, completing one revolution every 37 s.
Suppose the Ferris wheel begins to decelerate at the rate of 0.22 rad/s^2 when the passenger is at the top of the wheel.

(1) Find the magnitude of the passenger's acceleration at that time.
(2) Find the direction of the passenger's acceleration at that time.

a_c=r*w^2
a_t=r*alpha

## The Attempt at a Solution

I tried to find the acceleration at the top using those formulas and couldn't get it. I thought I should try total acceleration using a= sqrt(a_c^2+a_t^2) and got 0.260 m/s, but that was no good either. Honestly what is really confusing me is what exactly I'm finding; I've calculated several types of acceleration during this homework so I'm not sure what just the "passenger's acceleration" is. Because of part two I'm guessing I need to incorporate vectors somehow, but I can't figure out what approach to take.

Thanks for any help!

## Answers and Replies

Mentor
I thought I should try total acceleration using a= sqrt(a_c^2+a_t^2)
That's correct.
and got 0.260 m/s, but that was no good either.
Show your calculations for a_c and a_t.

fleabass123
I did:

a= sqrt(a_c^2+a_t^2)

a_c^2=r*w^2=9*.169815819^2=.259536711

(I calculated w by doing 2pi/37seconds)

a_t^2=r*theta=9*(-.22)^2=.4356

(I used the given deceleration as theta, but I'm not sure if that works)

so a= sqrt((.259536711^2)+(.4356^2))= .507

As you can see I got a different answer on this new attempt. I suppose I'll try that, but I'm still not confident my decision to use -.22 as my acceleration works.

Mentor
a_t^2=r*theta=9*(-.22)^2=.4356
I assume you meant this to be a_t, not a_t^2. Why did you square the angular acceleration? (Angular acceleration is usually represented by alpha, not theta.)

fleabass123
Yes, I meant alpha. And I squared it because I was accidentally looking at the formula for centripetal acceleration when I was writing the tangential acceleration formula. Is that squared term the root of my problems?

Mentor
Is that squared term the root of my problems?
Yes, it looks like the main problem in that last attempt.

fleabass123
That was right! Thanks so much. If you wouldn't mind I still can't get part two. I know that a_c and a_t are perpendicular to each other, so I was thinking I should use trigonometry to find the direction of acceleration. I tried arctan(a_c/a_t), but that was no good.

Mentor
I tried arctan(a_c/a_t), but that was no good.
That would work just fine, but you need to know how they want you to describe the direction. With respect to the vertical or the horizontal? (The angle that you calculated is with respect to what?)

fleabass123
"_____degrees below the direction of motion" is what the field looks like. I thought my trig function would work in that situation.

Mentor
"_____degrees below the direction of motion" is what the field looks like. I thought my trig function would work in that situation.
Yes, it should work. Show the details. What did you put for a_c, a_t, and theta?

fleabass123
I did "arctan theta=.259536711/1.98" and I got 7.47 for theta.

perrytones

Mentor
I did "arctan theta=.259536711/1.98" and I got 7.47 for theta.
Looks good, but what's the direction of motion? Remember that the acceleration is negative. 